-Video of Lecture from Fall Semester 2020. +The lectures will be recorded and updated videos will be posted after the lectures.
-The lectures will be recorded and updated videos will be posted after the lectures. +"Video of Lecture August 26, 2021":"https://www.uio.no/studier/emner/matnat/fys/FYS-STK4155/h21/forelesningsvideoer/LectureThursdayAugust26.mp4?vrtx=view-as-webpage
Zoom link for lectures: https://msu.zoom.us/j/93311529525?pwd=a1VXSzY4aTFWVy9Rb05mNDJTZ09lZz09 @@ -276,6 +381,9 @@ The lectures will be recorded and updated videos will be posted after the lectur
-In order to study various Machine Learning algorithms, we need to -access data. Acccessing data is an essential step in all machine -learning algorithms. In particular, setting up the so-called design -matrix (to be defined below) is often the first element we need in -order to perform our calculations. To set up the design matrix means -reading (and later, when the calculations are done, writing) data -in various formats, The formats span from reading files from disk, -loading data from databases and interacting with online sources -like web application programming interfaces (APIs). +We start with perhaps our simplest possible example, using Scikit-Learn to perform linear regression analysis on a data set produced by us.
-In handling various input formats, as discussed above, we will mainly stay with pandas, -a Python package which allows us, in a seamless and painless way, to -deal with a multitude of formats, from standard csv (comma separated -values) files, via excel, html to hdf5 formats. With pandas -and the DataFrame and Series functionalities we are able to convert text data -into the calculational formats we need for a specific algorithm. And our code is going to be -pretty close the basic mathematical expressions. +What follows is a simple Python code where we have defined a function +\( y \) in terms of the variable \( x \). Both are defined as vectors with \( 100 \) entries. +The numbers in the vector \( \hat{x} \) are given +by random numbers generated with a uniform distribution with entries +\( x_i \in [0,1] \) (more about probability distribution functions +later). These values are then used to define a function \( y(x) \) +(tabulated again as a vector) with a linear dependence on \( x \) plus a +random noise added via the normal distribution.
-Our first data set is going to be a classic from nuclear physics, namely all -available data on binding energies. Don't be intimidated if you are not familiar with nuclear physics. It serves merely as an example here of a data set. +The Numpy functions are imported used the import numpy as np +statement and the random number generator for the uniform distribution +is called using the function np.random.rand(), where we specificy +that we want \( 100 \) random variables. Using Numpy we define +automatically an array with the specified number of elements, \( 100 \) in +our case. With the Numpy function randn() we can compute random +numbers with the normal distribution (mean value \( \mu \) equal to zero and +variance \( \sigma^2 \) set to one) and produce the values of \( y \) assuming a linear +dependence as function of \( x \) + +$$ +y = 2x+N(0,1), +$$
-We will show some of the -strengths of packages like Scikit-Learn in fitting nuclear binding energies to -specific functions using linear regression first. Then, as a teaser, we will show you how -you can easily implement other algorithms like decision trees and random forests and neural networks. +where \( N(0,1) \) represents random numbers generated by the normal +distribution. From Scikit-Learn we import then the +LinearRegression functionality and make a prediction \( \tilde{y} = +\alpha + \beta x \) using the function fit(x,y). We call the set of +data \( (\hat{x},\hat{y}) \) for our training data. The Python package +scikit-learn has also a functionality which extracts the above +fitting parameters \( \alpha \) and \( \beta \) (see below). Later we will +distinguish between training data and test data.
-But before we really start with nuclear physics data, let's just look at some simpler polynomial fitting cases, such as, -(don't be offended) fitting straight lines! +For plotting we use the Python package +matplotlib which produces publication +quality figures. Feel free to explore the extensive +gallery of examples. In +this example we plot our original values of \( x \) and \( y \) as well as the +prediction ypredict (\( \tilde{y} \)), which attempts at fitting our +data with a straight line. + +
+The Python code follows here. +
+ + +
# Importing various packages
+import numpy as np
+import matplotlib.pyplot as plt
+from sklearn.linear_model import LinearRegression
+
+x = np.random.rand(100,1)
+y = 2*x+np.random.randn(100,1)
+linreg = LinearRegression()
+linreg.fit(x,y)
+xnew = np.array([[0],[1]])
+ypredict = linreg.predict(xnew)
+
+plt.plot(xnew, ypredict, "r-")
+plt.plot(x, y ,'ro')
+plt.axis([0,1.0,0, 5.0])
+plt.xlabel(r'$x$')
+plt.ylabel(r'$y$')
+plt.title(r'Simple Linear Regression')
+plt.show()
++This example serves several aims. It allows us to demonstrate several +aspects of data analysis and later machine learning algorithms. The +immediate visualization shows that our linear fit is not +impressive. It goes through the data points, but there are many +outliers which are not reproduced by our linear regression. We could +now play around with this small program and change for example the +factor in front of \( x \) and the normal distribution. Try to change the +function \( y \) to + +$$ +y = 10x+0.01 \times N(0,1), +$$ + +
+where \( x \) is defined as before. Does the fit look better? Indeed, by +reducing the role of the noise given by the normal distribution we see immediately that +our linear prediction seemingly reproduces better the training +set. However, this testing 'by the eye' is obviouly not satisfactory in the +long run. Here we have only defined the training data and our model, and +have not discussed a more rigorous approach to the cost function. + +
+We need more rigorous criteria in defining whether we have succeeded or +not in modeling our training data. You will be surprised to see that +many scientists seldomly venture beyond this 'by the eye' approach. A +standard approach for the cost function is the so-called \( \chi^2 \) +function (a variant of the mean-squared error (MSE)) + +$$ \chi^2 = \frac{1}{n} +\sum_{i=0}^{n-1}\frac{(y_i-\tilde{y}_i)^2}{\sigma_i^2}, +$$ + +
+where \( \sigma_i^2 \) is the variance (to be defined later) of the entry +\( y_i \). We may not know the explicit value of \( \sigma_i^2 \), it serves +however the aim of scaling the equations and make the cost function +dimensionless. + +
+Minimizing the cost function is a central aspect of +our discussions to come. Finding its minima as function of the model +parameters (\( \alpha \) and \( \beta \) in our case) will be a recurring +theme in these series of lectures. Essentially all machine learning +algorithms we will discuss center around the minimization of the +chosen cost function. This depends in turn on our specific +model for describing the data, a typical situation in supervised +learning. Automatizing the search for the minima of the cost function is a +central ingredient in all algorithms. Typical methods which are +employed are various variants of gradient methods. These will be +discussed in more detail later. Again, you'll be surprised to hear that +many practitioners minimize the above function ''by the eye', popularly dubbed as +'chi by the eye'. That is, change a parameter and see (visually and numerically) that +the \( \chi^2 \) function becomes smaller. + +
+There are many ways to define the cost function. A simpler approach is to look at the relative difference between the training data and the predicted data, that is we define +the relative error (why would we prefer the MSE instead of the relative error?) as + +$$ +\epsilon_{\mathrm{relative}}= \frac{\vert \hat{y} -\hat{\tilde{y}}\vert}{\vert \hat{y}\vert}. +$$ + +
+The squared cost function results in an arithmetic mean-unbiased +estimator, and the absolute-value cost function results in a +median-unbiased estimator (in the one-dimensional case, and a +geometric median-unbiased estimator for the multi-dimensional +case). The squared cost function has the disadvantage that it has the tendency +to be dominated by outliers. + +
+We can modify easily the above Python code and plot the relative error instead +
+ + +
import numpy as np
+import matplotlib.pyplot as plt
+from sklearn.linear_model import LinearRegression
+
+x = np.random.rand(100,1)
+y = 5*x+0.01*np.random.randn(100,1)
+linreg = LinearRegression()
+linreg.fit(x,y)
+ypredict = linreg.predict(x)
+
+plt.plot(x, np.abs(ypredict-y)/abs(y), "ro")
+plt.axis([0,1.0,0.0, 0.5])
+plt.xlabel(r'$x$')
+plt.ylabel(r'$\epsilon_{\mathrm{relative}}$')
+plt.title(r'Relative error')
+plt.show()
++Depending on the parameter in front of the normal distribution, we may +have a small or larger relative error. Try to play around with +different training data sets and study (graphically) the value of the +relative error. + +
+As mentioned above, Scikit-Learn has an impressive functionality. +We can for example extract the values of \( \alpha \) and \( \beta \) and +their error estimates, or the variance and standard deviation and many +other properties from the statistical data analysis. + +
+Here we show an +example of the functionality of Scikit-Learn. +
+ + +
import numpy as np
+import matplotlib.pyplot as plt
+from sklearn.linear_model import LinearRegression
+from sklearn.metrics import mean_squared_error, r2_score, mean_squared_log_error, mean_absolute_error
+
+x = np.random.rand(100,1)
+y = 2.0+ 5*x+0.5*np.random.randn(100,1)
+linreg = LinearRegression()
+linreg.fit(x,y)
+ypredict = linreg.predict(x)
+print('The intercept alpha: \n', linreg.intercept_)
+print('Coefficient beta : \n', linreg.coef_)
+# The mean squared error
+print("Mean squared error: %.2f" % mean_squared_error(y, ypredict))
+# Explained variance score: 1 is perfect prediction
+print('Variance score: %.2f' % r2_score(y, ypredict))
+# Mean squared log error
+print('Mean squared log error: %.2f' % mean_squared_log_error(y, ypredict) )
+# Mean absolute error
+print('Mean absolute error: %.2f' % mean_absolute_error(y, ypredict))
+plt.plot(x, ypredict, "r-")
+plt.plot(x, y ,'ro')
+plt.axis([0.0,1.0,1.5, 7.0])
+plt.xlabel(r'$x$')
+plt.ylabel(r'$y$')
+plt.title(r'Linear Regression fit ')
+plt.show()
++The function coef gives us the parameter \( \beta \) of our fit while intercept yields +\( \alpha \). Depending on the constant in front of the normal distribution, we get values near or far from \( \alpha =2 \) and \( \beta =5 \). Try to play around with different parameters in front of the normal distribution. The function meansquarederror gives us the mean square error, a risk metric corresponding to the expected value of the squared (quadratic) error or loss defined as +$$ MSE(\hat{y},\hat{\tilde{y}}) = \frac{1}{n} +\sum_{i=0}^{n-1}(y_i-\tilde{y}_i)^2, +$$ + +
+The smaller the value, the better the fit. Ideally we would like to +have an MSE equal zero. The attentive reader has probably recognized +this function as being similar to the \( \chi^2 \) function defined above. + +
+The r2score function computes \( R^2 \), the coefficient of +determination. It provides a measure of how well future samples are +likely to be predicted by the model. Best possible score is 1.0 and it +can be negative (because the model can be arbitrarily worse). A +constant model that always predicts the expected value of \( \hat{y} \), +disregarding the input features, would get a \( R^2 \) score of \( 0.0 \). + +
+If \( \tilde{\hat{y}}_i \) is the predicted value of the \( i-th \) sample and \( y_i \) is the corresponding true value, then the score \( R^2 \) is defined as +$$ +R^2(\hat{y}, \tilde{\hat{y}}) = 1 - \frac{\sum_{i=0}^{n - 1} (y_i - \tilde{y}_i)^2}{\sum_{i=0}^{n - 1} (y_i - \bar{y})^2}, +$$ + +where we have defined the mean value of \( \hat{y} \) as +$$ +\bar{y} = \frac{1}{n} \sum_{i=0}^{n - 1} y_i. +$$ + +Another quantity taht we will meet again in our discussions of regression analysis is + the mean absolute error (MAE), a risk metric corresponding to the expected value of the absolute error loss or what we call the \( l1 \)-norm loss. In our discussion above we presented the relative error. +The MAE is defined as follows +$$ +\text{MAE}(\hat{y}, \hat{\tilde{y}}) = \frac{1}{n} \sum_{i=0}^{n-1} \left| y_i - \tilde{y}_i \right|. +$$ + +We present the +squared logarithmic (quadratic) error +$$ +\text{MSLE}(\hat{y}, \hat{\tilde{y}}) = \frac{1}{n} \sum_{i=0}^{n - 1} (\log_e (1 + y_i) - \log_e (1 + \tilde{y}_i) )^2, +$$ + +
+where \( \log_e (x) \) stands for the natural logarithm of \( x \). This error +estimate is best to use when targets having exponential growth, such +as population counts, average sales of a commodity over a span of +years etc. + +
+Finally, another cost function is the Huber cost function used in robust regression. + +
+The rationale behind this possible cost function is its reduced +sensitivity to outliers in the data set. In our discussions on +dimensionality reduction and normalization of data we will meet other +ways of dealing with outliers. + +
+The Huber cost function is defined as +$$ +H_{\delta}(a)=\left\{\begin{array}\frac{1}{2}a^{2}&{\text{for }}|a|\leq \delta ,\\ \delta (|a|-\frac {1}{2}\delta ),&{\text{otherwise.}\end{array}\right. +$$ + +Here \( a=\boldsymbol{y} - \boldsymbol{\tilde{y}} \). +We will discuss in more +detail these and other functions in the various lectures. We conclude this part with another example. Instead of +a linear \( x \)-dependence we study now a cubic polynomial and use the polynomial regression analysis tools of scikit-learn. + +
+ + +
import matplotlib.pyplot as plt
+import numpy as np
+import random
+from sklearn.linear_model import Ridge
+from sklearn.preprocessing import PolynomialFeatures
+from sklearn.pipeline import make_pipeline
+from sklearn.linear_model import LinearRegression
+
+x=np.linspace(0.02,0.98,200)
+noise = np.asarray(random.sample((range(200)),200))
+y=x**3*noise
+yn=x**3*100
+poly3 = PolynomialFeatures(degree=3)
+X = poly3.fit_transform(x[:,np.newaxis])
+clf3 = LinearRegression()
+clf3.fit(X,y)
+
+Xplot=poly3.fit_transform(x[:,np.newaxis])
+poly3_plot=plt.plot(x, clf3.predict(Xplot), label='Cubic Fit')
+plt.plot(x,yn, color='red', label="True Cubic")
+plt.scatter(x, y, label='Data', color='orange', s=15)
+plt.legend()
+plt.show()
+
+def error(a):
+ for i in y:
+ err=(y-yn)/yn
+ return abs(np.sum(err))/len(err)
+
+print (error(y))
++Let us now dive into nuclear physics and remind ourselves briefly about some basic features about binding +energies. A basic quantity which can be measured for the ground +states of nuclei is the atomic mass \( M(N, Z) \) of the neutral atom with +atomic mass number \( A \) and charge \( Z \). The number of neutrons is \( N \). There are indeed several sophisticated experiments worldwide which allow us to measure this quantity to high precision (parts per million even). + +
+Atomic masses are usually tabulated in terms of the mass excess defined by +$$ +\Delta M(N, Z) = M(N, Z) - uA, +$$ + +where \( u \) is the Atomic Mass Unit +$$ +u = M(^{12}\mathrm{C})/12 = 931.4940954(57) \hspace{0.1cm} \mathrm{MeV}/c^2. +$$ + +The nucleon masses are +$$ +m_p = 1.00727646693(9)u, +$$ + +and +$$ +m_n = 939.56536(8)\hspace{0.1cm} \mathrm{MeV}/c^2 = 1.0086649156(6)u. +$$ + +
+In the 2016 mass evaluation of by W.J.Huang, G.Audi, M.Wang, F.G.Kondev, S.Naimi and X.Xu +there are data on masses and decays of 3437 nuclei. + +
+The nuclear binding energy is defined as the energy required to break +up a given nucleus into its constituent parts of \( N \) neutrons and \( Z \) +protons. In terms of the atomic masses \( M(N, Z) \) the binding energy is +defined by + +$$ +BE(N, Z) = ZM_H c^2 + Nm_n c^2 - M(N, Z)c^2 , +$$ + +where \( M_H \) is the mass of the hydrogen atom and \( m_n \) is the mass of the neutron. +In terms of the mass excess the binding energy is given by +$$ +BE(N, Z) = Z\Delta_H c^2 + N\Delta_n c^2 -\Delta(N, Z)c^2 , +$$ + +where \( \Delta_H c^2 = 7.2890 \) MeV and \( \Delta_n c^2 = 8.0713 \) MeV. + +
+A popular and physically intuitive model which can be used to parametrize +the experimental binding energies as function of \( A \), is the so-called +liquid drop model. The ansatz is based on the following expression + +$$ +BE(N,Z) = a_1A-a_2A^{2/3}-a_3\frac{Z^2}{A^{1/3}}-a_4\frac{(N-Z)^2}{A}, +$$ + +
+where \( A \) stands for the number of nucleons and the $a_i$s are parameters which are determined by a fit +to the experimental data. + +
+To arrive at the above expression we have assumed that we can make the following assumptions: + +
+Let us start with reading and organizing our data. +We start with the compilation of masses and binding energies from 2016. +After having downloaded this file to our own computer, we are now ready to read the file and start structuring our data. + +
+We start with preparing folders for storing our calculations and the data file over masses and binding energies. We import also various modules that we will find useful in order to present various Machine Learning methods. Here we focus mainly on the functionality of scikit-learn. +
+ + +
# Common imports
+import numpy as np
+import pandas as pd
+import matplotlib.pyplot as plt
+import sklearn.linear_model as skl
+from sklearn.model_selection import train_test_split
+from sklearn.metrics import mean_squared_error, r2_score, mean_absolute_error
+import os
+
+# Where to save the figures and data files
+PROJECT_ROOT_DIR = "Results"
+FIGURE_ID = "Results/FigureFiles"
+DATA_ID = "DataFiles/"
+
+if not os.path.exists(PROJECT_ROOT_DIR):
+ os.mkdir(PROJECT_ROOT_DIR)
+
+if not os.path.exists(FIGURE_ID):
+ os.makedirs(FIGURE_ID)
+
+if not os.path.exists(DATA_ID):
+ os.makedirs(DATA_ID)
+
+def image_path(fig_id):
+ return os.path.join(FIGURE_ID, fig_id)
+
+def data_path(dat_id):
+ return os.path.join(DATA_ID, dat_id)
+
+def save_fig(fig_id):
+ plt.savefig(image_path(fig_id) + ".png", format='png')
+
+infile = open(data_path("MassEval2016.dat"),'r')
++Before we proceed, we define also a function for making our plots. You can obviously avoid this and simply set up various matplotlib commands every time you need them. You may however find it convenient to collect all such commands in one function and simply call this function. +
+ + +
from pylab import plt, mpl
+plt.style.use('seaborn')
+mpl.rcParams['font.family'] = 'serif'
+
+def MakePlot(x,y, styles, labels, axlabels):
+ plt.figure(figsize=(10,6))
+ for i in range(len(x)):
+ plt.plot(x[i], y[i], styles[i], label = labels[i])
+ plt.xlabel(axlabels[0])
+ plt.ylabel(axlabels[1])
+ plt.legend(loc=0)
++Our next step is to read the data on experimental binding energies and +reorganize them as functions of the mass number \( A \), the number of +protons \( Z \) and neutrons \( N \) using pandas. Before we do this it is +always useful (unless you have a binary file or other types of compressed +data) to actually open the file and simply take a look at it! + +
+In particular, the program that outputs the final nuclear masses is written in Fortran with a specific format. It means that we need to figure out the format and which columns contain the data we are interested in. Pandas comes with a function that reads formatted output. After having admired the file, we are now ready to start massaging it with pandas. The file begins with some basic format information. +
+ + +
"""
+This is taken from the data file of the mass 2016 evaluation.
+All files are 3436 lines long with 124 character per line.
+ Headers are 39 lines long.
+ col 1 : Fortran character control: 1 = page feed 0 = line feed
+ format : a1,i3,i5,i5,i5,1x,a3,a4,1x,f13.5,f11.5,f11.3,f9.3,1x,a2,f11.3,f9.3,1x,i3,1x,f12.5,f11.5
+ These formats are reflected in the pandas widths variable below, see the statement
+ widths=(1,3,5,5,5,1,3,4,1,13,11,11,9,1,2,11,9,1,3,1,12,11,1),
+ Pandas has also a variable header, with length 39 in this case.
+"""
++The data we are interested in are in columns 2, 3, 4 and 11, giving us +the number of neutrons, protons, mass numbers and binding energies, +respectively. We add also for the sake of completeness the element name. The data are in fixed-width formatted lines and we will +covert them into the pandas DataFrame structure. + +
+ + +
# Read the experimental data with Pandas
+Masses = pd.read_fwf(infile, usecols=(2,3,4,6,11),
+ names=('N', 'Z', 'A', 'Element', 'Ebinding'),
+ widths=(1,3,5,5,5,1,3,4,1,13,11,11,9,1,2,11,9,1,3,1,12,11,1),
+ header=39,
+ index_col=False)
+
+# Extrapolated values are indicated by '#' in place of the decimal place, so
+# the Ebinding column won't be numeric. Coerce to float and drop these entries.
+Masses['Ebinding'] = pd.to_numeric(Masses['Ebinding'], errors='coerce')
+Masses = Masses.dropna()
+# Convert from keV to MeV.
+Masses['Ebinding'] /= 1000
+
+# Group the DataFrame by nucleon number, A.
+Masses = Masses.groupby('A')
+# Find the rows of the grouped DataFrame with the maximum binding energy.
+Masses = Masses.apply(lambda t: t[t.Ebinding==t.Ebinding.max()])
++We have now read in the data, grouped them according to the variables we are interested in. +We see how easy it is to reorganize the data using pandas. If we +were to do these operations in C/C++ or Fortran, we would have had to +write various functions/subroutines which perform the above +reorganizations for us. Having reorganized the data, we can now start +to make some simple fits using both the functionalities in numpy and +Scikit-Learn afterwards. + +
+Now we define five variables which contain +the number of nucleons \( A \), the number of protons \( Z \) and the number of neutrons \( N \), the element name and finally the energies themselves. +
+ + +
A = Masses['A']
+Z = Masses['Z']
+N = Masses['N']
+Element = Masses['Element']
+Energies = Masses['Ebinding']
+print(Masses)
++The next step, and we will define this mathematically later, is to set up the so-called design matrix. We will throughout call this matrix \( \boldsymbol{X} \). +It has dimensionality \( p\times n \), where \( n \) is the number of data points and \( p \) are the so-called predictors. In our case here they are given by the number of polynomials in \( A \) we wish to include in the fit. +
+ + +
# Now we set up the design matrix X
+X = np.zeros((len(A),5))
+X[:,0] = 1
+X[:,1] = A
+X[:,2] = A**(2.0/3.0)
+X[:,3] = A**(-1.0/3.0)
+X[:,4] = A**(-1.0)
++With scikitlearn we are now ready to use linear regression and fit our data. +
+ + +
clf = skl.LinearRegression().fit(X, Energies)
+fity = clf.predict(X)
++Pretty simple! +Now we can print measures of how our fit is doing, the coefficients from the fits and plot the final fit together with our data. +
+ + +
# The mean squared error
+print("Mean squared error: %.2f" % mean_squared_error(Energies, fity))
+# Explained variance score: 1 is perfect prediction
+print('Variance score: %.2f' % r2_score(Energies, fity))
+# Mean absolute error
+print('Mean absolute error: %.2f' % mean_absolute_error(Energies, fity))
+print(clf.coef_, clf.intercept_)
+
+Masses['Eapprox'] = fity
+# Generate a plot comparing the experimental with the fitted values values.
+fig, ax = plt.subplots()
+ax.set_xlabel(r'$A = N + Z$')
+ax.set_ylabel(r'$E_\mathrm{bind}\,/\mathrm{MeV}$')
+ax.plot(Masses['A'], Masses['Ebinding'], alpha=0.7, lw=2,
+ label='Ame2016')
+ax.plot(Masses['A'], Masses['Eapprox'], alpha=0.7, lw=2, c='m',
+ label='Fit')
+ax.legend()
+save_fig("Masses2016")
+plt.show()
++As a teaser, let us now see how we can do this with decision trees using scikit-learn. Later we will switch to so-called random forests! + +
+ + +
#Decision Tree Regression
+from sklearn.tree import DecisionTreeRegressor
+regr_1=DecisionTreeRegressor(max_depth=5)
+regr_2=DecisionTreeRegressor(max_depth=7)
+regr_3=DecisionTreeRegressor(max_depth=9)
+regr_1.fit(X, Energies)
+regr_2.fit(X, Energies)
+regr_3.fit(X, Energies)
+
+
+y_1 = regr_1.predict(X)
+y_2 = regr_2.predict(X)
+y_3=regr_3.predict(X)
+Masses['Eapprox'] = y_3
+# Plot the results
+plt.figure()
+plt.plot(A, Energies, color="blue", label="Data", linewidth=2)
+plt.plot(A, y_1, color="red", label="max_depth=5", linewidth=2)
+plt.plot(A, y_2, color="green", label="max_depth=7", linewidth=2)
+plt.plot(A, y_3, color="m", label="max_depth=9", linewidth=2)
+
+plt.xlabel("$A$")
+plt.ylabel("$E$[MeV]")
+plt.title("Decision Tree Regression")
+plt.legend()
+save_fig("Masses2016Trees")
+plt.show()
+print(Masses)
+print(np.mean( (Energies-y_1)**2))
++ + +
from sklearn.neural_network import MLPRegressor
+from sklearn.metrics import accuracy_score
+import seaborn as sns
+
+X_train = X
+Y_train = Energies
+n_hidden_neurons = 100
+epochs = 100
+# store models for later use
+eta_vals = np.logspace(-5, 1, 7)
+lmbd_vals = np.logspace(-5, 1, 7)
+# store the models for later use
+DNN_scikit = np.zeros((len(eta_vals), len(lmbd_vals)), dtype=object)
+train_accuracy = np.zeros((len(eta_vals), len(lmbd_vals)))
+sns.set()
+for i, eta in enumerate(eta_vals):
+ for j, lmbd in enumerate(lmbd_vals):
+ dnn = MLPRegressor(hidden_layer_sizes=(n_hidden_neurons), activation='logistic',
+ alpha=lmbd, learning_rate_init=eta, max_iter=epochs)
+ dnn.fit(X_train, Y_train)
+ DNN_scikit[i][j] = dnn
+ train_accuracy[i][j] = dnn.score(X_train, Y_train)
+
+fig, ax = plt.subplots(figsize = (10, 10))
+sns.heatmap(train_accuracy, annot=True, ax=ax, cmap="viridis")
+ax.set_title("Training Accuracy")
+ax.set_ylabel("$\eta$")
+ax.set_xlabel("$\lambda$")
+plt.show()
++The aim behind these introductory words was to present to you various +Python libraries and their functionalities, in particular libraries like +numpy, pandas, xarray and matplotlib and other that make our life much easier +in handling various data sets and visualizing data. + +
+Furthermore, +Scikit-Learn allows us with few lines of code to implement popular +Machine Learning algorithms for supervised learning. Later we will meet Tensorflow, a powerful library for deep learning. +Now it is time to dive more into the details of various methods. We will start with linear regression and try to take a deeper look at what it entails.
@@ -315,6 +1057,14 @@ But before we really start with nuclear physics data, let's just look at some si
+Fitting a continuous function with linear parameterization in terms of the parameters \( \boldsymbol{\beta} \). + +
@@ -280,6 +403,15 @@ MathJax.Hub.Config({
-We start with perhaps our simplest possible example, using Scikit-Learn to perform linear regression analysis on a data set produced by us. +Regression modeling deals with the description of the sampling distribution of a given random variable \( y \) and how it varies as function of another variable or a set of such variables \( \boldsymbol{x} =[x_0, x_1,\dots, x_{n-1}]^T \). +The first variable is called the dependent, the outcome or the response variable while the set of variables \( \boldsymbol{x} \) is called the independent variable, or the predictor variable or the explanatory variable.
-What follows is a simple Python code where we have defined a function -\( y \) in terms of the variable \( x \). Both are defined as vectors with \( 100 \) entries. -The numbers in the vector \( \hat{x} \) are given -by random numbers generated with a uniform distribution with entries -\( x_i \in [0,1] \) (more about probability distribution functions -later). These values are then used to define a function \( y(x) \) -(tabulated again as a vector) with a linear dependence on \( x \) plus a -random noise added via the normal distribution. - -
-The Numpy functions are imported used the import numpy as np -statement and the random number generator for the uniform distribution -is called using the function np.random.rand(), where we specificy -that we want \( 100 \) random variables. Using Numpy we define -automatically an array with the specified number of elements, \( 100 \) in -our case. With the Numpy function randn() we can compute random -numbers with the normal distribution (mean value \( \mu \) equal to zero and -variance \( \sigma^2 \) set to one) and produce the values of \( y \) assuming a linear -dependence as function of \( x \) - -$$ -y = 2x+N(0,1), -$$ - -
-where \( N(0,1) \) represents random numbers generated by the normal -distribution. From Scikit-Learn we import then the -LinearRegression functionality and make a prediction \( \tilde{y} = -\alpha + \beta x \) using the function fit(x,y). We call the set of -data \( (\hat{x},\hat{y}) \) for our training data. The Python package -scikit-learn has also a functionality which extracts the above -fitting parameters \( \alpha \) and \( \beta \) (see below). Later we will -distinguish between training data and test data. - -
-For plotting we use the Python package -matplotlib which produces publication -quality figures. Feel free to explore the extensive -gallery of examples. In -this example we plot our original values of \( x \) and \( y \) as well as the -prediction ypredict (\( \tilde{y} \)), which attempts at fitting our -data with a straight line. - -
-The Python code follows here. -
- - -
# Importing various packages
-import numpy as np
-import matplotlib.pyplot as plt
-from sklearn.linear_model import LinearRegression
-
-x = np.random.rand(100,1)
-y = 2*x+np.random.randn(100,1)
-linreg = LinearRegression()
-linreg.fit(x,y)
-xnew = np.array([[0],[1]])
-ypredict = linreg.predict(xnew)
-
-plt.plot(xnew, ypredict, "r-")
-plt.plot(x, y ,'ro')
-plt.axis([0,1.0,0, 5.0])
-plt.xlabel(r'$x$')
-plt.ylabel(r'$y$')
-plt.title(r'Simple Linear Regression')
-plt.show()
--This example serves several aims. It allows us to demonstrate several -aspects of data analysis and later machine learning algorithms. The -immediate visualization shows that our linear fit is not -impressive. It goes through the data points, but there are many -outliers which are not reproduced by our linear regression. We could -now play around with this small program and change for example the -factor in front of \( x \) and the normal distribution. Try to change the -function \( y \) to - -$$ -y = 10x+0.01 \times N(0,1), -$$ - -
-where \( x \) is defined as before. Does the fit look better? Indeed, by -reducing the role of the noise given by the normal distribution we see immediately that -our linear prediction seemingly reproduces better the training -set. However, this testing 'by the eye' is obviouly not satisfactory in the -long run. Here we have only defined the training data and our model, and -have not discussed a more rigorous approach to the cost function. - -
-We need more rigorous criteria in defining whether we have succeeded or -not in modeling our training data. You will be surprised to see that -many scientists seldomly venture beyond this 'by the eye' approach. A -standard approach for the cost function is the so-called \( \chi^2 \) -function (a variant of the mean-squared error (MSE)) - -$$ \chi^2 = \frac{1}{n} -\sum_{i=0}^{n-1}\frac{(y_i-\tilde{y}_i)^2}{\sigma_i^2}, -$$ - -
-where \( \sigma_i^2 \) is the variance (to be defined later) of the entry -\( y_i \). We may not know the explicit value of \( \sigma_i^2 \), it serves -however the aim of scaling the equations and make the cost function -dimensionless. - -
-Minimizing the cost function is a central aspect of -our discussions to come. Finding its minima as function of the model -parameters (\( \alpha \) and \( \beta \) in our case) will be a recurring -theme in these series of lectures. Essentially all machine learning -algorithms we will discuss center around the minimization of the -chosen cost function. This depends in turn on our specific -model for describing the data, a typical situation in supervised -learning. Automatizing the search for the minima of the cost function is a -central ingredient in all algorithms. Typical methods which are -employed are various variants of gradient methods. These will be -discussed in more detail later. Again, you'll be surprised to hear that -many practitioners minimize the above function ''by the eye', popularly dubbed as -'chi by the eye'. That is, change a parameter and see (visually and numerically) that -the \( \chi^2 \) function becomes smaller. - -
-There are many ways to define the cost function. A simpler approach is to look at the relative difference between the training data and the predicted data, that is we define -the relative error (why would we prefer the MSE instead of the relative error?) as - -$$ -\epsilon_{\mathrm{relative}}= \frac{\vert \hat{y} -\hat{\tilde{y}}\vert}{\vert \hat{y}\vert}. -$$ - -
-The squared cost function results in an arithmetic mean-unbiased -estimator, and the absolute-value cost function results in a -median-unbiased estimator (in the one-dimensional case, and a -geometric median-unbiased estimator for the multi-dimensional -case). The squared cost function has the disadvantage that it has the tendency -to be dominated by outliers. - -
-We can modify easily the above Python code and plot the relative error instead -
- - -
import numpy as np
-import matplotlib.pyplot as plt
-from sklearn.linear_model import LinearRegression
-
-x = np.random.rand(100,1)
-y = 5*x+0.01*np.random.randn(100,1)
-linreg = LinearRegression()
-linreg.fit(x,y)
-ypredict = linreg.predict(x)
-
-plt.plot(x, np.abs(ypredict-y)/abs(y), "ro")
-plt.axis([0,1.0,0.0, 0.5])
-plt.xlabel(r'$x$')
-plt.ylabel(r'$\epsilon_{\mathrm{relative}}$')
-plt.title(r'Relative error')
-plt.show()
--Depending on the parameter in front of the normal distribution, we may -have a small or larger relative error. Try to play around with -different training data sets and study (graphically) the value of the -relative error. - -
-As mentioned above, Scikit-Learn has an impressive functionality. -We can for example extract the values of \( \alpha \) and \( \beta \) and -their error estimates, or the variance and standard deviation and many -other properties from the statistical data analysis. - -
-Here we show an -example of the functionality of Scikit-Learn. -
- - -
import numpy as np
-import matplotlib.pyplot as plt
-from sklearn.linear_model import LinearRegression
-from sklearn.metrics import mean_squared_error, r2_score, mean_squared_log_error, mean_absolute_error
-
-x = np.random.rand(100,1)
-y = 2.0+ 5*x+0.5*np.random.randn(100,1)
-linreg = LinearRegression()
-linreg.fit(x,y)
-ypredict = linreg.predict(x)
-print('The intercept alpha: \n', linreg.intercept_)
-print('Coefficient beta : \n', linreg.coef_)
-# The mean squared error
-print("Mean squared error: %.2f" % mean_squared_error(y, ypredict))
-# Explained variance score: 1 is perfect prediction
-print('Variance score: %.2f' % r2_score(y, ypredict))
-# Mean squared log error
-print('Mean squared log error: %.2f' % mean_squared_log_error(y, ypredict) )
-# Mean absolute error
-print('Mean absolute error: %.2f' % mean_absolute_error(y, ypredict))
-plt.plot(x, ypredict, "r-")
-plt.plot(x, y ,'ro')
-plt.axis([0.0,1.0,1.5, 7.0])
-plt.xlabel(r'$x$')
-plt.ylabel(r'$y$')
-plt.title(r'Linear Regression fit ')
-plt.show()
--The function coef gives us the parameter \( \beta \) of our fit while intercept yields -\( \alpha \). Depending on the constant in front of the normal distribution, we get values near or far from \( \alpha =2 \) and \( \beta =5 \). Try to play around with different parameters in front of the normal distribution. The function meansquarederror gives us the mean square error, a risk metric corresponding to the expected value of the squared (quadratic) error or loss defined as -$$ MSE(\hat{y},\hat{\tilde{y}}) = \frac{1}{n} -\sum_{i=0}^{n-1}(y_i-\tilde{y}_i)^2, -$$ - -
-The smaller the value, the better the fit. Ideally we would like to -have an MSE equal zero. The attentive reader has probably recognized -this function as being similar to the \( \chi^2 \) function defined above. - -
-The r2score function computes \( R^2 \), the coefficient of -determination. It provides a measure of how well future samples are -likely to be predicted by the model. Best possible score is 1.0 and it -can be negative (because the model can be arbitrarily worse). A -constant model that always predicts the expected value of \( \hat{y} \), -disregarding the input features, would get a \( R^2 \) score of \( 0.0 \). - -
-If \( \tilde{\hat{y}}_i \) is the predicted value of the \( i-th \) sample and \( y_i \) is the corresponding true value, then the score \( R^2 \) is defined as -$$ -R^2(\hat{y}, \tilde{\hat{y}}) = 1 - \frac{\sum_{i=0}^{n - 1} (y_i - \tilde{y}_i)^2}{\sum_{i=0}^{n - 1} (y_i - \bar{y})^2}, -$$ - -where we have defined the mean value of \( \hat{y} \) as -$$ -\bar{y} = \frac{1}{n} \sum_{i=0}^{n - 1} y_i. -$$ - -Another quantity taht we will meet again in our discussions of regression analysis is - the mean absolute error (MAE), a risk metric corresponding to the expected value of the absolute error loss or what we call the \( l1 \)-norm loss. In our discussion above we presented the relative error. -The MAE is defined as follows -$$ -\text{MAE}(\hat{y}, \hat{\tilde{y}}) = \frac{1}{n} \sum_{i=0}^{n-1} \left| y_i - \tilde{y}_i \right|. -$$ - -We present the -squared logarithmic (quadratic) error -$$ -\text{MSLE}(\hat{y}, \hat{\tilde{y}}) = \frac{1}{n} \sum_{i=0}^{n - 1} (\log_e (1 + y_i) - \log_e (1 + \tilde{y}_i) )^2, -$$ - -
-where \( \log_e (x) \) stands for the natural logarithm of \( x \). This error -estimate is best to use when targets having exponential growth, such -as population counts, average sales of a commodity over a span of -years etc. - -
-Finally, another cost function is the Huber cost function used in robust regression. - -
-The rationale behind this possible cost function is its reduced -sensitivity to outliers in the data set. In our discussions on -dimensionality reduction and normalization of data we will meet other -ways of dealing with outliers. - -
-The Huber cost function is defined as -$$ -H_{\delta}(a)=\left\{\begin{array}\frac{1}{2}a^{2}&{\text{for }}|a|\leq \delta ,\\ \delta (|a|-\frac {1}{2}\delta ),&{\text{otherwise.}\end{array}\right. -$$ - -Here \( a=\boldsymbol{y} - \boldsymbol{\tilde{y}} \). -We will discuss in more -detail these and other functions in the various lectures. We conclude this part with another example. Instead of -a linear \( x \)-dependence we study now a cubic polynomial and use the polynomial regression analysis tools of scikit-learn. - -
- - -
import matplotlib.pyplot as plt
-import numpy as np
-import random
-from sklearn.linear_model import Ridge
-from sklearn.preprocessing import PolynomialFeatures
-from sklearn.pipeline import make_pipeline
-from sklearn.linear_model import LinearRegression
-
-x=np.linspace(0.02,0.98,200)
-noise = np.asarray(random.sample((range(200)),200))
-y=x**3*noise
-yn=x**3*100
-poly3 = PolynomialFeatures(degree=3)
-X = poly3.fit_transform(x[:,np.newaxis])
-clf3 = LinearRegression()
-clf3.fit(X,y)
-
-Xplot=poly3.fit_transform(x[:,np.newaxis])
-poly3_plot=plt.plot(x, clf3.predict(Xplot), label='Cubic Fit')
-plt.plot(x,yn, color='red', label="True Cubic")
-plt.scatter(x, y, label='Data', color='orange', s=15)
-plt.legend()
-plt.show()
-
-def error(a):
- for i in y:
- err=(y-yn)/yn
- return abs(np.sum(err))/len(err)
-
-print (error(y))
--Let us now dive into nuclear physics and remind ourselves briefly about some basic features about binding -energies. A basic quantity which can be measured for the ground -states of nuclei is the atomic mass \( M(N, Z) \) of the neutral atom with -atomic mass number \( A \) and charge \( Z \). The number of neutrons is \( N \). There are indeed several sophisticated experiments worldwide which allow us to measure this quantity to high precision (parts per million even). - -
-Atomic masses are usually tabulated in terms of the mass excess defined by -$$ -\Delta M(N, Z) = M(N, Z) - uA, -$$ - -where \( u \) is the Atomic Mass Unit -$$ -u = M(^{12}\mathrm{C})/12 = 931.4940954(57) \hspace{0.1cm} \mathrm{MeV}/c^2. -$$ - -The nucleon masses are -$$ -m_p = 1.00727646693(9)u, -$$ - -and -$$ -m_n = 939.56536(8)\hspace{0.1cm} \mathrm{MeV}/c^2 = 1.0086649156(6)u. -$$ - -
-In the 2016 mass evaluation of by W.J.Huang, G.Audi, M.Wang, F.G.Kondev, S.Naimi and X.Xu -there are data on masses and decays of 3437 nuclei. - -
-The nuclear binding energy is defined as the energy required to break -up a given nucleus into its constituent parts of \( N \) neutrons and \( Z \) -protons. In terms of the atomic masses \( M(N, Z) \) the binding energy is -defined by - -$$ -BE(N, Z) = ZM_H c^2 + Nm_n c^2 - M(N, Z)c^2 , -$$ - -where \( M_H \) is the mass of the hydrogen atom and \( m_n \) is the mass of the neutron. -In terms of the mass excess the binding energy is given by -$$ -BE(N, Z) = Z\Delta_H c^2 + N\Delta_n c^2 -\Delta(N, Z)c^2 , -$$ - -where \( \Delta_H c^2 = 7.2890 \) MeV and \( \Delta_n c^2 = 8.0713 \) MeV. - -
-A popular and physically intuitive model which can be used to parametrize -the experimental binding energies as function of \( A \), is the so-called -liquid drop model. The ansatz is based on the following expression - -$$ -BE(N,Z) = a_1A-a_2A^{2/3}-a_3\frac{Z^2}{A^{1/3}}-a_4\frac{(N-Z)^2}{A}, -$$ - -
-where \( A \) stands for the number of nucleons and the $a_i$s are parameters which are determined by a fit -to the experimental data. - -
-To arrive at the above expression we have assumed that we can make the following assumptions: +A regression model aims at finding a likelihood function \( p(\boldsymbol{y}\vert \boldsymbol{x}) \), that is the conditional distribution for \( \boldsymbol{y} \) with a given \( \boldsymbol{x} \). The estimation of \( p(\boldsymbol{y}\vert \boldsymbol{x}) \) is made using a data set with
-Let us start with reading and organizing our data. -We start with the compilation of masses and binding energies from 2016. -After having downloaded this file to our own computer, we are now ready to read the file and start structuring our data. - -
-We start with preparing folders for storing our calculations and the data file over masses and binding energies. We import also various modules that we will find useful in order to present various Machine Learning methods. Here we focus mainly on the functionality of scikit-learn. -
- - -
# Common imports
-import numpy as np
-import pandas as pd
-import matplotlib.pyplot as plt
-import sklearn.linear_model as skl
-from sklearn.model_selection import train_test_split
-from sklearn.metrics import mean_squared_error, r2_score, mean_absolute_error
-import os
-
-# Where to save the figures and data files
-PROJECT_ROOT_DIR = "Results"
-FIGURE_ID = "Results/FigureFiles"
-DATA_ID = "DataFiles/"
-
-if not os.path.exists(PROJECT_ROOT_DIR):
- os.mkdir(PROJECT_ROOT_DIR)
-
-if not os.path.exists(FIGURE_ID):
- os.makedirs(FIGURE_ID)
-
-if not os.path.exists(DATA_ID):
- os.makedirs(DATA_ID)
-
-def image_path(fig_id):
- return os.path.join(FIGURE_ID, fig_id)
-
-def data_path(dat_id):
- return os.path.join(DATA_ID, dat_id)
-
-def save_fig(fig_id):
- plt.savefig(image_path(fig_id) + ".png", format='png')
-
-infile = open(data_path("MassEval2016.dat"),'r')
--Before we proceed, we define also a function for making our plots. You can obviously avoid this and simply set up various matplotlib commands every time you need them. You may however find it convenient to collect all such commands in one function and simply call this function. -
- - -
from pylab import plt, mpl
-plt.style.use('seaborn')
-mpl.rcParams['font.family'] = 'serif'
-
-def MakePlot(x,y, styles, labels, axlabels):
- plt.figure(figsize=(10,6))
- for i in range(len(x)):
- plt.plot(x[i], y[i], styles[i], label = labels[i])
- plt.xlabel(axlabels[0])
- plt.ylabel(axlabels[1])
- plt.legend(loc=0)
--Our next step is to read the data on experimental binding energies and -reorganize them as functions of the mass number \( A \), the number of -protons \( Z \) and neutrons \( N \) using pandas. Before we do this it is -always useful (unless you have a binary file or other types of compressed -data) to actually open the file and simply take a look at it! - -
-In particular, the program that outputs the final nuclear masses is written in Fortran with a specific format. It means that we need to figure out the format and which columns contain the data we are interested in. Pandas comes with a function that reads formatted output. After having admired the file, we are now ready to start massaging it with pandas. The file begins with some basic format information. -
- - -
"""
-This is taken from the data file of the mass 2016 evaluation.
-All files are 3436 lines long with 124 character per line.
- Headers are 39 lines long.
- col 1 : Fortran character control: 1 = page feed 0 = line feed
- format : a1,i3,i5,i5,i5,1x,a3,a4,1x,f13.5,f11.5,f11.3,f9.3,1x,a2,f11.3,f9.3,1x,i3,1x,f12.5,f11.5
- These formats are reflected in the pandas widths variable below, see the statement
- widths=(1,3,5,5,5,1,3,4,1,13,11,11,9,1,2,11,9,1,3,1,12,11,1),
- Pandas has also a variable header, with length 39 in this case.
-"""
--The data we are interested in are in columns 2, 3, 4 and 11, giving us -the number of neutrons, protons, mass numbers and binding energies, -respectively. We add also for the sake of completeness the element name. The data are in fixed-width formatted lines and we will -covert them into the pandas DataFrame structure. - -
- - -
# Read the experimental data with Pandas
-Masses = pd.read_fwf(infile, usecols=(2,3,4,6,11),
- names=('N', 'Z', 'A', 'Element', 'Ebinding'),
- widths=(1,3,5,5,5,1,3,4,1,13,11,11,9,1,2,11,9,1,3,1,12,11,1),
- header=39,
- index_col=False)
-
-# Extrapolated values are indicated by '#' in place of the decimal place, so
-# the Ebinding column won't be numeric. Coerce to float and drop these entries.
-Masses['Ebinding'] = pd.to_numeric(Masses['Ebinding'], errors='coerce')
-Masses = Masses.dropna()
-# Convert from keV to MeV.
-Masses['Ebinding'] /= 1000
-
-# Group the DataFrame by nucleon number, A.
-Masses = Masses.groupby('A')
-# Find the rows of the grouped DataFrame with the maximum binding energy.
-Masses = Masses.apply(lambda t: t[t.Ebinding==t.Ebinding.max()])
--We have now read in the data, grouped them according to the variables we are interested in. -We see how easy it is to reorganize the data using pandas. If we -were to do these operations in C/C++ or Fortran, we would have had to -write various functions/subroutines which perform the above -reorganizations for us. Having reorganized the data, we can now start -to make some simple fits using both the functionalities in numpy and -Scikit-Learn afterwards. - -
-Now we define five variables which contain -the number of nucleons \( A \), the number of protons \( Z \) and the number of neutrons \( N \), the element name and finally the energies themselves. -
- - -
A = Masses['A']
-Z = Masses['Z']
-N = Masses['N']
-Element = Masses['Element']
-Energies = Masses['Ebinding']
-print(Masses)
--The next step, and we will define this mathematically later, is to set up the so-called design matrix. We will throughout call this matrix \( \boldsymbol{X} \). -It has dimensionality \( p\times n \), where \( n \) is the number of data points and \( p \) are the so-called predictors. In our case here they are given by the number of polynomials in \( A \) we wish to include in the fit. -
- - -
# Now we set up the design matrix X
-X = np.zeros((len(A),5))
-X[:,0] = 1
-X[:,1] = A
-X[:,2] = A**(2.0/3.0)
-X[:,3] = A**(-1.0/3.0)
-X[:,4] = A**(-1.0)
--With scikitlearn we are now ready to use linear regression and fit our data. -
- - -
clf = skl.LinearRegression().fit(X, Energies)
-fity = clf.predict(X)
--Pretty simple! -Now we can print measures of how our fit is doing, the coefficients from the fits and plot the final fit together with our data. -
- - -
# The mean squared error
-print("Mean squared error: %.2f" % mean_squared_error(Energies, fity))
-# Explained variance score: 1 is perfect prediction
-print('Variance score: %.2f' % r2_score(Energies, fity))
-# Mean absolute error
-print('Mean absolute error: %.2f' % mean_absolute_error(Energies, fity))
-print(clf.coef_, clf.intercept_)
-
-Masses['Eapprox'] = fity
-# Generate a plot comparing the experimental with the fitted values values.
-fig, ax = plt.subplots()
-ax.set_xlabel(r'$A = N + Z$')
-ax.set_ylabel(r'$E_\mathrm{bind}\,/\mathrm{MeV}$')
-ax.plot(Masses['A'], Masses['Ebinding'], alpha=0.7, lw=2,
- label='Ame2016')
-ax.plot(Masses['A'], Masses['Eapprox'], alpha=0.7, lw=2, c='m',
- label='Fit')
-ax.legend()
-save_fig("Masses2016")
-plt.show()
--As a teaser, let us now see how we can do this with decision trees using scikit-learn. Later we will switch to so-called random forests! - -
- - -
#Decision Tree Regression
-from sklearn.tree import DecisionTreeRegressor
-regr_1=DecisionTreeRegressor(max_depth=5)
-regr_2=DecisionTreeRegressor(max_depth=7)
-regr_3=DecisionTreeRegressor(max_depth=9)
-regr_1.fit(X, Energies)
-regr_2.fit(X, Energies)
-regr_3.fit(X, Energies)
-
-
-y_1 = regr_1.predict(X)
-y_2 = regr_2.predict(X)
-y_3=regr_3.predict(X)
-Masses['Eapprox'] = y_3
-# Plot the results
-plt.figure()
-plt.plot(A, Energies, color="blue", label="Data", linewidth=2)
-plt.plot(A, y_1, color="red", label="max_depth=5", linewidth=2)
-plt.plot(A, y_2, color="green", label="max_depth=7", linewidth=2)
-plt.plot(A, y_3, color="m", label="max_depth=9", linewidth=2)
-
-plt.xlabel("$A$")
-plt.ylabel("$E$[MeV]")
-plt.title("Decision Tree Regression")
-plt.legend()
-save_fig("Masses2016Trees")
-plt.show()
-print(Masses)
-print(np.mean( (Energies-y_1)**2))
-- - -
from sklearn.neural_network import MLPRegressor
-from sklearn.metrics import accuracy_score
-import seaborn as sns
-
-X_train = X
-Y_train = Energies
-n_hidden_neurons = 100
-epochs = 100
-# store models for later use
-eta_vals = np.logspace(-5, 1, 7)
-lmbd_vals = np.logspace(-5, 1, 7)
-# store the models for later use
-DNN_scikit = np.zeros((len(eta_vals), len(lmbd_vals)), dtype=object)
-train_accuracy = np.zeros((len(eta_vals), len(lmbd_vals)))
-sns.set()
-for i, eta in enumerate(eta_vals):
- for j, lmbd in enumerate(lmbd_vals):
- dnn = MLPRegressor(hidden_layer_sizes=(n_hidden_neurons), activation='logistic',
- alpha=lmbd, learning_rate_init=eta, max_iter=epochs)
- dnn.fit(X_train, Y_train)
- DNN_scikit[i][j] = dnn
- train_accuracy[i][j] = dnn.score(X_train, Y_train)
-
-fig, ax = plt.subplots(figsize = (10, 10))
-sns.heatmap(train_accuracy, annot=True, ax=ax, cmap="viridis")
-ax.set_title("Training Accuracy")
-ax.set_ylabel("$\eta$")
-ax.set_xlabel("$\lambda$")
-plt.show()
--The aim behind these introductory words was to present to you various -Python libraries and their functionalities, in particular libraries like -numpy, pandas, xarray and matplotlib and other that make our life much easier -in handling various data sets and visualizing data. - -
-Furthermore, -Scikit-Learn allows us with few lines of code to implement popular -Machine Learning algorithms for supervised learning. Later we will meet Tensorflow, a powerful library for deep learning. -Now it is time to dive more into the details of various methods. We will start with linear regression and try to take a deeper look at what it entails.
@@ -950,6 +405,16 @@ Now it is time to dive more into the details of various methods. We will start w
- - -
-The first exercise here is of a mere technical art. We want you to have - -
-If you have Python installed (we recommend Python3) and you feel -pretty familiar with installing different packages, we recommend that -you install the following Python packages via pip as - -
-We will come back to tensorflow later. - -
-For Python3, replace pip with pip3. - -
-For OSX users we recommend, after having installed Xcode, to -install brew. Brew allows for a seamless installation of additional -software via for example - -
-We recommend using Anaconda if you are not too familiar with setting paths in a terminal environment. - -
- - -
- - -
-We will generate our own dataset for a function \( y(x) \) where \( x \in [0,1] \) and defined by random numbers computed with the uniform distribution. The function \( y \) is a quadratic polynomial in \( x \) with added stochastic noise according to the normal distribution \( \cal {N}(0,1) \). -The following simple Python instructions define our \( x \) and \( y \) values (with 100 data points). -
- - -
x = np.random.rand(100,1)
-y = 2.0+5*x*x+0.1*np.random.randn(100,1)
-- - -
- - -Solution. - -
-The code here is an example of where we define our own design matrix and fit parameters \( \beta \). -
- - -
import os
-import numpy as np
-import pandas as pd
-import matplotlib.pyplot as plt
-from sklearn.model_selection import train_test_split
-
-def save_fig(fig_id):
- plt.savefig(image_path(fig_id) + ".png", format='png')
-
-def R2(y_data, y_model):
- return 1 - np.sum((y_data - y_model) ** 2) / np.sum((y_data - np.mean(y_data)) ** 2)
-def MSE(y_data,y_model):
- n = np.size(y_model)
- return np.sum((y_data-y_model)**2)/n
-
-x = np.random.rand(100)
-y = 2.0+5*x*x+0.1*np.random.randn(100)
-
-
-# The design matrix now as function of a given polynomial
-X = np.zeros((len(x),3))
-X[:,0] = 1.0
-X[:,1] = x
-X[:,2] = x**2
-# We split the data in test and training data
-X_train, X_test, y_train, y_test = train_test_split(X, y, test_size=0.2)
-# matrix inversion to find beta
-beta = np.linalg.inv(X_train.T @ X_train) @ X_train.T @ y_train
-print(beta)
-# and then make the prediction
-ytilde = X_train @ beta
-print("Training R2")
-print(R2(y_train,ytilde))
-print("Training MSE")
-print(MSE(y_train,ytilde))
-ypredict = X_test @ beta
-print("Test R2")
-print(R2(y_test,ypredict))
-print("Test MSE")
-print(MSE(y_test,ypredict))
--
- - -
- - -
- - -
-A much used approach before starting to train the data is to preprocess our -data. Normally the data may need a rescaling and/or may be sensitive -to extreme values. Scaling the data renders our inputs much more -suitable for the algorithms we want to employ. - -
-Scikit-Learn has several functions which allow us to rescale the -data, normally resulting in much better results in terms of various -accuracy scores. The StandardScaler function in Scikit-Learn -ensures that for each feature/predictor we study the mean value is -zero and the variance is one (every column in the design/feature -matrix). This scaling has the drawback that it does not ensure that -we have a particular maximum or minimum in our data set. Another -function included in Scikit-Learn is the MinMaxScaler which -ensures that all features are exactly between \( 0 \) and \( 1 \). The - -
-The Normalizer scales each data -point such that the feature vector has a euclidean length of one. In other words, it -projects a data point on the circle (or sphere in the case of higher dimensions) with a -radius of 1. This means every data point is scaled by a different number (by the -inverse of it’s length). -This normalization is often used when only the direction (or angle) of the data matters, -not the length of the feature vector. - -
-The RobustScaler works similarly to the StandardScaler in that it -ensures statistical properties for each feature that guarantee that -they are on the same scale. However, the RobustScaler uses the median -and quartiles, instead of mean and variance. This makes the -RobustScaler ignore data points that are very different from the rest -(like measurement errors). These odd data points are also called -outliers, and might often lead to trouble for other scaling -techniques. - -
-It also common to split the data in a training set and a testing set. A typical split is to use \( 80\% \) of the data for training and the rest -for testing. This can be done as follows with our design matrix \( \boldsymbol{X} \) and data \( \boldsymbol{y} \) (remember to import scikit-learn) -
- - -
# split in training and test data
-X_train, X_test, y_train, y_test = train_test_split(X,y,test_size=0.2)
--Then we can use the standard scaler to scale our data as -
- - -
scaler = StandardScaler()
-scaler.fit(X_train)
-X_train_scaled = scaler.transform(X_train)
-X_test_scaled = scaler.transform(X_test)
--In this exercise we want you to to compute the MSE for the training -data and the test data as function of the complexity of a polynomial, -that is the degree of a given polynomial. We want you also to compute the \( R2 \) score as function of the complexity of the model for both training data and test data. You should also run the calculation with and without scaling. - -
-One of -the aims is to reproduce Figure 2.11 of Hastie et al. - -
-Our data is defined by \( x\in [-3,3] \) with a total of for example \( 100 \) data points. -
- - -
np.random.seed()
-n = 100
-maxdegree = 14
-# Make data set.
-x = np.linspace(-3, 3, n).reshape(-1, 1)
-y = np.exp(-x**2) + 1.5 * np.exp(-(x-2)**2)+ np.random.normal(0, 0.1, x.shape)
--where \( y \) is the function we want to fit with a given polynomial. - -
-a) -Write a first code which sets up a design matrix \( X \) defined by a fifth-order polynomial. Scale your data and split it in training and test data. - -
-b) -Perform an ordinary least squares and compute the means squared error and the \( R2 \) factor for the training data and the test data, with and without scaling. - -
-c) -Add now a model which allows you to make polynomials up to degree \( 15 \). Perform a standard OLS fitting of the training data and compute the MSE and \( R2 \) for the training and test data and plot both test and training data MSE and \( R2 \) as functions of the polynomial degree. Compare what you see with Figure 2.11 of Hastie et al. Comment your results. For which polynomial degree do you find an optimal MSE (smallest value)? - -
- - -
- - -Solution. - -
-Here you simply need to change the degree of the polynomial in the above code to \( n=15 \). -
- - -
- -
-Video of Lecture from Fall Semester 2020. +The lectures will be recorded and updated videos will be posted after the lectures.
-The lectures will be recorded and updated videos will be posted after the lectures. +"Video of Lecture August 26, 2021":"https://www.uio.no/studier/emner/matnat/fys/FYS-STK4155/h21/forelesningsvideoer/LectureThursdayAugust26.mp4?vrtx=view-as-webpage
Zoom link for lectures: https://msu.zoom.us/j/93311529525?pwd=a1VXSzY4aTFWVy9Rb05mNDJTZ09lZz09 @@ -218,6 +218,9 @@ The lectures will be recorded and updated videos will be posted after the lectur
+
+Video of Lecture from Fall Semester 2020.
@@ -1444,50 +1447,6 @@ For multidimensional arrays, we recommend strongly Reading Data and fitting
-
-
-In order to study various Machine Learning algorithms, we need to
-access data. Acccessing data is an essential step in all machine
-learning algorithms. In particular, setting up the so-called design
-matrix (to be defined below) is often the first element we need in
-order to perform our calculations. To set up the design matrix means
-reading (and later, when the calculations are done, writing) data
-in various formats, The formats span from reading files from disk,
-loading data from databases and interacting with online sources
-like web application programming interfaces (APIs).
-
-
-In handling various input formats, as discussed above, we will mainly stay with pandas,
-a Python package which allows us, in a seamless and painless way, to
-deal with a multitude of formats, from standard csv (comma separated
-values) files, via excel, html to hdf5 formats. With pandas
-and the DataFrame and Series functionalities we are able to convert text data
-into the calculational formats we need for a specific algorithm. And our code is going to be
-pretty close the basic mathematical expressions.
-
-
-Our first data set is going to be a classic from nuclear physics, namely all
-available data on binding energies. Don't be intimidated if you are not familiar with nuclear physics. It serves merely as an example here of a data set.
-
-
-We will show some of the
-strengths of packages like Scikit-Learn in fitting nuclear binding energies to
-specific functions using linear regression first. Then, as a teaser, we will show you how
-you can easily implement other algorithms like decision trees and random forests and neural networks.
-
-
-But before we really start with nuclear physics data, let's just look at some simpler polynomial fitting cases, such as,
-(don't be offended) fitting straight lines!
-
-
-
-
+Fitting a continuous function with linear parameterization in terms of the parameters \( \boldsymbol{\beta} \).
+
+
+
+For more discussions of Ridge and Lasso regression, Wessel van Wieringen's article is highly recommended.
+Similarly, Mehta et al's article is also recommended.
+
+Regression modeling deals with the description of the sampling distribution of a given random variable \( y \) and how it varies as function of another variable or a set of such variables \( \boldsymbol{x} =[x_0, x_1,\dots, x_{n-1}]^T \).
+The first variable is called the dependent, the outcome or the response variable while the set of variables \( \boldsymbol{x} \) is called the independent variable, or the predictor variable or the explanatory variable.
+
+
+A regression model aims at finding a likelihood function \( p(\boldsymbol{y}\vert \boldsymbol{x}) \), that is the conditional distribution for \( \boldsymbol{y} \) with a given \( \boldsymbol{x} \). The estimation of \( p(\boldsymbol{y}\vert \boldsymbol{x}) \) is made using a data set with
+
+
+
+ The goal of the regression analysis is to extract/exploit relationship between \( \boldsymbol{y} \) and \( \boldsymbol{x} \) in or to infer causal dependencies, approximations to the likelihood functions, functional relationships and to make predictions, making fits and many other things.
+
+Consider an experiment in which \( p \) characteristics of \( n \) samples are
+measured. The data from this experiment, for various explanatory variables \( p \) are normally represented by a matrix
+\( \mathbf{X} \).
+
+
+The matrix \( \mathbf{X} \) is called the design
+matrix. Additional information of the samples is available in the
+form of \( \boldsymbol{y} \) (also as above). The variable \( \boldsymbol{y} \) is
+generally referred to as the response variable. The aim of
+regression analysis is to explain \( \boldsymbol{y} \) in terms of
+\( \boldsymbol{X} \) through a functional relationship like \( y_i =
+f(\mathbf{X}_{i,\ast}) \). When no prior knowledge on the form of
+\( f(\cdot) \) is available, it is common to assume a linear relationship
+between \( \boldsymbol{X} \) and \( \boldsymbol{y} \). This assumption gives rise to
+the linear regression model where \( \boldsymbol{\beta} = [\beta_0, \ldots,
+\beta_{p-1}]^{T} \) are the regression parameters.
+
+
+Linear regression gives us a set of analytical equations for the parameters \( \beta_j \).
+
+
+
+In order to understand the relation among the predictors \( p \), the set of data \( n \) and the target (outcome, output etc) \( \boldsymbol{y} \),
+consider the model we discussed for describing nuclear binding energies.
+
+
+There we assumed that we could parametrize the data using a polynomial approximation based on the liquid drop model.
+Assuming
+
+Here the predictors are based on a model we have made. A popular data set which is widely encountered in ML applications is the
+so-called credit card default data from Taiwan. The data set contains data on \( n=30000 \) credit card holders with predictors like gender, marital status, age, profession, education, etc. In total there are \( 24 \) such predictors or attributes leading to a design matrix of dimensionality \( 24 \times 30000 \). This is however a classification problem and we will come back to it when we discuss Logistic Regression.
+
+
+
+Before we proceed let us study a case from linear algebra where we aim at fitting a set of data \( \boldsymbol{y}=[y_0,y_1,\dots,y_{n-1}] \). We could think of these data as a result of an experiment or a complicated numerical experiment. These data are functions of a series of variables \( \boldsymbol{x}=[x_0,x_1,\dots,x_{n-1}] \), that is \( y_i = y(x_i) \) with \( i=0,1,2,\dots,n-1 \). The variables \( x_i \) could represent physical quantities like time, temperature, position etc. We assume that \( y(x) \) is a smooth function.
+
+
+Since obtaining these data points may not be trivial, we want to use these data to fit a function which can allow us to make predictions for values of \( y \) which are not in the present set. The perhaps simplest approach is to assume we can parametrize our function in terms of a polynomial of degree \( n-1 \) with \( n \) points, that is
+
+For every set of values \( y_i,x_i \) we have thus the corresponding set of equations
+
+Defining the vectors
+
+We are obviously not limited to the above polynomial expansions. We
+could replace the various powers of \( x \) with elements of Fourier
+series or instead of \( x_i^j \) we could have \( \cos{(j x_i)} \) or \( \sin{(j
+x_i)} \), or time series or other orthogonal functions. For every set
+of values \( y_i,x_i \) we can then generalize the equations to
+
+
+Note that we have \( p=n \) here. The matrix is symmetric. This is generally not the case!
+
+We redefine in turn the matrix \( \boldsymbol{X} \) as
+
+We have defined the matrix \( \boldsymbol{X} \) via the equations
+
+As we noted above, we stayed with a system with the design matrix
+ \( \boldsymbol{X}\in {\mathbb{R}}^{n\times n} \), that is we have \( p=n \). For reasons to come later (algorithmic arguments) we will hereafter define
+our matrix as \( \boldsymbol{X}\in {\mathbb{R}}^{n\times p} \), with the predictors refering to the column numbers and the entries \( n \) being the row elements.
+
+
+
+In our introductory notes we looked at the so-called liquid drop model. Let us remind ourselves about what we did by looking at the code.
+
+
+We restate the parts of the code we are most interested in.
+
+
+
+
+With \( \boldsymbol{\beta}\in {\mathbb{R}}^{p\times 1} \), it means that we will hereafter write our equations for the approximation as
+
+With the above we use the design matrix to define the approximation \( \boldsymbol{\tilde{y}} \) via the unknown quantity \( \boldsymbol{\beta} \) as
+
+It is also common to define
+the function \( C \) as
+
+
+The function
+
+where \( \langle y_i \rangle \) is the mean value. Keep in mind also that
+till now we have treated \( y_i \) as the exact value. Normally, the
+response (dependent or outcome) variable \( y_i \) the outcome of a
+numerical experiment or another type of experiment and is thus only an
+approximation to the true value. It is then always accompanied by an
+error estimate, often limited to a statistical error estimate given by
+the standard deviation discussed earlier. In the discussion here we
+will treat \( y_i \) as our exact value for the response variable.
+
+
+In order to find the parameters \( \beta_i \) we will then minimize the spread of \( C(\boldsymbol{\beta}) \), that is we are going to solve the problem
+
+We can rewrite
+
+We note also that since our design matrix is defined as \( \boldsymbol{X}\in
+{\mathbb{R}}^{n\times p} \), the product \( \boldsymbol{X}^T\boldsymbol{X} \in
+{\mathbb{R}}^{p\times p} \). In the above case we have that \( p \ll n \),
+in our case \( p=5 \) meaning that we end up with inverting a small
+\( 5\times 5 \) matrix. This is a rather common situation, in many cases we end up with low-dimensional
+matrices to invert. The methods discussed here and for many other
+supervised learning algorithms like classification with logistic
+regression or support vector machines, exhibit dimensionalities which
+allow for the usage of direct linear algebra methods such as LU decomposition or Singular Value Decomposition (SVD) for finding the inverse of the matrix
+\( \boldsymbol{X}^T\boldsymbol{X} \).
+
+
+Small question: Do you think the example we have at hand here (the nuclear binding energies) can lead to problems in inverting the matrix \( \boldsymbol{X}^T\boldsymbol{X} \)? What kind of problems can we expect?
+
+The following matrix and vector relation will be useful here and for the rest of the course. Vectors are always written as boldfaced lower case letters and
+matrices as upper case boldfaced letters.
+
+
+The residuals \( \boldsymbol{\epsilon} \) are in turn given by
+
+Let us now return to our nuclear binding energies and simply code the above equations.
+
+It is rather straightforward to implement the matrix inversion and obtain the parameters \( \boldsymbol{\beta} \). After having defined the matrix \( \boldsymbol{X} \) we simply need to
+write
+
+
+
+
+Alternatively, you can use the least squares functionality in Numpy as
+
+
+
+
+And finally we plot our fit with and compare with data
+
+
+
+
+We can easily test our fit by computing the \( R2 \) score that we discussed in connection with the functionality of Scikit-Learn in the introductory slides.
+Since we are not using Scikit-Learn here we can define our own \( R2 \) function as
+
+
+
+
+and we would be using it as
+
+
+
+
+We can easily add our MSE score as
+
+
+
+
+and finally the relative error as
+
+
+
+
+Normally, the response (dependent or outcome) variable \( y_i \) is the
+outcome of a numerical experiment or another type of experiment and is
+thus only an approximation to the true value. It is then always
+accompanied by an error estimate, often limited to a statistical error
+estimate given by the standard deviation discussed earlier. In the
+discussion here we will treat \( y_i \) as our exact value for the
+response variable.
+
+
+Introducing the standard deviation \( \sigma_i \) for each measurement
+\( y_i \), we define now the \( \chi^2 \) function (omitting the \( 1/n \) term)
+as
+
+
+In order to find the parameters \( \beta_i \) we will then minimize the spread of \( \chi^2(\boldsymbol{\beta}) \) by requiring
+
+We can rewrite
+
+If we then introduce the matrix
+
+The first step here is to approximate the function \( y \) with a first-order polynomial, that is we write
+
+For a linear fit (a first-order polynomial) we don't need to invert a matrix!!
+Defining
+
+we obtain
+
+
+This approach (different linear and non-linear regression) suffers
+often from both being underdetermined and overdetermined in the
+unknown coefficients \( \beta_i \). A better approach is to use the
+Singular Value Decomposition (SVD) method discussed next week.
+
+
+
+Before we continue, let us introduce yet another example. We are going to fit the
+nuclear equation of state using results from many-body calculations.
+The equation of state we have made available here, as function of
+density, has been derived using modern nucleon-nucleon potentials with
+the addition of three-body
+forces. This
+time the file is presented as a standard csv file.
+
+
+The beginning of the Python code here is similar to what you have seen
+before, with the same initializations and declarations. We use also
+pandas again, rather extensively in order to organize our data.
+
+
+The difference now is that we use Scikit-Learn's regression tools
+instead of our own matrix inversion implementation. Furthermore, we
+sneak in Ridge regression (to be discussed below) which includes a
+hyperparameter \( \lambda \), also to be explained below.
+
+
+
+
+The above simple polynomial in density \( \rho \) gives an excellent fit
+to the data.
+
+
+We note also that there is a small deviation between the
+standard OLS and the Ridge regression at higher densities. We discuss this in more detail
+below.
+
+It is normal in essentially all Machine Learning studies to split the
+data in a training set and a test set (sometimes also an additional
+validation set). Scikit-Learn has an own function for this. There
+is no explicit recipe for how much data should be included as training
+data and say test data. An accepted rule of thumb is to use
+approximately \( 2/3 \) to \( 4/5 \) of the data as training data. We will
+postpone a discussion of this splitting to the end of these notes and
+our discussion of the so-called bias-variance tradeoff. Here we
+limit ourselves to repeat the above equation of state fitting example
+but now splitting the data into a training set and a test set.
+
+
+
+
+
-Video of Lecture from Fall Semester 2020.
+The lectures will be recorded and updated videos will be posted after the lectures.
-The lectures will be recorded and updated videos will be posted after the lectures.
+"Video of Lecture August 26, 2021":"https://www.uio.no/studier/emner/matnat/fys/FYS-STK4155/h21/forelesningsvideoer/LectureThursdayAugust26.mp4?vrtx=view-as-webpage
Zoom link for lectures: https://msu.zoom.us/j/93311529525?pwd=a1VXSzY4aTFWVy9Rb05mNDJTZ09lZz09
@@ -283,6 +363,9 @@ The lectures will be recorded and updated videos will be posted after the lectur
-In order to study various Machine Learning algorithms, we need to
-access data. Acccessing data is an essential step in all machine
-learning algorithms. In particular, setting up the so-called design
-matrix (to be defined below) is often the first element we need in
-order to perform our calculations. To set up the design matrix means
-reading (and later, when the calculations are done, writing) data
-in various formats, The formats span from reading files from disk,
-loading data from databases and interacting with online sources
-like web application programming interfaces (APIs).
-
-
-In handling various input formats, as discussed above, we will mainly stay with pandas,
-a Python package which allows us, in a seamless and painless way, to
-deal with a multitude of formats, from standard csv (comma separated
-values) files, via excel, html to hdf5 formats. With pandas
-and the DataFrame and Series functionalities we are able to convert text data
-into the calculational formats we need for a specific algorithm. And our code is going to be
-pretty close the basic mathematical expressions.
-
-
-Our first data set is going to be a classic from nuclear physics, namely all
-available data on binding energies. Don't be intimidated if you are not familiar with nuclear physics. It serves merely as an example here of a data set.
-
-
-We will show some of the
-strengths of packages like Scikit-Learn in fitting nuclear binding energies to
-specific functions using linear regression first. Then, as a teaser, we will show you how
-you can easily implement other algorithms like decision trees and random forests and neural networks.
-
-
-But before we really start with nuclear physics data, let's just look at some simpler polynomial fitting cases, such as,
-(don't be offended) fitting straight lines!
-
-
-
-
@@ -2188,6 +2227,1065 @@ Now it is time to dive more into the details of various methods. We will start w
+Fitting a continuous function with linear parameterization in terms of the parameters \( \boldsymbol{\beta} \).
+
+
+
+
+
+Regression modeling deals with the description of the sampling distribution of a given random variable \( y \) and how it varies as function of another variable or a set of such variables \( \boldsymbol{x} =[x_0, x_1,\dots, x_{n-1}]^T \).
+The first variable is called the dependent, the outcome or the response variable while the set of variables \( \boldsymbol{x} \) is called the independent variable, or the predictor variable or the explanatory variable.
+
+
+A regression model aims at finding a likelihood function \( p(\boldsymbol{y}\vert \boldsymbol{x}) \), that is the conditional distribution for \( \boldsymbol{y} \) with a given \( \boldsymbol{x} \). The estimation of \( p(\boldsymbol{y}\vert \boldsymbol{x}) \) is made using a data set with
+
+
+
+
+
+Consider an experiment in which \( p \) characteristics of \( n \) samples are
+measured. The data from this experiment, for various explanatory variables \( p \) are normally represented by a matrix
+\( \mathbf{X} \).
+
+
+The matrix \( \mathbf{X} \) is called the design
+matrix. Additional information of the samples is available in the
+form of \( \boldsymbol{y} \) (also as above). The variable \( \boldsymbol{y} \) is
+generally referred to as the response variable. The aim of
+regression analysis is to explain \( \boldsymbol{y} \) in terms of
+\( \boldsymbol{X} \) through a functional relationship like \( y_i =
+f(\mathbf{X}_{i,\ast}) \). When no prior knowledge on the form of
+\( f(\cdot) \) is available, it is common to assume a linear relationship
+between \( \boldsymbol{X} \) and \( \boldsymbol{y} \). This assumption gives rise to
+the linear regression model where \( \boldsymbol{\beta} = [\beta_0, \ldots,
+\beta_{p-1}]^{T} \) are the regression parameters.
+
+
+Linear regression gives us a set of analytical equations for the parameters \( \beta_j \).
+
+
+
+
+In order to understand the relation among the predictors \( p \), the set of data \( n \) and the target (outcome, output etc) \( \boldsymbol{y} \),
+consider the model we discussed for describing nuclear binding energies.
+
+
+There we assumed that we could parametrize the data using a polynomial approximation based on the liquid drop model.
+Assuming
+$$
+BE(A) = a_0+a_1A+a_2A^{2/3}+a_3A^{-1/3}+a_4A^{-1},
+$$
+
+we have five predictors, that is the intercept, the \( A \) dependent term, the \( A^{2/3} \) term and the \( A^{-1/3} \) and \( A^{-1} \) terms.
+This gives \( p=0,1,2,3,4 \). Furthermore we have \( n \) entries for each predictor. It means that our design matrix is a
+\( p\times n \) matrix \( \boldsymbol{X} \).
+
+
+Here the predictors are based on a model we have made. A popular data set which is widely encountered in ML applications is the
+so-called credit card default data from Taiwan. The data set contains data on \( n=30000 \) credit card holders with predictors like gender, marital status, age, profession, education, etc. In total there are \( 24 \) such predictors or attributes leading to a design matrix of dimensionality \( 24 \times 30000 \). This is however a classification problem and we will come back to it when we discuss Logistic Regression.
+
+
+
+
+Before we proceed let us study a case from linear algebra where we aim at fitting a set of data \( \boldsymbol{y}=[y_0,y_1,\dots,y_{n-1}] \). We could think of these data as a result of an experiment or a complicated numerical experiment. These data are functions of a series of variables \( \boldsymbol{x}=[x_0,x_1,\dots,x_{n-1}] \), that is \( y_i = y(x_i) \) with \( i=0,1,2,\dots,n-1 \). The variables \( x_i \) could represent physical quantities like time, temperature, position etc. We assume that \( y(x) \) is a smooth function.
+
+
+Since obtaining these data points may not be trivial, we want to use these data to fit a function which can allow us to make predictions for values of \( y \) which are not in the present set. The perhaps simplest approach is to assume we can parametrize our function in terms of a polynomial of degree \( n-1 \) with \( n \) points, that is
+$$
+y=y(x) \rightarrow y(x_i)=\tilde{y}_i+\epsilon_i=\sum_{j=0}^{n-1} \beta_j x_i^j+\epsilon_i,
+$$
+
+where \( \epsilon_i \) is the error in our approximation.
+
+
+
+
+For every set of values \( y_i,x_i \) we have thus the corresponding set of equations
+$$
+\begin{align*}
+y_0&=\beta_0+\beta_1x_0^1+\beta_2x_0^2+\dots+\beta_{n-1}x_0^{n-1}+\epsilon_0\\
+y_1&=\beta_0+\beta_1x_1^1+\beta_2x_1^2+\dots+\beta_{n-1}x_1^{n-1}+\epsilon_1\\
+y_2&=\beta_0+\beta_1x_2^1+\beta_2x_2^2+\dots+\beta_{n-1}x_2^{n-1}+\epsilon_2\\
+\dots & \dots \\
+y_{n-1}&=\beta_0+\beta_1x_{n-1}^1+\beta_2x_{n-1}^2+\dots+\beta_{n-1}x_{n-1}^{n-1}+\epsilon_{n-1}.\\
+\end{align*}
+$$
+
+
+Defining the vectors
+$$
+\boldsymbol{y} = [y_0,y_1, y_2,\dots, y_{n-1}]^T,
+$$
+
+and
+$$
+\boldsymbol{\beta} = [\beta_0,\beta_1, \beta_2,\dots, \beta_{n-1}]^T,
+$$
+
+and
+$$
+\boldsymbol{\epsilon} = [\epsilon_0,\epsilon_1, \epsilon_2,\dots, \epsilon_{n-1}]^T,
+$$
+
+and the design matrix
+$$
+\boldsymbol{X}=
+\begin{bmatrix}
+1& x_{0}^1 &x_{0}^2& \dots & \dots &x_{0}^{n-1}\\
+1& x_{1}^1 &x_{1}^2& \dots & \dots &x_{1}^{n-1}\\
+1& x_{2}^1 &x_{2}^2& \dots & \dots &x_{2}^{n-1}\\
+\dots& \dots &\dots& \dots & \dots &\dots\\
+1& x_{n-1}^1 &x_{n-1}^2& \dots & \dots &x_{n-1}^{n-1}\\
+\end{bmatrix}
+$$
+
+we can rewrite our equations as
+$$
+\boldsymbol{y} = \boldsymbol{X}\boldsymbol{\beta}+\boldsymbol{\epsilon}.
+$$
+
+The above design matrix is called a Vandermonde matrix.
+
+
+
+
+We are obviously not limited to the above polynomial expansions. We
+could replace the various powers of \( x \) with elements of Fourier
+series or instead of \( x_i^j \) we could have \( \cos{(j x_i)} \) or \( \sin{(j
+x_i)} \), or time series or other orthogonal functions. For every set
+of values \( y_i,x_i \) we can then generalize the equations to
+
+$$
+\begin{align*}
+y_0&=\beta_0x_{00}+\beta_1x_{01}+\beta_2x_{02}+\dots+\beta_{n-1}x_{0n-1}+\epsilon_0\\
+y_1&=\beta_0x_{10}+\beta_1x_{11}+\beta_2x_{12}+\dots+\beta_{n-1}x_{1n-1}+\epsilon_1\\
+y_2&=\beta_0x_{20}+\beta_1x_{21}+\beta_2x_{22}+\dots+\beta_{n-1}x_{2n-1}+\epsilon_2\\
+\dots & \dots \\
+y_{i}&=\beta_0x_{i0}+\beta_1x_{i1}+\beta_2x_{i2}+\dots+\beta_{n-1}x_{in-1}+\epsilon_i\\
+\dots & \dots \\
+y_{n-1}&=\beta_0x_{n-1,0}+\beta_1x_{n-1,2}+\beta_2x_{n-1,2}+\dots+\beta_{n-1}x_{n-1,n-1}+\epsilon_{n-1}.\\
+\end{align*}
+$$
+
+
+Note that we have \( p=n \) here. The matrix is symmetric. This is generally not the case!
+
+
+We redefine in turn the matrix \( \boldsymbol{X} \) as
+$$
+\boldsymbol{X}=
+\begin{bmatrix}
+x_{00}& x_{01} &x_{02}& \dots & \dots &x_{0,n-1}\\
+x_{10}& x_{11} &x_{12}& \dots & \dots &x_{1,n-1}\\
+x_{20}& x_{21} &x_{22}& \dots & \dots &x_{2,n-1}\\
+\dots& \dots &\dots& \dots & \dots &\dots\\
+x_{n-1,0}& x_{n-1,1} &x_{n-1,2}& \dots & \dots &x_{n-1,n-1}\\
+\end{bmatrix}
+$$
+
+and without loss of generality we rewrite again our equations as
+$$
+\boldsymbol{y} = \boldsymbol{X}\boldsymbol{\beta}+\boldsymbol{\epsilon}.
+$$
+
+The left-hand side of this equation is kwown. Our error vector \( \boldsymbol{\epsilon} \) and the parameter vector \( \boldsymbol{\beta} \) are our unknow quantities. How can we obtain the optimal set of \( \beta_i \) values?
+
+
+We have defined the matrix \( \boldsymbol{X} \) via the equations
+$$
+\begin{align*}
+y_0&=\beta_0x_{00}+\beta_1x_{01}+\beta_2x_{02}+\dots+\beta_{n-1}x_{0n-1}+\epsilon_0\\
+y_1&=\beta_0x_{10}+\beta_1x_{11}+\beta_2x_{12}+\dots+\beta_{n-1}x_{1n-1}+\epsilon_1\\
+y_2&=\beta_0x_{20}+\beta_1x_{21}+\beta_2x_{22}+\dots+\beta_{n-1}x_{2n-1}+\epsilon_1\\
+\dots & \dots \\
+y_{i}&=\beta_0x_{i0}+\beta_1x_{i1}+\beta_2x_{i2}+\dots+\beta_{n-1}x_{in-1}+\epsilon_1\\
+\dots & \dots \\
+y_{n-1}&=\beta_0x_{n-1,0}+\beta_1x_{n-1,2}+\beta_2x_{n-1,2}+\dots+\beta_{n-1}x_{n-1,n-1}+\epsilon_{n-1}.\\
+\end{align*}
+$$
+
+
+As we noted above, we stayed with a system with the design matrix
+ \( \boldsymbol{X}\in {\mathbb{R}}^{n\times n} \), that is we have \( p=n \). For reasons to come later (algorithmic arguments) we will hereafter define
+our matrix as \( \boldsymbol{X}\in {\mathbb{R}}^{n\times p} \), with the predictors refering to the column numbers and the entries \( n \) being the row elements.
+
+
+
+
+In our introductory notes we looked at the so-called liquid drop model. Let us remind ourselves about what we did by looking at the code.
+
+
+We restate the parts of the code we are most interested in.
+
+
+
+
+With \( \boldsymbol{\beta}\in {\mathbb{R}}^{p\times 1} \), it means that we will hereafter write our equations for the approximation as
+$$
+\boldsymbol{\tilde{y}}= \boldsymbol{X}\boldsymbol{\beta},
+$$
+
+throughout these lectures.
+
+
+
+With the above we use the design matrix to define the approximation \( \boldsymbol{\tilde{y}} \) via the unknown quantity \( \boldsymbol{\beta} \) as
+$$
+\boldsymbol{\tilde{y}}= \boldsymbol{X}\boldsymbol{\beta},
+$$
+
+and in order to find the optimal parameters \( \beta_i \) instead of solving the above linear algebra problem, we define a function which gives a measure of the spread between the values \( y_i \) (which represent hopefully the exact values) and the parameterized values \( \tilde{y}_i \), namely
+$$
+C(\boldsymbol{\beta})=\frac{1}{n}\sum_{i=0}^{n-1}\left(y_i-\tilde{y}_i\right)^2=\frac{1}{n}\left\{\left(\boldsymbol{y}-\boldsymbol{\tilde{y}}\right)^T\left(\boldsymbol{y}-\boldsymbol{\tilde{y}}\right)\right\},
+$$
+
+or using the matrix \( \boldsymbol{X} \) and in a more compact matrix-vector notation as
+$$
+C(\boldsymbol{\beta})=\frac{1}{n}\left\{\left(\boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\right)^T\left(\boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\right)\right\}.
+$$
+
+This function is one possible way to define the so-called cost function.
+
+
+It is also common to define
+the function \( C \) as
+
+$$
+C(\boldsymbol{\beta})=\frac{1}{2n}\sum_{i=0}^{n-1}\left(y_i-\tilde{y}_i\right)^2,
+$$
+
+since when taking the first derivative with respect to the unknown parameters \( \beta \), the factor of \( 2 \) cancels out.
+
+
+
+
+The function
+$$
+C(\boldsymbol{\beta})=\frac{1}{n}\left\{\left(\boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\right)^T\left(\boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\right)\right\},
+$$
+
+can be linked to the variance of the quantity \( y_i \) if we interpret the latter as the mean value.
+When linking (see the discussion below) with the maximum likelihood approach below, we will indeed interpret \( y_i \) as a mean value
+$$
+y_{i}=\langle y_i \rangle = \beta_0x_{i,0}+\beta_1x_{i,1}+\beta_2x_{i,2}+\dots+\beta_{n-1}x_{i,n-1}+\epsilon_i,
+$$
+
+
+where \( \langle y_i \rangle \) is the mean value. Keep in mind also that
+till now we have treated \( y_i \) as the exact value. Normally, the
+response (dependent or outcome) variable \( y_i \) the outcome of a
+numerical experiment or another type of experiment and is thus only an
+approximation to the true value. It is then always accompanied by an
+error estimate, often limited to a statistical error estimate given by
+the standard deviation discussed earlier. In the discussion here we
+will treat \( y_i \) as our exact value for the response variable.
+
+
+In order to find the parameters \( \beta_i \) we will then minimize the spread of \( C(\boldsymbol{\beta}) \), that is we are going to solve the problem
+$$
+{\displaystyle \min_{\boldsymbol{\beta}\in
+{\mathbb{R}}^{p}}}\frac{1}{n}\left\{\left(\boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\right)^T\left(\boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\right)\right\}.
+$$
+
+In practical terms it means we will require
+$$
+\frac{\partial C(\boldsymbol{\beta})}{\partial \beta_j} = \frac{\partial }{\partial \beta_j}\left[ \frac{1}{n}\sum_{i=0}^{n-1}\left(y_i-\beta_0x_{i,0}-\beta_1x_{i,1}-\beta_2x_{i,2}-\dots-\beta_{n-1}x_{i,n-1}\right)^2\right]=0,
+$$
+
+which results in
+$$
+\frac{\partial C(\boldsymbol{\beta})}{\partial \beta_j} = -\frac{2}{n}\left[ \sum_{i=0}^{n-1}x_{ij}\left(y_i-\beta_0x_{i,0}-\beta_1x_{i,1}-\beta_2x_{i,2}-\dots-\beta_{n-1}x_{i,n-1}\right)\right]=0,
+$$
+
+or in a matrix-vector form as
+$$
+\frac{\partial C(\boldsymbol{\beta})}{\partial \boldsymbol{\beta}} = 0 = \boldsymbol{X}^T\left( \boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\right).
+$$
+
+
+
+
+We can rewrite
+$$
+\frac{\partial C(\boldsymbol{\beta})}{\partial \boldsymbol{\beta}} = 0 = \boldsymbol{X}^T\left( \boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\right),
+$$
+
+as
+$$
+\boldsymbol{X}^T\boldsymbol{y} = \boldsymbol{X}^T\boldsymbol{X}\boldsymbol{\beta},
+$$
+
+and if the matrix \( \boldsymbol{X}^T\boldsymbol{X} \) is invertible we have the solution
+$$
+\boldsymbol{\beta} =\left(\boldsymbol{X}^T\boldsymbol{X}\right)^{-1}\boldsymbol{X}^T\boldsymbol{y}.
+$$
+
+
+We note also that since our design matrix is defined as \( \boldsymbol{X}\in
+{\mathbb{R}}^{n\times p} \), the product \( \boldsymbol{X}^T\boldsymbol{X} \in
+{\mathbb{R}}^{p\times p} \). In the above case we have that \( p \ll n \),
+in our case \( p=5 \) meaning that we end up with inverting a small
+\( 5\times 5 \) matrix. This is a rather common situation, in many cases we end up with low-dimensional
+matrices to invert. The methods discussed here and for many other
+supervised learning algorithms like classification with logistic
+regression or support vector machines, exhibit dimensionalities which
+allow for the usage of direct linear algebra methods such as LU decomposition or Singular Value Decomposition (SVD) for finding the inverse of the matrix
+\( \boldsymbol{X}^T\boldsymbol{X} \).
+
+
+Small question: Do you think the example we have at hand here (the nuclear binding energies) can lead to problems in inverting the matrix \( \boldsymbol{X}^T\boldsymbol{X} \)? What kind of problems can we expect?
+
+
+The following matrix and vector relation will be useful here and for the rest of the course. Vectors are always written as boldfaced lower case letters and
+matrices as upper case boldfaced letters.
+
+$$
+\frac{\partial (\boldsymbol{b}^T\boldsymbol{a})}{\partial \boldsymbol{a}} = \boldsymbol{b},
+$$
+
+$$
+\frac{\partial (\boldsymbol{a}^T\boldsymbol{A}\boldsymbol{a})}{\partial \boldsymbol{a}} = (\boldsymbol{A}+\boldsymbol{A}^T)\boldsymbol{a},
+$$
+
+$$
+\frac{\partial tr(\boldsymbol{B}\boldsymbol{A})}{\partial \boldsymbol{A}} = \boldsymbol{B}^T,
+$$
+
+$$
+\frac{\partial \log{\vert\boldsymbol{A}\vert}}{\partial \boldsymbol{A}} = (\boldsymbol{A}^{-1})^T.
+$$
+
+
+The residuals \( \boldsymbol{\epsilon} \) are in turn given by
+$$
+\boldsymbol{\epsilon} = \boldsymbol{y}-\boldsymbol{\tilde{y}} = \boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta},
+$$
+
+and with
+$$
+\boldsymbol{X}^T\left( \boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\right)= 0,
+$$
+
+we have
+$$
+\boldsymbol{X}^T\boldsymbol{\epsilon}=\boldsymbol{X}^T\left( \boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\right)= 0,
+$$
+
+meaning that the solution for \( \boldsymbol{\beta} \) is the one which minimizes the residuals. Later we will link this with the maximum likelihood approach.
+
+
+
+Let us now return to our nuclear binding energies and simply code the above equations.
+
+
+
+It is rather straightforward to implement the matrix inversion and obtain the parameters \( \boldsymbol{\beta} \). After having defined the matrix \( \boldsymbol{X} \) we simply need to
+write
+
+
+
+
+Alternatively, you can use the least squares functionality in Numpy as
+
+
+
+
+And finally we plot our fit with and compare with data
+
+
+
+
+
+We can easily test our fit by computing the \( R2 \) score that we discussed in connection with the functionality of Scikit-Learn in the introductory slides.
+Since we are not using Scikit-Learn here we can define our own \( R2 \) function as
+
+
+
+
+and we would be using it as
+
+
+
+
+We can easily add our MSE score as
+
+
+
+
+and finally the relative error as
+
+
+
+
+
+
+
+Normally, the response (dependent or outcome) variable \( y_i \) is the
+outcome of a numerical experiment or another type of experiment and is
+thus only an approximation to the true value. It is then always
+accompanied by an error estimate, often limited to a statistical error
+estimate given by the standard deviation discussed earlier. In the
+discussion here we will treat \( y_i \) as our exact value for the
+response variable.
+
+
+Introducing the standard deviation \( \sigma_i \) for each measurement
+\( y_i \), we define now the \( \chi^2 \) function (omitting the \( 1/n \) term)
+as
+
+$$
+\chi^2(\boldsymbol{\beta})=\frac{1}{n}\sum_{i=0}^{n-1}\frac{\left(y_i-\tilde{y}_i\right)^2}{\sigma_i^2}=\frac{1}{n}\left\{\left(\boldsymbol{y}-\boldsymbol{\tilde{y}}\right)^T\frac{1}{\boldsymbol{\Sigma^2}}\left(\boldsymbol{y}-\boldsymbol{\tilde{y}}\right)\right\},
+$$
+
+where the matrix \( \boldsymbol{\Sigma} \) is a diagonal matrix with \( \sigma_i \) as matrix elements.
+
+
+
+
+
+
+In order to find the parameters \( \beta_i \) we will then minimize the spread of \( \chi^2(\boldsymbol{\beta}) \) by requiring
+$$
+\frac{\partial \chi^2(\boldsymbol{\beta})}{\partial \beta_j} = \frac{\partial }{\partial \beta_j}\left[ \frac{1}{n}\sum_{i=0}^{n-1}\left(\frac{y_i-\beta_0x_{i,0}-\beta_1x_{i,1}-\beta_2x_{i,2}-\dots-\beta_{n-1}x_{i,n-1}}{\sigma_i}\right)^2\right]=0,
+$$
+
+which results in
+$$
+\frac{\partial \chi^2(\boldsymbol{\beta})}{\partial \beta_j} = -\frac{2}{n}\left[ \sum_{i=0}^{n-1}\frac{x_{ij}}{\sigma_i}\left(\frac{y_i-\beta_0x_{i,0}-\beta_1x_{i,1}-\beta_2x_{i,2}-\dots-\beta_{n-1}x_{i,n-1}}{\sigma_i}\right)\right]=0,
+$$
+
+or in a matrix-vector form as
+$$
+\frac{\partial \chi^2(\boldsymbol{\beta})}{\partial \boldsymbol{\beta}} = 0 = \boldsymbol{A}^T\left( \boldsymbol{b}-\boldsymbol{A}\boldsymbol{\beta}\right).
+$$
+
+where we have defined the matrix \( \boldsymbol{A} =\boldsymbol{X}/\boldsymbol{\Sigma} \) with matrix elements \( a_{ij} = x_{ij}/\sigma_i \) and the vector \( \boldsymbol{b} \) with elements \( b_i = y_i/\sigma_i \).
+
+
+
+
+We can rewrite
+$$
+\frac{\partial \chi^2(\boldsymbol{\beta})}{\partial \boldsymbol{\beta}} = 0 = \boldsymbol{A}^T\left( \boldsymbol{b}-\boldsymbol{A}\boldsymbol{\beta}\right),
+$$
+
+as
+$$
+\boldsymbol{A}^T\boldsymbol{b} = \boldsymbol{A}^T\boldsymbol{A}\boldsymbol{\beta},
+$$
+
+and if the matrix \( \boldsymbol{A}^T\boldsymbol{A} \) is invertible we have the solution
+$$
+\boldsymbol{\beta} =\left(\boldsymbol{A}^T\boldsymbol{A}\right)^{-1}\boldsymbol{A}^T\boldsymbol{b}.
+$$
+
+
+
+
+If we then introduce the matrix
+$$
+\boldsymbol{H} = \left(\boldsymbol{A}^T\boldsymbol{A}\right)^{-1},
+$$
+
+we have then the following expression for the parameters \( \beta_j \) (the matrix elements of \( \boldsymbol{H} \) are \( h_{ij} \))
+$$
+\beta_j = \sum_{k=0}^{p-1}h_{jk}\sum_{i=0}^{n-1}\frac{y_i}{\sigma_i}\frac{x_{ik}}{\sigma_i} = \sum_{k=0}^{p-1}h_{jk}\sum_{i=0}^{n-1}b_ia_{ik}
+$$
+
+We state without proof the expression for the uncertainty in the parameters \( \beta_j \) as (we leave this as an exercise)
+$$
+\sigma^2(\beta_j) = \sum_{i=0}^{n-1}\sigma_i^2\left( \frac{\partial \beta_j}{\partial y_i}\right)^2,
+$$
+
+resulting in
+$$
+\sigma^2(\beta_j) = \left(\sum_{k=0}^{p-1}h_{jk}\sum_{i=0}^{n-1}a_{ik}\right)\left(\sum_{l=0}^{p-1}h_{jl}\sum_{m=0}^{n-1}a_{ml}\right) = h_{jj}!
+$$
+
+
+The first step here is to approximate the function \( y \) with a first-order polynomial, that is we write
+$$
+y=y(x) \rightarrow y(x_i) \approx \beta_0+\beta_1 x_i.
+$$
+
+By computing the derivatives of \( \chi^2 \) with respect to \( \beta_0 \) and \( \beta_1 \) show that these are given by
+$$
+\frac{\partial \chi^2(\boldsymbol{\beta})}{\partial \beta_0} = -2\left[ \frac{1}{n}\sum_{i=0}^{n-1}\left(\frac{y_i-\beta_0-\beta_1x_{i}}{\sigma_i^2}\right)\right]=0,
+$$
+
+and
+$$
+\frac{\partial \chi^2(\boldsymbol{\beta})}{\partial \beta_1} = -\frac{2}{n}\left[ \sum_{i=0}^{n-1}x_i\left(\frac{y_i-\beta_0-\beta_1x_{i}}{\sigma_i^2}\right)\right]=0.
+$$
+
+
+
+
+For a linear fit (a first-order polynomial) we don't need to invert a matrix!!
+Defining
+$$
+\gamma = \sum_{i=0}^{n-1}\frac{1}{\sigma_i^2},
+$$
+
+
+$$
+\gamma_x = \sum_{i=0}^{n-1}\frac{x_{i}}{\sigma_i^2},
+$$
+
+
+$$
+\gamma_y = \sum_{i=0}^{n-1}\left(\frac{y_i}{\sigma_i^2}\right),
+$$
+
+
+$$
+\gamma_{xx} = \sum_{i=0}^{n-1}\frac{x_ix_{i}}{\sigma_i^2},
+$$
+
+
+$$
+\gamma_{xy} = \sum_{i=0}^{n-1}\frac{y_ix_{i}}{\sigma_i^2},
+$$
+
+
+we obtain
+
+$$
+\beta_0 = \frac{\gamma_{xx}\gamma_y-\gamma_x\gamma_y}{\gamma\gamma_{xx}-\gamma_x^2},
+$$
+
+
+$$
+\beta_1 = \frac{\gamma_{xy}\gamma-\gamma_x\gamma_y}{\gamma\gamma_{xx}-\gamma_x^2}.
+$$
+
+
+This approach (different linear and non-linear regression) suffers
+often from both being underdetermined and overdetermined in the
+unknown coefficients \( \beta_i \). A better approach is to use the
+Singular Value Decomposition (SVD) method discussed next week.
+
+
+
+
+Before we continue, let us introduce yet another example. We are going to fit the
+nuclear equation of state using results from many-body calculations.
+The equation of state we have made available here, as function of
+density, has been derived using modern nucleon-nucleon potentials with
+the addition of three-body
+forces. This
+time the file is presented as a standard csv file.
+
+
+The beginning of the Python code here is similar to what you have seen
+before, with the same initializations and declarations. We use also
+pandas again, rather extensively in order to organize our data.
+
+
+The difference now is that we use Scikit-Learn's regression tools
+instead of our own matrix inversion implementation. Furthermore, we
+sneak in Ridge regression (to be discussed below) which includes a
+hyperparameter \( \lambda \), also to be explained below.
+
+
+
+
+
+
+The above simple polynomial in density \( \rho \) gives an excellent fit
+to the data.
+
+
+We note also that there is a small deviation between the
+standard OLS and the Ridge regression at higher densities. We discuss this in more detail
+below.
+
+
+
+It is normal in essentially all Machine Learning studies to split the
+data in a training set and a test set (sometimes also an additional
+validation set). Scikit-Learn has an own function for this. There
+is no explicit recipe for how much data should be included as training
+data and say test data. An accepted rule of thumb is to use
+approximately \( 2/3 \) to \( 4/5 \) of the data as training data. We will
+postpone a discussion of this splitting to the end of these notes and
+our discussion of the so-called bias-variance tradeoff. Here we
+limit ourselves to repeat the above equation of state fitting example
+but now splitting the data into a training set and a test set.
+
+
+
+
+
+
-Video of Lecture from Fall Semester 2020.
+The lectures will be recorded and updated videos will be posted after the lectures.
-The lectures will be recorded and updated videos will be posted after the lectures.
+"Video of Lecture August 26, 2021":"https://www.uio.no/studier/emner/matnat/fys/FYS-STK4155/h21/forelesningsvideoer/LectureThursdayAugust26.mp4?vrtx=view-as-webpage
Zoom link for lectures: https://msu.zoom.us/j/93311529525?pwd=a1VXSzY4aTFWVy9Rb05mNDJTZ09lZz09
@@ -288,6 +368,9 @@ The lectures will be recorded and updated videos will be posted after the lectur
-In order to study various Machine Learning algorithms, we need to
-access data. Acccessing data is an essential step in all machine
-learning algorithms. In particular, setting up the so-called design
-matrix (to be defined below) is often the first element we need in
-order to perform our calculations. To set up the design matrix means
-reading (and later, when the calculations are done, writing) data
-in various formats, The formats span from reading files from disk,
-loading data from databases and interacting with online sources
-like web application programming interfaces (APIs).
-
-
-In handling various input formats, as discussed above, we will mainly stay with pandas,
-a Python package which allows us, in a seamless and painless way, to
-deal with a multitude of formats, from standard csv (comma separated
-values) files, via excel, html to hdf5 formats. With pandas
-and the DataFrame and Series functionalities we are able to convert text data
-into the calculational formats we need for a specific algorithm. And our code is going to be
-pretty close the basic mathematical expressions.
-
-
-Our first data set is going to be a classic from nuclear physics, namely all
-available data on binding energies. Don't be intimidated if you are not familiar with nuclear physics. It serves merely as an example here of a data set.
-
-
-We will show some of the
-strengths of packages like Scikit-Learn in fitting nuclear binding energies to
-specific functions using linear regression first. Then, as a teaser, we will show you how
-you can easily implement other algorithms like decision trees and random forests and neural networks.
-
-
-But before we really start with nuclear physics data, let's just look at some simpler polynomial fitting cases, such as,
-(don't be offended) fitting straight lines!
-
-
-
-
@@ -2193,6 +2232,1065 @@ Now it is time to dive more into the details of various methods. We will start w
+Fitting a continuous function with linear parameterization in terms of the parameters \( \boldsymbol{\beta} \).
+
+
+
+
+
+Regression modeling deals with the description of the sampling distribution of a given random variable \( y \) and how it varies as function of another variable or a set of such variables \( \boldsymbol{x} =[x_0, x_1,\dots, x_{n-1}]^T \).
+The first variable is called the dependent, the outcome or the response variable while the set of variables \( \boldsymbol{x} \) is called the independent variable, or the predictor variable or the explanatory variable.
+
+
+A regression model aims at finding a likelihood function \( p(\boldsymbol{y}\vert \boldsymbol{x}) \), that is the conditional distribution for \( \boldsymbol{y} \) with a given \( \boldsymbol{x} \). The estimation of \( p(\boldsymbol{y}\vert \boldsymbol{x}) \) is made using a data set with
+
+
+
+
+
+Consider an experiment in which \( p \) characteristics of \( n \) samples are
+measured. The data from this experiment, for various explanatory variables \( p \) are normally represented by a matrix
+\( \mathbf{X} \).
+
+
+The matrix \( \mathbf{X} \) is called the design
+matrix. Additional information of the samples is available in the
+form of \( \boldsymbol{y} \) (also as above). The variable \( \boldsymbol{y} \) is
+generally referred to as the response variable. The aim of
+regression analysis is to explain \( \boldsymbol{y} \) in terms of
+\( \boldsymbol{X} \) through a functional relationship like \( y_i =
+f(\mathbf{X}_{i,\ast}) \). When no prior knowledge on the form of
+\( f(\cdot) \) is available, it is common to assume a linear relationship
+between \( \boldsymbol{X} \) and \( \boldsymbol{y} \). This assumption gives rise to
+the linear regression model where \( \boldsymbol{\beta} = [\beta_0, \ldots,
+\beta_{p-1}]^{T} \) are the regression parameters.
+
+
+Linear regression gives us a set of analytical equations for the parameters \( \beta_j \).
+
+
+
+
+In order to understand the relation among the predictors \( p \), the set of data \( n \) and the target (outcome, output etc) \( \boldsymbol{y} \),
+consider the model we discussed for describing nuclear binding energies.
+
+
+There we assumed that we could parametrize the data using a polynomial approximation based on the liquid drop model.
+Assuming
+$$
+BE(A) = a_0+a_1A+a_2A^{2/3}+a_3A^{-1/3}+a_4A^{-1},
+$$
+
+we have five predictors, that is the intercept, the \( A \) dependent term, the \( A^{2/3} \) term and the \( A^{-1/3} \) and \( A^{-1} \) terms.
+This gives \( p=0,1,2,3,4 \). Furthermore we have \( n \) entries for each predictor. It means that our design matrix is a
+\( p\times n \) matrix \( \boldsymbol{X} \).
+
+
+Here the predictors are based on a model we have made. A popular data set which is widely encountered in ML applications is the
+so-called credit card default data from Taiwan. The data set contains data on \( n=30000 \) credit card holders with predictors like gender, marital status, age, profession, education, etc. In total there are \( 24 \) such predictors or attributes leading to a design matrix of dimensionality \( 24 \times 30000 \). This is however a classification problem and we will come back to it when we discuss Logistic Regression.
+
+
+
+
+Before we proceed let us study a case from linear algebra where we aim at fitting a set of data \( \boldsymbol{y}=[y_0,y_1,\dots,y_{n-1}] \). We could think of these data as a result of an experiment or a complicated numerical experiment. These data are functions of a series of variables \( \boldsymbol{x}=[x_0,x_1,\dots,x_{n-1}] \), that is \( y_i = y(x_i) \) with \( i=0,1,2,\dots,n-1 \). The variables \( x_i \) could represent physical quantities like time, temperature, position etc. We assume that \( y(x) \) is a smooth function.
+
+
+Since obtaining these data points may not be trivial, we want to use these data to fit a function which can allow us to make predictions for values of \( y \) which are not in the present set. The perhaps simplest approach is to assume we can parametrize our function in terms of a polynomial of degree \( n-1 \) with \( n \) points, that is
+$$
+y=y(x) \rightarrow y(x_i)=\tilde{y}_i+\epsilon_i=\sum_{j=0}^{n-1} \beta_j x_i^j+\epsilon_i,
+$$
+
+where \( \epsilon_i \) is the error in our approximation.
+
+
+
+
+For every set of values \( y_i,x_i \) we have thus the corresponding set of equations
+$$
+\begin{align*}
+y_0&=\beta_0+\beta_1x_0^1+\beta_2x_0^2+\dots+\beta_{n-1}x_0^{n-1}+\epsilon_0\\
+y_1&=\beta_0+\beta_1x_1^1+\beta_2x_1^2+\dots+\beta_{n-1}x_1^{n-1}+\epsilon_1\\
+y_2&=\beta_0+\beta_1x_2^1+\beta_2x_2^2+\dots+\beta_{n-1}x_2^{n-1}+\epsilon_2\\
+\dots & \dots \\
+y_{n-1}&=\beta_0+\beta_1x_{n-1}^1+\beta_2x_{n-1}^2+\dots+\beta_{n-1}x_{n-1}^{n-1}+\epsilon_{n-1}.\\
+\end{align*}
+$$
+
+
+Defining the vectors
+$$
+\boldsymbol{y} = [y_0,y_1, y_2,\dots, y_{n-1}]^T,
+$$
+
+and
+$$
+\boldsymbol{\beta} = [\beta_0,\beta_1, \beta_2,\dots, \beta_{n-1}]^T,
+$$
+
+and
+$$
+\boldsymbol{\epsilon} = [\epsilon_0,\epsilon_1, \epsilon_2,\dots, \epsilon_{n-1}]^T,
+$$
+
+and the design matrix
+$$
+\boldsymbol{X}=
+\begin{bmatrix}
+1& x_{0}^1 &x_{0}^2& \dots & \dots &x_{0}^{n-1}\\
+1& x_{1}^1 &x_{1}^2& \dots & \dots &x_{1}^{n-1}\\
+1& x_{2}^1 &x_{2}^2& \dots & \dots &x_{2}^{n-1}\\
+\dots& \dots &\dots& \dots & \dots &\dots\\
+1& x_{n-1}^1 &x_{n-1}^2& \dots & \dots &x_{n-1}^{n-1}\\
+\end{bmatrix}
+$$
+
+we can rewrite our equations as
+$$
+\boldsymbol{y} = \boldsymbol{X}\boldsymbol{\beta}+\boldsymbol{\epsilon}.
+$$
+
+The above design matrix is called a Vandermonde matrix.
+
+
+
+
+We are obviously not limited to the above polynomial expansions. We
+could replace the various powers of \( x \) with elements of Fourier
+series or instead of \( x_i^j \) we could have \( \cos{(j x_i)} \) or \( \sin{(j
+x_i)} \), or time series or other orthogonal functions. For every set
+of values \( y_i,x_i \) we can then generalize the equations to
+
+$$
+\begin{align*}
+y_0&=\beta_0x_{00}+\beta_1x_{01}+\beta_2x_{02}+\dots+\beta_{n-1}x_{0n-1}+\epsilon_0\\
+y_1&=\beta_0x_{10}+\beta_1x_{11}+\beta_2x_{12}+\dots+\beta_{n-1}x_{1n-1}+\epsilon_1\\
+y_2&=\beta_0x_{20}+\beta_1x_{21}+\beta_2x_{22}+\dots+\beta_{n-1}x_{2n-1}+\epsilon_2\\
+\dots & \dots \\
+y_{i}&=\beta_0x_{i0}+\beta_1x_{i1}+\beta_2x_{i2}+\dots+\beta_{n-1}x_{in-1}+\epsilon_i\\
+\dots & \dots \\
+y_{n-1}&=\beta_0x_{n-1,0}+\beta_1x_{n-1,2}+\beta_2x_{n-1,2}+\dots+\beta_{n-1}x_{n-1,n-1}+\epsilon_{n-1}.\\
+\end{align*}
+$$
+
+
+Note that we have \( p=n \) here. The matrix is symmetric. This is generally not the case!
+
+
+We redefine in turn the matrix \( \boldsymbol{X} \) as
+$$
+\boldsymbol{X}=
+\begin{bmatrix}
+x_{00}& x_{01} &x_{02}& \dots & \dots &x_{0,n-1}\\
+x_{10}& x_{11} &x_{12}& \dots & \dots &x_{1,n-1}\\
+x_{20}& x_{21} &x_{22}& \dots & \dots &x_{2,n-1}\\
+\dots& \dots &\dots& \dots & \dots &\dots\\
+x_{n-1,0}& x_{n-1,1} &x_{n-1,2}& \dots & \dots &x_{n-1,n-1}\\
+\end{bmatrix}
+$$
+
+and without loss of generality we rewrite again our equations as
+$$
+\boldsymbol{y} = \boldsymbol{X}\boldsymbol{\beta}+\boldsymbol{\epsilon}.
+$$
+
+The left-hand side of this equation is kwown. Our error vector \( \boldsymbol{\epsilon} \) and the parameter vector \( \boldsymbol{\beta} \) are our unknow quantities. How can we obtain the optimal set of \( \beta_i \) values?
+
+
+We have defined the matrix \( \boldsymbol{X} \) via the equations
+$$
+\begin{align*}
+y_0&=\beta_0x_{00}+\beta_1x_{01}+\beta_2x_{02}+\dots+\beta_{n-1}x_{0n-1}+\epsilon_0\\
+y_1&=\beta_0x_{10}+\beta_1x_{11}+\beta_2x_{12}+\dots+\beta_{n-1}x_{1n-1}+\epsilon_1\\
+y_2&=\beta_0x_{20}+\beta_1x_{21}+\beta_2x_{22}+\dots+\beta_{n-1}x_{2n-1}+\epsilon_1\\
+\dots & \dots \\
+y_{i}&=\beta_0x_{i0}+\beta_1x_{i1}+\beta_2x_{i2}+\dots+\beta_{n-1}x_{in-1}+\epsilon_1\\
+\dots & \dots \\
+y_{n-1}&=\beta_0x_{n-1,0}+\beta_1x_{n-1,2}+\beta_2x_{n-1,2}+\dots+\beta_{n-1}x_{n-1,n-1}+\epsilon_{n-1}.\\
+\end{align*}
+$$
+
+
+As we noted above, we stayed with a system with the design matrix
+ \( \boldsymbol{X}\in {\mathbb{R}}^{n\times n} \), that is we have \( p=n \). For reasons to come later (algorithmic arguments) we will hereafter define
+our matrix as \( \boldsymbol{X}\in {\mathbb{R}}^{n\times p} \), with the predictors refering to the column numbers and the entries \( n \) being the row elements.
+
+
+
+
+In our introductory notes we looked at the so-called liquid drop model. Let us remind ourselves about what we did by looking at the code.
+
+
+We restate the parts of the code we are most interested in.
+
+
+
+
+With \( \boldsymbol{\beta}\in {\mathbb{R}}^{p\times 1} \), it means that we will hereafter write our equations for the approximation as
+$$
+\boldsymbol{\tilde{y}}= \boldsymbol{X}\boldsymbol{\beta},
+$$
+
+throughout these lectures.
+
+
+
+With the above we use the design matrix to define the approximation \( \boldsymbol{\tilde{y}} \) via the unknown quantity \( \boldsymbol{\beta} \) as
+$$
+\boldsymbol{\tilde{y}}= \boldsymbol{X}\boldsymbol{\beta},
+$$
+
+and in order to find the optimal parameters \( \beta_i \) instead of solving the above linear algebra problem, we define a function which gives a measure of the spread between the values \( y_i \) (which represent hopefully the exact values) and the parameterized values \( \tilde{y}_i \), namely
+$$
+C(\boldsymbol{\beta})=\frac{1}{n}\sum_{i=0}^{n-1}\left(y_i-\tilde{y}_i\right)^2=\frac{1}{n}\left\{\left(\boldsymbol{y}-\boldsymbol{\tilde{y}}\right)^T\left(\boldsymbol{y}-\boldsymbol{\tilde{y}}\right)\right\},
+$$
+
+or using the matrix \( \boldsymbol{X} \) and in a more compact matrix-vector notation as
+$$
+C(\boldsymbol{\beta})=\frac{1}{n}\left\{\left(\boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\right)^T\left(\boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\right)\right\}.
+$$
+
+This function is one possible way to define the so-called cost function.
+
+
+It is also common to define
+the function \( C \) as
+
+$$
+C(\boldsymbol{\beta})=\frac{1}{2n}\sum_{i=0}^{n-1}\left(y_i-\tilde{y}_i\right)^2,
+$$
+
+since when taking the first derivative with respect to the unknown parameters \( \beta \), the factor of \( 2 \) cancels out.
+
+
+
+
+The function
+$$
+C(\boldsymbol{\beta})=\frac{1}{n}\left\{\left(\boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\right)^T\left(\boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\right)\right\},
+$$
+
+can be linked to the variance of the quantity \( y_i \) if we interpret the latter as the mean value.
+When linking (see the discussion below) with the maximum likelihood approach below, we will indeed interpret \( y_i \) as a mean value
+$$
+y_{i}=\langle y_i \rangle = \beta_0x_{i,0}+\beta_1x_{i,1}+\beta_2x_{i,2}+\dots+\beta_{n-1}x_{i,n-1}+\epsilon_i,
+$$
+
+
+where \( \langle y_i \rangle \) is the mean value. Keep in mind also that
+till now we have treated \( y_i \) as the exact value. Normally, the
+response (dependent or outcome) variable \( y_i \) the outcome of a
+numerical experiment or another type of experiment and is thus only an
+approximation to the true value. It is then always accompanied by an
+error estimate, often limited to a statistical error estimate given by
+the standard deviation discussed earlier. In the discussion here we
+will treat \( y_i \) as our exact value for the response variable.
+
+
+In order to find the parameters \( \beta_i \) we will then minimize the spread of \( C(\boldsymbol{\beta}) \), that is we are going to solve the problem
+$$
+{\displaystyle \min_{\boldsymbol{\beta}\in
+{\mathbb{R}}^{p}}}\frac{1}{n}\left\{\left(\boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\right)^T\left(\boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\right)\right\}.
+$$
+
+In practical terms it means we will require
+$$
+\frac{\partial C(\boldsymbol{\beta})}{\partial \beta_j} = \frac{\partial }{\partial \beta_j}\left[ \frac{1}{n}\sum_{i=0}^{n-1}\left(y_i-\beta_0x_{i,0}-\beta_1x_{i,1}-\beta_2x_{i,2}-\dots-\beta_{n-1}x_{i,n-1}\right)^2\right]=0,
+$$
+
+which results in
+$$
+\frac{\partial C(\boldsymbol{\beta})}{\partial \beta_j} = -\frac{2}{n}\left[ \sum_{i=0}^{n-1}x_{ij}\left(y_i-\beta_0x_{i,0}-\beta_1x_{i,1}-\beta_2x_{i,2}-\dots-\beta_{n-1}x_{i,n-1}\right)\right]=0,
+$$
+
+or in a matrix-vector form as
+$$
+\frac{\partial C(\boldsymbol{\beta})}{\partial \boldsymbol{\beta}} = 0 = \boldsymbol{X}^T\left( \boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\right).
+$$
+
+
+
+
+We can rewrite
+$$
+\frac{\partial C(\boldsymbol{\beta})}{\partial \boldsymbol{\beta}} = 0 = \boldsymbol{X}^T\left( \boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\right),
+$$
+
+as
+$$
+\boldsymbol{X}^T\boldsymbol{y} = \boldsymbol{X}^T\boldsymbol{X}\boldsymbol{\beta},
+$$
+
+and if the matrix \( \boldsymbol{X}^T\boldsymbol{X} \) is invertible we have the solution
+$$
+\boldsymbol{\beta} =\left(\boldsymbol{X}^T\boldsymbol{X}\right)^{-1}\boldsymbol{X}^T\boldsymbol{y}.
+$$
+
+
+We note also that since our design matrix is defined as \( \boldsymbol{X}\in
+{\mathbb{R}}^{n\times p} \), the product \( \boldsymbol{X}^T\boldsymbol{X} \in
+{\mathbb{R}}^{p\times p} \). In the above case we have that \( p \ll n \),
+in our case \( p=5 \) meaning that we end up with inverting a small
+\( 5\times 5 \) matrix. This is a rather common situation, in many cases we end up with low-dimensional
+matrices to invert. The methods discussed here and for many other
+supervised learning algorithms like classification with logistic
+regression or support vector machines, exhibit dimensionalities which
+allow for the usage of direct linear algebra methods such as LU decomposition or Singular Value Decomposition (SVD) for finding the inverse of the matrix
+\( \boldsymbol{X}^T\boldsymbol{X} \).
+
+
+Small question: Do you think the example we have at hand here (the nuclear binding energies) can lead to problems in inverting the matrix \( \boldsymbol{X}^T\boldsymbol{X} \)? What kind of problems can we expect?
+
+
+The following matrix and vector relation will be useful here and for the rest of the course. Vectors are always written as boldfaced lower case letters and
+matrices as upper case boldfaced letters.
+
+$$
+\frac{\partial (\boldsymbol{b}^T\boldsymbol{a})}{\partial \boldsymbol{a}} = \boldsymbol{b},
+$$
+
+$$
+\frac{\partial (\boldsymbol{a}^T\boldsymbol{A}\boldsymbol{a})}{\partial \boldsymbol{a}} = (\boldsymbol{A}+\boldsymbol{A}^T)\boldsymbol{a},
+$$
+
+$$
+\frac{\partial tr(\boldsymbol{B}\boldsymbol{A})}{\partial \boldsymbol{A}} = \boldsymbol{B}^T,
+$$
+
+$$
+\frac{\partial \log{\vert\boldsymbol{A}\vert}}{\partial \boldsymbol{A}} = (\boldsymbol{A}^{-1})^T.
+$$
+
+
+The residuals \( \boldsymbol{\epsilon} \) are in turn given by
+$$
+\boldsymbol{\epsilon} = \boldsymbol{y}-\boldsymbol{\tilde{y}} = \boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta},
+$$
+
+and with
+$$
+\boldsymbol{X}^T\left( \boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\right)= 0,
+$$
+
+we have
+$$
+\boldsymbol{X}^T\boldsymbol{\epsilon}=\boldsymbol{X}^T\left( \boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\right)= 0,
+$$
+
+meaning that the solution for \( \boldsymbol{\beta} \) is the one which minimizes the residuals. Later we will link this with the maximum likelihood approach.
+
+
+
+Let us now return to our nuclear binding energies and simply code the above equations.
+
+
+
+It is rather straightforward to implement the matrix inversion and obtain the parameters \( \boldsymbol{\beta} \). After having defined the matrix \( \boldsymbol{X} \) we simply need to
+write
+
+
+
+
+Alternatively, you can use the least squares functionality in Numpy as
+
+
+
+
+And finally we plot our fit with and compare with data
+
+
+
+
+
+We can easily test our fit by computing the \( R2 \) score that we discussed in connection with the functionality of Scikit-Learn in the introductory slides.
+Since we are not using Scikit-Learn here we can define our own \( R2 \) function as
+
+
+
+
+and we would be using it as
+
+
+
+
+We can easily add our MSE score as
+
+
+
+
+and finally the relative error as
+
+
+
+
+
+
+
+Normally, the response (dependent or outcome) variable \( y_i \) is the
+outcome of a numerical experiment or another type of experiment and is
+thus only an approximation to the true value. It is then always
+accompanied by an error estimate, often limited to a statistical error
+estimate given by the standard deviation discussed earlier. In the
+discussion here we will treat \( y_i \) as our exact value for the
+response variable.
+
+
+Introducing the standard deviation \( \sigma_i \) for each measurement
+\( y_i \), we define now the \( \chi^2 \) function (omitting the \( 1/n \) term)
+as
+
+$$
+\chi^2(\boldsymbol{\beta})=\frac{1}{n}\sum_{i=0}^{n-1}\frac{\left(y_i-\tilde{y}_i\right)^2}{\sigma_i^2}=\frac{1}{n}\left\{\left(\boldsymbol{y}-\boldsymbol{\tilde{y}}\right)^T\frac{1}{\boldsymbol{\Sigma^2}}\left(\boldsymbol{y}-\boldsymbol{\tilde{y}}\right)\right\},
+$$
+
+where the matrix \( \boldsymbol{\Sigma} \) is a diagonal matrix with \( \sigma_i \) as matrix elements.
+
+
+
+
+
+
+In order to find the parameters \( \beta_i \) we will then minimize the spread of \( \chi^2(\boldsymbol{\beta}) \) by requiring
+$$
+\frac{\partial \chi^2(\boldsymbol{\beta})}{\partial \beta_j} = \frac{\partial }{\partial \beta_j}\left[ \frac{1}{n}\sum_{i=0}^{n-1}\left(\frac{y_i-\beta_0x_{i,0}-\beta_1x_{i,1}-\beta_2x_{i,2}-\dots-\beta_{n-1}x_{i,n-1}}{\sigma_i}\right)^2\right]=0,
+$$
+
+which results in
+$$
+\frac{\partial \chi^2(\boldsymbol{\beta})}{\partial \beta_j} = -\frac{2}{n}\left[ \sum_{i=0}^{n-1}\frac{x_{ij}}{\sigma_i}\left(\frac{y_i-\beta_0x_{i,0}-\beta_1x_{i,1}-\beta_2x_{i,2}-\dots-\beta_{n-1}x_{i,n-1}}{\sigma_i}\right)\right]=0,
+$$
+
+or in a matrix-vector form as
+$$
+\frac{\partial \chi^2(\boldsymbol{\beta})}{\partial \boldsymbol{\beta}} = 0 = \boldsymbol{A}^T\left( \boldsymbol{b}-\boldsymbol{A}\boldsymbol{\beta}\right).
+$$
+
+where we have defined the matrix \( \boldsymbol{A} =\boldsymbol{X}/\boldsymbol{\Sigma} \) with matrix elements \( a_{ij} = x_{ij}/\sigma_i \) and the vector \( \boldsymbol{b} \) with elements \( b_i = y_i/\sigma_i \).
+
+
+
+
+We can rewrite
+$$
+\frac{\partial \chi^2(\boldsymbol{\beta})}{\partial \boldsymbol{\beta}} = 0 = \boldsymbol{A}^T\left( \boldsymbol{b}-\boldsymbol{A}\boldsymbol{\beta}\right),
+$$
+
+as
+$$
+\boldsymbol{A}^T\boldsymbol{b} = \boldsymbol{A}^T\boldsymbol{A}\boldsymbol{\beta},
+$$
+
+and if the matrix \( \boldsymbol{A}^T\boldsymbol{A} \) is invertible we have the solution
+$$
+\boldsymbol{\beta} =\left(\boldsymbol{A}^T\boldsymbol{A}\right)^{-1}\boldsymbol{A}^T\boldsymbol{b}.
+$$
+
+
+
+
+If we then introduce the matrix
+$$
+\boldsymbol{H} = \left(\boldsymbol{A}^T\boldsymbol{A}\right)^{-1},
+$$
+
+we have then the following expression for the parameters \( \beta_j \) (the matrix elements of \( \boldsymbol{H} \) are \( h_{ij} \))
+$$
+\beta_j = \sum_{k=0}^{p-1}h_{jk}\sum_{i=0}^{n-1}\frac{y_i}{\sigma_i}\frac{x_{ik}}{\sigma_i} = \sum_{k=0}^{p-1}h_{jk}\sum_{i=0}^{n-1}b_ia_{ik}
+$$
+
+We state without proof the expression for the uncertainty in the parameters \( \beta_j \) as (we leave this as an exercise)
+$$
+\sigma^2(\beta_j) = \sum_{i=0}^{n-1}\sigma_i^2\left( \frac{\partial \beta_j}{\partial y_i}\right)^2,
+$$
+
+resulting in
+$$
+\sigma^2(\beta_j) = \left(\sum_{k=0}^{p-1}h_{jk}\sum_{i=0}^{n-1}a_{ik}\right)\left(\sum_{l=0}^{p-1}h_{jl}\sum_{m=0}^{n-1}a_{ml}\right) = h_{jj}!
+$$
+
+
+The first step here is to approximate the function \( y \) with a first-order polynomial, that is we write
+$$
+y=y(x) \rightarrow y(x_i) \approx \beta_0+\beta_1 x_i.
+$$
+
+By computing the derivatives of \( \chi^2 \) with respect to \( \beta_0 \) and \( \beta_1 \) show that these are given by
+$$
+\frac{\partial \chi^2(\boldsymbol{\beta})}{\partial \beta_0} = -2\left[ \frac{1}{n}\sum_{i=0}^{n-1}\left(\frac{y_i-\beta_0-\beta_1x_{i}}{\sigma_i^2}\right)\right]=0,
+$$
+
+and
+$$
+\frac{\partial \chi^2(\boldsymbol{\beta})}{\partial \beta_1} = -\frac{2}{n}\left[ \sum_{i=0}^{n-1}x_i\left(\frac{y_i-\beta_0-\beta_1x_{i}}{\sigma_i^2}\right)\right]=0.
+$$
+
+
+
+
+For a linear fit (a first-order polynomial) we don't need to invert a matrix!!
+Defining
+$$
+\gamma = \sum_{i=0}^{n-1}\frac{1}{\sigma_i^2},
+$$
+
+
+$$
+\gamma_x = \sum_{i=0}^{n-1}\frac{x_{i}}{\sigma_i^2},
+$$
+
+
+$$
+\gamma_y = \sum_{i=0}^{n-1}\left(\frac{y_i}{\sigma_i^2}\right),
+$$
+
+
+$$
+\gamma_{xx} = \sum_{i=0}^{n-1}\frac{x_ix_{i}}{\sigma_i^2},
+$$
+
+
+$$
+\gamma_{xy} = \sum_{i=0}^{n-1}\frac{y_ix_{i}}{\sigma_i^2},
+$$
+
+
+we obtain
+
+$$
+\beta_0 = \frac{\gamma_{xx}\gamma_y-\gamma_x\gamma_y}{\gamma\gamma_{xx}-\gamma_x^2},
+$$
+
+
+$$
+\beta_1 = \frac{\gamma_{xy}\gamma-\gamma_x\gamma_y}{\gamma\gamma_{xx}-\gamma_x^2}.
+$$
+
+
+This approach (different linear and non-linear regression) suffers
+often from both being underdetermined and overdetermined in the
+unknown coefficients \( \beta_i \). A better approach is to use the
+Singular Value Decomposition (SVD) method discussed next week.
+
+
+
+
+Before we continue, let us introduce yet another example. We are going to fit the
+nuclear equation of state using results from many-body calculations.
+The equation of state we have made available here, as function of
+density, has been derived using modern nucleon-nucleon potentials with
+the addition of three-body
+forces. This
+time the file is presented as a standard csv file.
+
+
+The beginning of the Python code here is similar to what you have seen
+before, with the same initializations and declarations. We use also
+pandas again, rather extensively in order to organize our data.
+
+
+The difference now is that we use Scikit-Learn's regression tools
+instead of our own matrix inversion implementation. Furthermore, we
+sneak in Ridge regression (to be discussed below) which includes a
+hyperparameter \( \lambda \), also to be explained below.
+
+
+
+
+
+
+The above simple polynomial in density \( \rho \) gives an excellent fit
+to the data.
+
+
+We note also that there is a small deviation between the
+standard OLS and the Ridge regression at higher densities. We discuss this in more detail
+below.
+
+
+
+It is normal in essentially all Machine Learning studies to split the
+data in a training set and a test set (sometimes also an additional
+validation set). Scikit-Learn has an own function for this. There
+is no explicit recipe for how much data should be included as training
+data and say test data. An accepted rule of thumb is to use
+approximately \( 2/3 \) to \( 4/5 \) of the data as training data. We will
+postpone a discussion of this splitting to the end of these notes and
+our discussion of the so-called bias-variance tradeoff. Here we
+limit ourselves to repeat the above equation of state fitting example
+but now splitting the data into a training set and a test set.
+
+
+
+
+
+Friday August 27
-Simple linear regression model using scikit-learn
@@ -2201,6 +2160,1136 @@ Now it is time to dive more into the details of various methods. We will start w
Why Linear Regression (aka Ordinary Least Squares and family)
+
+
+
+Regression analysis, overarching aims
+
+
+Regression analysis, overarching aims II
+Examples
+
+$$
+BE(A) = a_0+a_1A+a_2A^{2/3}+a_3A^{-1/3}+a_4A^{-1},
+$$
+
+
+we have five predictors, that is the intercept, the \( A \) dependent term, the \( A^{2/3} \) term and the \( A^{-1/3} \) and \( A^{-1} \) terms.
+This gives \( p=0,1,2,3,4 \). Furthermore we have \( n \) entries for each predictor. It means that our design matrix is a
+\( p\times n \) matrix \( \boldsymbol{X} \).
+
+General linear models
+
+$$
+y=y(x) \rightarrow y(x_i)=\tilde{y}_i+\epsilon_i=\sum_{j=0}^{n-1} \beta_j x_i^j+\epsilon_i,
+$$
+
+
+where \( \epsilon_i \) is the error in our approximation.
+
+
+Rewriting the fitting procedure as a linear algebra problem
+
+$$
+\begin{align*}
+y_0&=\beta_0+\beta_1x_0^1+\beta_2x_0^2+\dots+\beta_{n-1}x_0^{n-1}+\epsilon_0\\
+y_1&=\beta_0+\beta_1x_1^1+\beta_2x_1^2+\dots+\beta_{n-1}x_1^{n-1}+\epsilon_1\\
+y_2&=\beta_0+\beta_1x_2^1+\beta_2x_2^2+\dots+\beta_{n-1}x_2^{n-1}+\epsilon_2\\
+\dots & \dots \\
+y_{n-1}&=\beta_0+\beta_1x_{n-1}^1+\beta_2x_{n-1}^2+\dots+\beta_{n-1}x_{n-1}^{n-1}+\epsilon_{n-1}.\\
+\end{align*}
+$$
+
+Rewriting the fitting procedure as a linear algebra problem, more details
+
+$$
+\boldsymbol{y} = [y_0,y_1, y_2,\dots, y_{n-1}]^T,
+$$
+
+
+and
+
+$$
+\boldsymbol{\beta} = [\beta_0,\beta_1, \beta_2,\dots, \beta_{n-1}]^T,
+$$
+
+
+and
+
+$$
+\boldsymbol{\epsilon} = [\epsilon_0,\epsilon_1, \epsilon_2,\dots, \epsilon_{n-1}]^T,
+$$
+
+
+and the design matrix
+
+$$
+\boldsymbol{X}=
+\begin{bmatrix}
+1& x_{0}^1 &x_{0}^2& \dots & \dots &x_{0}^{n-1}\\
+1& x_{1}^1 &x_{1}^2& \dots & \dots &x_{1}^{n-1}\\
+1& x_{2}^1 &x_{2}^2& \dots & \dots &x_{2}^{n-1}\\
+\dots& \dots &\dots& \dots & \dots &\dots\\
+1& x_{n-1}^1 &x_{n-1}^2& \dots & \dots &x_{n-1}^{n-1}\\
+\end{bmatrix}
+$$
+
+
+we can rewrite our equations as
+
+$$
+\boldsymbol{y} = \boldsymbol{X}\boldsymbol{\beta}+\boldsymbol{\epsilon}.
+$$
+
+
+The above design matrix is called a Vandermonde matrix.
+Generalizing the fitting procedure as a linear algebra problem
+
+$$
+\begin{align*}
+y_0&=\beta_0x_{00}+\beta_1x_{01}+\beta_2x_{02}+\dots+\beta_{n-1}x_{0n-1}+\epsilon_0\\
+y_1&=\beta_0x_{10}+\beta_1x_{11}+\beta_2x_{12}+\dots+\beta_{n-1}x_{1n-1}+\epsilon_1\\
+y_2&=\beta_0x_{20}+\beta_1x_{21}+\beta_2x_{22}+\dots+\beta_{n-1}x_{2n-1}+\epsilon_2\\
+\dots & \dots \\
+y_{i}&=\beta_0x_{i0}+\beta_1x_{i1}+\beta_2x_{i2}+\dots+\beta_{n-1}x_{in-1}+\epsilon_i\\
+\dots & \dots \\
+y_{n-1}&=\beta_0x_{n-1,0}+\beta_1x_{n-1,2}+\beta_2x_{n-1,2}+\dots+\beta_{n-1}x_{n-1,n-1}+\epsilon_{n-1}.\\
+\end{align*}
+$$
+
+
+Generalizing the fitting procedure as a linear algebra problem
+
+$$
+\boldsymbol{X}=
+\begin{bmatrix}
+x_{00}& x_{01} &x_{02}& \dots & \dots &x_{0,n-1}\\
+x_{10}& x_{11} &x_{12}& \dots & \dots &x_{1,n-1}\\
+x_{20}& x_{21} &x_{22}& \dots & \dots &x_{2,n-1}\\
+\dots& \dots &\dots& \dots & \dots &\dots\\
+x_{n-1,0}& x_{n-1,1} &x_{n-1,2}& \dots & \dots &x_{n-1,n-1}\\
+\end{bmatrix}
+$$
+
+
+and without loss of generality we rewrite again our equations as
+
+$$
+\boldsymbol{y} = \boldsymbol{X}\boldsymbol{\beta}+\boldsymbol{\epsilon}.
+$$
+
+
+The left-hand side of this equation is kwown. Our error vector \( \boldsymbol{\epsilon} \) and the parameter vector \( \boldsymbol{\beta} \) are our unknow quantities. How can we obtain the optimal set of \( \beta_i \) values?
+Optimizing our parameters
+
+$$
+\begin{align*}
+y_0&=\beta_0x_{00}+\beta_1x_{01}+\beta_2x_{02}+\dots+\beta_{n-1}x_{0n-1}+\epsilon_0\\
+y_1&=\beta_0x_{10}+\beta_1x_{11}+\beta_2x_{12}+\dots+\beta_{n-1}x_{1n-1}+\epsilon_1\\
+y_2&=\beta_0x_{20}+\beta_1x_{21}+\beta_2x_{22}+\dots+\beta_{n-1}x_{2n-1}+\epsilon_1\\
+\dots & \dots \\
+y_{i}&=\beta_0x_{i0}+\beta_1x_{i1}+\beta_2x_{i2}+\dots+\beta_{n-1}x_{in-1}+\epsilon_1\\
+\dots & \dots \\
+y_{n-1}&=\beta_0x_{n-1,0}+\beta_1x_{n-1,2}+\beta_2x_{n-1,2}+\dots+\beta_{n-1}x_{n-1,n-1}+\epsilon_{n-1}.\\
+\end{align*}
+$$
+
+
+Our model for the nuclear binding energies
+
+# Common imports
+import numpy as np
+import pandas as pd
+import matplotlib.pyplot as plt
+from IPython.display import display
+import os
+
+# Where to save the figures and data files
+PROJECT_ROOT_DIR = "Results"
+FIGURE_ID = "Results/FigureFiles"
+DATA_ID = "DataFiles/"
+
+if not os.path.exists(PROJECT_ROOT_DIR):
+ os.mkdir(PROJECT_ROOT_DIR)
+
+if not os.path.exists(FIGURE_ID):
+ os.makedirs(FIGURE_ID)
+
+if not os.path.exists(DATA_ID):
+ os.makedirs(DATA_ID)
+
+def image_path(fig_id):
+ return os.path.join(FIGURE_ID, fig_id)
+
+def data_path(dat_id):
+ return os.path.join(DATA_ID, dat_id)
+
+def save_fig(fig_id):
+ plt.savefig(image_path(fig_id) + ".png", format='png')
+
+infile = open(data_path("MassEval2016.dat"),'r')
+
+
+# Read the experimental data with Pandas
+Masses = pd.read_fwf(infile, usecols=(2,3,4,6,11),
+ names=('N', 'Z', 'A', 'Element', 'Ebinding'),
+ widths=(1,3,5,5,5,1,3,4,1,13,11,11,9,1,2,11,9,1,3,1,12,11,1),
+ header=39,
+ index_col=False)
+
+# Extrapolated values are indicated by '#' in place of the decimal place, so
+# the Ebinding column won't be numeric. Coerce to float and drop these entries.
+Masses['Ebinding'] = pd.to_numeric(Masses['Ebinding'], errors='coerce')
+Masses = Masses.dropna()
+# Convert from keV to MeV.
+Masses['Ebinding'] /= 1000
+
+# Group the DataFrame by nucleon number, A.
+Masses = Masses.groupby('A')
+# Find the rows of the grouped DataFrame with the maximum binding energy.
+Masses = Masses.apply(lambda t: t[t.Ebinding==t.Ebinding.max()])
+A = Masses['A']
+Z = Masses['Z']
+N = Masses['N']
+Element = Masses['Element']
+Energies = Masses['Ebinding']
+
+# Now we set up the design matrix X
+X = np.zeros((len(A),5))
+X[:,0] = 1
+X[:,1] = A
+X[:,2] = A**(2.0/3.0)
+X[:,3] = A**(-1.0/3.0)
+X[:,4] = A**(-1.0)
+# Then nice printout using pandas
+DesignMatrix = pd.DataFrame(X)
+DesignMatrix.index = A
+DesignMatrix.columns = ['1', 'A', 'A^(2/3)', 'A^(-1/3)', '1/A']
+display(DesignMatrix)
+
+$$
+\boldsymbol{\tilde{y}}= \boldsymbol{X}\boldsymbol{\beta},
+$$
+
+
+throughout these lectures.
+Optimizing our parameters, more details
+
+$$
+\boldsymbol{\tilde{y}}= \boldsymbol{X}\boldsymbol{\beta},
+$$
+
+
+and in order to find the optimal parameters \( \beta_i \) instead of solving the above linear algebra problem, we define a function which gives a measure of the spread between the values \( y_i \) (which represent hopefully the exact values) and the parameterized values \( \tilde{y}_i \), namely
+
+$$
+C(\boldsymbol{\beta})=\frac{1}{n}\sum_{i=0}^{n-1}\left(y_i-\tilde{y}_i\right)^2=\frac{1}{n}\left\{\left(\boldsymbol{y}-\boldsymbol{\tilde{y}}\right)^T\left(\boldsymbol{y}-\boldsymbol{\tilde{y}}\right)\right\},
+$$
+
+
+or using the matrix \( \boldsymbol{X} \) and in a more compact matrix-vector notation as
+
+$$
+C(\boldsymbol{\beta})=\frac{1}{n}\left\{\left(\boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\right)^T\left(\boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\right)\right\}.
+$$
+
+
+This function is one possible way to define the so-called cost function.
+
+
+$$
+C(\boldsymbol{\beta})=\frac{1}{2n}\sum_{i=0}^{n-1}\left(y_i-\tilde{y}_i\right)^2,
+$$
+
+
+since when taking the first derivative with respect to the unknown parameters \( \beta \), the factor of \( 2 \) cancels out.
+Interpretations and optimizing our parameters
+
+$$
+C(\boldsymbol{\beta})=\frac{1}{n}\left\{\left(\boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\right)^T\left(\boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\right)\right\},
+$$
+
+
+can be linked to the variance of the quantity \( y_i \) if we interpret the latter as the mean value.
+When linking (see the discussion below) with the maximum likelihood approach below, we will indeed interpret \( y_i \) as a mean value
+
+$$
+y_{i}=\langle y_i \rangle = \beta_0x_{i,0}+\beta_1x_{i,1}+\beta_2x_{i,2}+\dots+\beta_{n-1}x_{i,n-1}+\epsilon_i,
+$$
+
+
+
+$$
+{\displaystyle \min_{\boldsymbol{\beta}\in
+{\mathbb{R}}^{p}}}\frac{1}{n}\left\{\left(\boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\right)^T\left(\boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\right)\right\}.
+$$
+
+
+In practical terms it means we will require
+
+$$
+\frac{\partial C(\boldsymbol{\beta})}{\partial \beta_j} = \frac{\partial }{\partial \beta_j}\left[ \frac{1}{n}\sum_{i=0}^{n-1}\left(y_i-\beta_0x_{i,0}-\beta_1x_{i,1}-\beta_2x_{i,2}-\dots-\beta_{n-1}x_{i,n-1}\right)^2\right]=0,
+$$
+
+
+which results in
+
+$$
+\frac{\partial C(\boldsymbol{\beta})}{\partial \beta_j} = -\frac{2}{n}\left[ \sum_{i=0}^{n-1}x_{ij}\left(y_i-\beta_0x_{i,0}-\beta_1x_{i,1}-\beta_2x_{i,2}-\dots-\beta_{n-1}x_{i,n-1}\right)\right]=0,
+$$
+
+
+or in a matrix-vector form as
+
+$$
+\frac{\partial C(\boldsymbol{\beta})}{\partial \boldsymbol{\beta}} = 0 = \boldsymbol{X}^T\left( \boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\right).
+$$
+
+
+
+Interpretations and optimizing our parameters
+
+$$
+\frac{\partial C(\boldsymbol{\beta})}{\partial \boldsymbol{\beta}} = 0 = \boldsymbol{X}^T\left( \boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\right),
+$$
+
+
+as
+
+$$
+\boldsymbol{X}^T\boldsymbol{y} = \boldsymbol{X}^T\boldsymbol{X}\boldsymbol{\beta},
+$$
+
+
+and if the matrix \( \boldsymbol{X}^T\boldsymbol{X} \) is invertible we have the solution
+
+$$
+\boldsymbol{\beta} =\left(\boldsymbol{X}^T\boldsymbol{X}\right)^{-1}\boldsymbol{X}^T\boldsymbol{y}.
+$$
+
+
+Some useful matrix and vector expressions
+
+
+$$
+\frac{\partial (\boldsymbol{b}^T\boldsymbol{a})}{\partial \boldsymbol{a}} = \boldsymbol{b},
+$$
+
+
+
+$$
+\frac{\partial (\boldsymbol{a}^T\boldsymbol{A}\boldsymbol{a})}{\partial \boldsymbol{a}} = (\boldsymbol{A}+\boldsymbol{A}^T)\boldsymbol{a},
+$$
+
+
+
+$$
+\frac{\partial tr(\boldsymbol{B}\boldsymbol{A})}{\partial \boldsymbol{A}} = \boldsymbol{B}^T,
+$$
+
+
+
+$$
+\frac{\partial \log{\vert\boldsymbol{A}\vert}}{\partial \boldsymbol{A}} = (\boldsymbol{A}^{-1})^T.
+$$
+
+Interpretations and optimizing our parameters
+
+$$
+\boldsymbol{\epsilon} = \boldsymbol{y}-\boldsymbol{\tilde{y}} = \boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta},
+$$
+
+
+and with
+
+$$
+\boldsymbol{X}^T\left( \boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\right)= 0,
+$$
+
+
+we have
+
+$$
+\boldsymbol{X}^T\boldsymbol{\epsilon}=\boldsymbol{X}^T\left( \boldsymbol{y}-\boldsymbol{X}\boldsymbol{\beta}\right)= 0,
+$$
+
+
+meaning that the solution for \( \boldsymbol{\beta} \) is the one which minimizes the residuals. Later we will link this with the maximum likelihood approach.
+
+
+Own code for Ordinary Least Squares
+
+# matrix inversion to find beta
+beta = np.linalg.inv(X.T.dot(X)).dot(X.T).dot(Energies)
+# and then make the prediction
+ytilde = X @ beta
+
fit = np.linalg.lstsq(X, Energies, rcond =None)[0]
+ytildenp = np.dot(fit,X.T)
+
Masses['Eapprox'] = ytilde
+# Generate a plot comparing the experimental with the fitted values values.
+fig, ax = plt.subplots()
+ax.set_xlabel(r'$A = N + Z$')
+ax.set_ylabel(r'$E_\mathrm{bind}\,/\mathrm{MeV}$')
+ax.plot(Masses['A'], Masses['Ebinding'], alpha=0.7, lw=2,
+ label='Ame2016')
+ax.plot(Masses['A'], Masses['Eapprox'], alpha=0.7, lw=2, c='m',
+ label='Fit')
+ax.legend()
+save_fig("Masses2016OLS")
+plt.show()
+
Adding error analysis and training set up
+
+def R2(y_data, y_model):
+ return 1 - np.sum((y_data - y_model) ** 2) / np.sum((y_data - np.mean(y_data)) ** 2)
+
print(R2(Energies,ytilde))
+def MSE(y_data,y_model):
+ n = np.size(y_model)
+ return np.sum((y_data-y_model)**2)/n
+
+print(MSE(Energies,ytilde))
+
def RelativeError(y_data,y_model):
+ return abs((y_data-y_model)/y_data)
+print(RelativeError(Energies, ytilde))
+
The \( \chi^2 \) function
+
+$$
+\chi^2(\boldsymbol{\beta})=\frac{1}{n}\sum_{i=0}^{n-1}\frac{\left(y_i-\tilde{y}_i\right)^2}{\sigma_i^2}=\frac{1}{n}\left\{\left(\boldsymbol{y}-\boldsymbol{\tilde{y}}\right)^T\frac{1}{\boldsymbol{\Sigma^2}}\left(\boldsymbol{y}-\boldsymbol{\tilde{y}}\right)\right\},
+$$
+
+
+where the matrix \( \boldsymbol{\Sigma} \) is a diagonal matrix with \( \sigma_i \) as matrix elements.
+
+
+The \( \chi^2 \) function
+
+$$
+\frac{\partial \chi^2(\boldsymbol{\beta})}{\partial \beta_j} = \frac{\partial }{\partial \beta_j}\left[ \frac{1}{n}\sum_{i=0}^{n-1}\left(\frac{y_i-\beta_0x_{i,0}-\beta_1x_{i,1}-\beta_2x_{i,2}-\dots-\beta_{n-1}x_{i,n-1}}{\sigma_i}\right)^2\right]=0,
+$$
+
+
+which results in
+
+$$
+\frac{\partial \chi^2(\boldsymbol{\beta})}{\partial \beta_j} = -\frac{2}{n}\left[ \sum_{i=0}^{n-1}\frac{x_{ij}}{\sigma_i}\left(\frac{y_i-\beta_0x_{i,0}-\beta_1x_{i,1}-\beta_2x_{i,2}-\dots-\beta_{n-1}x_{i,n-1}}{\sigma_i}\right)\right]=0,
+$$
+
+
+or in a matrix-vector form as
+
+$$
+\frac{\partial \chi^2(\boldsymbol{\beta})}{\partial \boldsymbol{\beta}} = 0 = \boldsymbol{A}^T\left( \boldsymbol{b}-\boldsymbol{A}\boldsymbol{\beta}\right).
+$$
+
+
+where we have defined the matrix \( \boldsymbol{A} =\boldsymbol{X}/\boldsymbol{\Sigma} \) with matrix elements \( a_{ij} = x_{ij}/\sigma_i \) and the vector \( \boldsymbol{b} \) with elements \( b_i = y_i/\sigma_i \).
+The \( \chi^2 \) function
+
+$$
+\frac{\partial \chi^2(\boldsymbol{\beta})}{\partial \boldsymbol{\beta}} = 0 = \boldsymbol{A}^T\left( \boldsymbol{b}-\boldsymbol{A}\boldsymbol{\beta}\right),
+$$
+
+
+as
+
+$$
+\boldsymbol{A}^T\boldsymbol{b} = \boldsymbol{A}^T\boldsymbol{A}\boldsymbol{\beta},
+$$
+
+
+and if the matrix \( \boldsymbol{A}^T\boldsymbol{A} \) is invertible we have the solution
+
+$$
+\boldsymbol{\beta} =\left(\boldsymbol{A}^T\boldsymbol{A}\right)^{-1}\boldsymbol{A}^T\boldsymbol{b}.
+$$
+
+The \( \chi^2 \) function
+
+$$
+\boldsymbol{H} = \left(\boldsymbol{A}^T\boldsymbol{A}\right)^{-1},
+$$
+
+
+we have then the following expression for the parameters \( \beta_j \) (the matrix elements of \( \boldsymbol{H} \) are \( h_{ij} \))
+
+$$
+\beta_j = \sum_{k=0}^{p-1}h_{jk}\sum_{i=0}^{n-1}\frac{y_i}{\sigma_i}\frac{x_{ik}}{\sigma_i} = \sum_{k=0}^{p-1}h_{jk}\sum_{i=0}^{n-1}b_ia_{ik}
+$$
+
+
+We state without proof the expression for the uncertainty in the parameters \( \beta_j \) as (we leave this as an exercise)
+
+$$
+\sigma^2(\beta_j) = \sum_{i=0}^{n-1}\sigma_i^2\left( \frac{\partial \beta_j}{\partial y_i}\right)^2,
+$$
+
+
+resulting in
+
+$$
+\sigma^2(\beta_j) = \left(\sum_{k=0}^{p-1}h_{jk}\sum_{i=0}^{n-1}a_{ik}\right)\left(\sum_{l=0}^{p-1}h_{jl}\sum_{m=0}^{n-1}a_{ml}\right) = h_{jj}!
+$$
+
+The \( \chi^2 \) function
+
+$$
+y=y(x) \rightarrow y(x_i) \approx \beta_0+\beta_1 x_i.
+$$
+
+
+By computing the derivatives of \( \chi^2 \) with respect to \( \beta_0 \) and \( \beta_1 \) show that these are given by
+
+$$
+\frac{\partial \chi^2(\boldsymbol{\beta})}{\partial \beta_0} = -2\left[ \frac{1}{n}\sum_{i=0}^{n-1}\left(\frac{y_i-\beta_0-\beta_1x_{i}}{\sigma_i^2}\right)\right]=0,
+$$
+
+
+and
+
+$$
+\frac{\partial \chi^2(\boldsymbol{\beta})}{\partial \beta_1} = -\frac{2}{n}\left[ \sum_{i=0}^{n-1}x_i\left(\frac{y_i-\beta_0-\beta_1x_{i}}{\sigma_i^2}\right)\right]=0.
+$$
+
+The \( \chi^2 \) function
+
+$$
+\gamma = \sum_{i=0}^{n-1}\frac{1}{\sigma_i^2},
+$$
+
+
+
+$$
+\gamma_x = \sum_{i=0}^{n-1}\frac{x_{i}}{\sigma_i^2},
+$$
+
+
+
+$$
+\gamma_y = \sum_{i=0}^{n-1}\left(\frac{y_i}{\sigma_i^2}\right),
+$$
+
+
+
+$$
+\gamma_{xx} = \sum_{i=0}^{n-1}\frac{x_ix_{i}}{\sigma_i^2},
+$$
+
+
+
+$$
+\gamma_{xy} = \sum_{i=0}^{n-1}\frac{y_ix_{i}}{\sigma_i^2},
+$$
+
+
+
+$$
+\beta_0 = \frac{\gamma_{xx}\gamma_y-\gamma_x\gamma_y}{\gamma\gamma_{xx}-\gamma_x^2},
+$$
+
+
+
+$$
+\beta_1 = \frac{\gamma_{xy}\gamma-\gamma_x\gamma_y}{\gamma\gamma_{xx}-\gamma_x^2}.
+$$
+
+
+Fitting an Equation of State for Dense Nuclear Matter
+
+The code
+
+# Common imports
+import os
+import numpy as np
+import pandas as pd
+import matplotlib.pyplot as plt
+import matplotlib.pyplot as plt
+import sklearn.linear_model as skl
+from sklearn.metrics import mean_squared_error, r2_score, mean_absolute_error
+
+# Where to save the figures and data files
+PROJECT_ROOT_DIR = "Results"
+FIGURE_ID = "Results/FigureFiles"
+DATA_ID = "DataFiles/"
+
+if not os.path.exists(PROJECT_ROOT_DIR):
+ os.mkdir(PROJECT_ROOT_DIR)
+
+if not os.path.exists(FIGURE_ID):
+ os.makedirs(FIGURE_ID)
+
+if not os.path.exists(DATA_ID):
+ os.makedirs(DATA_ID)
+
+def image_path(fig_id):
+ return os.path.join(FIGURE_ID, fig_id)
+
+def data_path(dat_id):
+ return os.path.join(DATA_ID, dat_id)
+
+def save_fig(fig_id):
+ plt.savefig(image_path(fig_id) + ".png", format='png')
+
+infile = open(data_path("EoS.csv"),'r')
+
+# Read the EoS data as csv file and organize the data into two arrays with density and energies
+EoS = pd.read_csv(infile, names=('Density', 'Energy'))
+EoS['Energy'] = pd.to_numeric(EoS['Energy'], errors='coerce')
+EoS = EoS.dropna()
+Energies = EoS['Energy']
+Density = EoS['Density']
+# The design matrix now as function of various polytrops
+X = np.zeros((len(Density),4))
+X[:,3] = Density**(4.0/3.0)
+X[:,2] = Density
+X[:,1] = Density**(2.0/3.0)
+X[:,0] = 1
+
+# We use now Scikit-Learn's linear regressor and ridge regressor
+# OLS part
+clf = skl.LinearRegression().fit(X, Energies)
+ytilde = clf.predict(X)
+EoS['Eols'] = ytilde
+# The mean squared error
+print("Mean squared error: %.2f" % mean_squared_error(Energies, ytilde))
+# Explained variance score: 1 is perfect prediction
+print('Variance score: %.2f' % r2_score(Energies, ytilde))
+# Mean absolute error
+print('Mean absolute error: %.2f' % mean_absolute_error(Energies, ytilde))
+print(clf.coef_, clf.intercept_)
+
+# The Ridge regression with a hyperparameter lambda = 0.1
+_lambda = 0.1
+clf_ridge = skl.Ridge(alpha=_lambda).fit(X, Energies)
+yridge = clf_ridge.predict(X)
+EoS['Eridge'] = yridge
+# The mean squared error
+print("Mean squared error: %.2f" % mean_squared_error(Energies, yridge))
+# Explained variance score: 1 is perfect prediction
+print('Variance score: %.2f' % r2_score(Energies, yridge))
+# Mean absolute error
+print('Mean absolute error: %.2f' % mean_absolute_error(Energies, yridge))
+print(clf_ridge.coef_, clf_ridge.intercept_)
+
+fig, ax = plt.subplots()
+ax.set_xlabel(r'$\rho[\mathrm{fm}^{-3}]$')
+ax.set_ylabel(r'Energy per particle')
+ax.plot(EoS['Density'], EoS['Energy'], alpha=0.7, lw=2,
+ label='Theoretical data')
+ax.plot(EoS['Density'], EoS['Eols'], alpha=0.7, lw=2, c='m',
+ label='OLS')
+ax.plot(EoS['Density'], EoS['Eridge'], alpha=0.7, lw=2, c='g',
+ label='Ridge $\lambda = 0.1$')
+ax.legend()
+save_fig("EoSfitting")
+plt.show()
+
Splitting our Data in Training and Test data
+
+import os
+import numpy as np
+import pandas as pd
+import matplotlib.pyplot as plt
+from sklearn.model_selection import train_test_split
+# Where to save the figures and data files
+PROJECT_ROOT_DIR = "Results"
+FIGURE_ID = "Results/FigureFiles"
+DATA_ID = "DataFiles/"
+
+if not os.path.exists(PROJECT_ROOT_DIR):
+ os.mkdir(PROJECT_ROOT_DIR)
+
+if not os.path.exists(FIGURE_ID):
+ os.makedirs(FIGURE_ID)
+
+if not os.path.exists(DATA_ID):
+ os.makedirs(DATA_ID)
+
+def image_path(fig_id):
+ return os.path.join(FIGURE_ID, fig_id)
+
+def data_path(dat_id):
+ return os.path.join(DATA_ID, dat_id)
+
+def save_fig(fig_id):
+ plt.savefig(image_path(fig_id) + ".png", format='png')
+
+def R2(y_data, y_model):
+ return 1 - np.sum((y_data - y_model) ** 2) / np.sum((y_data - np.mean(y_data)) ** 2)
+def MSE(y_data,y_model):
+ n = np.size(y_model)
+ return np.sum((y_data-y_model)**2)/n
+
+infile = open(data_path("EoS.csv"),'r')
+
+# Read the EoS data as csv file and organized into two arrays with density and energies
+EoS = pd.read_csv(infile, names=('Density', 'Energy'))
+EoS['Energy'] = pd.to_numeric(EoS['Energy'], errors='coerce')
+EoS = EoS.dropna()
+Energies = EoS['Energy']
+Density = EoS['Density']
+# The design matrix now as function of various polytrops
+X = np.zeros((len(Density),5))
+X[:,0] = 1
+X[:,1] = Density**(2.0/3.0)
+X[:,2] = Density
+X[:,3] = Density**(4.0/3.0)
+X[:,4] = Density**(5.0/3.0)
+# We split the data in test and training data
+X_train, X_test, y_train, y_test = train_test_split(X, Energies, test_size=0.2)
+# matrix inversion to find beta
+beta = np.linalg.inv(X_train.T.dot(X_train)).dot(X_train.T).dot(y_train)
+# and then make the prediction
+ytilde = X_train @ beta
+print("Training R2")
+print(R2(y_train,ytilde))
+print("Training MSE")
+print(MSE(y_train,ytilde))
+ypredict = X_test @ beta
+print("Test R2")
+print(R2(y_test,ypredict))
+print("Test MSE")
+print(MSE(y_test,ypredict))
+
Exercises for week 36
Here are three possible exercises for week 36 and the lab sessions of Wednesday September 1..
diff --git a/doc/pub/week34/html/week34-solarized.html b/doc/pub/week34/html/week34-solarized.html
index f2bd8c572..08191ef55 100644
--- a/doc/pub/week34/html/week34-solarized.html
+++ b/doc/pub/week34/html/week34-solarized.html
@@ -145,8 +145,6 @@ div { text-align: justify; text-justify: inter-word; }
('Matrices in Python', 2, None, 'matrices-in-python'),
('Meet the Pandas', 2, None, 'meet-the-pandas'),
('Friday August 27', 2, None, 'friday-august-27'),
- ('Reading Data and fitting', 2, None, 'reading-data-and-fitting'),
- ('Friday August 27', 2, None, 'friday-august-27'),
('Simple linear regression model using _scikit-learn_',
3,
None,
@@ -166,6 +164,88 @@ div { text-align: justify; text-justify: inter-word; }
None,
'and-what-about-using-neural-networks'),
('A first summary', 2, None, 'a-first-summary'),
+ ('Why Linear Regression (aka Ordinary Least Squares and family)',
+ 2,
+ None,
+ 'why-linear-regression-aka-ordinary-least-squares-and-family'),
+ ('Regression analysis, overarching aims',
+ 2,
+ None,
+ 'regression-analysis-overarching-aims'),
+ ('Regression analysis, overarching aims II',
+ 2,
+ None,
+ 'regression-analysis-overarching-aims-ii'),
+ ('Examples', 2, None, 'examples'),
+ ('General linear models', 2, None, 'general-linear-models'),
+ ('Rewriting the fitting procedure as a linear algebra problem',
+ 2,
+ None,
+ 'rewriting-the-fitting-procedure-as-a-linear-algebra-problem'),
+ ('Rewriting the fitting procedure as a linear algebra problem, '
+ 'more details',
+ 2,
+ None,
+ 'rewriting-the-fitting-procedure-as-a-linear-algebra-problem-more-details'),
+ ('Generalizing the fitting procedure as a linear algebra problem',
+ 2,
+ None,
+ 'generalizing-the-fitting-procedure-as-a-linear-algebra-problem'),
+ ('Generalizing the fitting procedure as a linear algebra problem',
+ 2,
+ None,
+ 'generalizing-the-fitting-procedure-as-a-linear-algebra-problem'),
+ ('Optimizing our parameters',
+ 2,
+ None,
+ 'optimizing-our-parameters'),
+ ('Our model for the nuclear binding energies',
+ 2,
+ None,
+ 'our-model-for-the-nuclear-binding-energies'),
+ ('Optimizing our parameters, more details',
+ 2,
+ None,
+ 'optimizing-our-parameters-more-details'),
+ ('Interpretations and optimizing our parameters',
+ 2,
+ None,
+ 'interpretations-and-optimizing-our-parameters'),
+ ('Interpretations and optimizing our parameters',
+ 2,
+ None,
+ 'interpretations-and-optimizing-our-parameters'),
+ ('Some useful matrix and vector expressions',
+ 2,
+ None,
+ 'some-useful-matrix-and-vector-expressions'),
+ ('Interpretations and optimizing our parameters',
+ 2,
+ None,
+ 'interpretations-and-optimizing-our-parameters'),
+ ('Own code for Ordinary Least Squares',
+ 2,
+ None,
+ 'own-code-for-ordinary-least-squares'),
+ ('Adding error analysis and training set up',
+ 2,
+ None,
+ 'adding-error-analysis-and-training-set-up'),
+ ('The $\\chi^2$ function', 2, None, 'the-chi-2-function'),
+ ('The $\\chi^2$ function', 2, None, 'the-chi-2-function'),
+ ('The $\\chi^2$ function', 2, None, 'the-chi-2-function'),
+ ('The $\\chi^2$ function', 2, None, 'the-chi-2-function'),
+ ('The $\\chi^2$ function', 2, None, 'the-chi-2-function'),
+ ('The $\\chi^2$ function', 2, None, 'the-chi-2-function'),
+ ('Fitting an Equation of State for Dense Nuclear Matter',
+ 2,
+ None,
+ 'fitting-an-equation-of-state-for-dense-nuclear-matter'),
+ ('The code', 2, None, 'the-code'),
+ ('Splitting our Data in Training and Test data',
+ 2,
+ None,
+ 'splitting-our-data-in-training-and-test-data'),
('Exercises for week 36', 2, None, 'exercises-for-week-36'),
('Exercise 1: Setting up various Python environments',
2,
@@ -270,10 +350,10 @@ Reading recommendations this week: Refresh linear algebra, GBC chapters 1 and 2.
Thursday August 26
Lectures and ComputerLab
@@ -1468,50 +1551,6 @@ For multidimensional arrays, we recommend strongly Reading Data and fitting
-
-
-
-Friday August 27
-
-
-
Simple linear regression model using scikit-learn
+Why Linear Regression (aka Ordinary Least Squares and family)
+
+
+
+
+For more discussions of Ridge and Lasso regression, Wessel van Wieringen's article is highly recommended.
+Similarly, Mehta et al's article is also recommended.
+
+
+
+Regression analysis, overarching aims
+
+
+
+ The goal of the regression analysis is to extract/exploit relationship between \( \boldsymbol{y} \) and \( \boldsymbol{x} \) in or to infer causal dependencies, approximations to the likelihood functions, functional relationships and to make predictions, making fits and many other things.
+
+
+Regression analysis, overarching aims II
+
+
+Examples
+
+
+General linear models
+
+
+Rewriting the fitting procedure as a linear algebra problem
+
+
+Rewriting the fitting procedure as a linear algebra problem, more details
+
+
+Generalizing the fitting procedure as a linear algebra problem
+
+
+Generalizing the fitting procedure as a linear algebra problem
+
+
+Optimizing our parameters
+
+
+Our model for the nuclear binding energies
+
+# Common imports
+import numpy as np
+import pandas as pd
+import matplotlib.pyplot as plt
+from IPython.display import display
+import os
+
+# Where to save the figures and data files
+PROJECT_ROOT_DIR = "Results"
+FIGURE_ID = "Results/FigureFiles"
+DATA_ID = "DataFiles/"
+
+if not os.path.exists(PROJECT_ROOT_DIR):
+ os.mkdir(PROJECT_ROOT_DIR)
+
+if not os.path.exists(FIGURE_ID):
+ os.makedirs(FIGURE_ID)
+
+if not os.path.exists(DATA_ID):
+ os.makedirs(DATA_ID)
+
+def image_path(fig_id):
+ return os.path.join(FIGURE_ID, fig_id)
+
+def data_path(dat_id):
+ return os.path.join(DATA_ID, dat_id)
+
+def save_fig(fig_id):
+ plt.savefig(image_path(fig_id) + ".png", format='png')
+
+infile = open(data_path("MassEval2016.dat"),'r')
+
+
+# Read the experimental data with Pandas
+Masses = pd.read_fwf(infile, usecols=(2,3,4,6,11),
+ names=('N', 'Z', 'A', 'Element', 'Ebinding'),
+ widths=(1,3,5,5,5,1,3,4,1,13,11,11,9,1,2,11,9,1,3,1,12,11,1),
+ header=39,
+ index_col=False)
+
+# Extrapolated values are indicated by '#' in place of the decimal place, so
+# the Ebinding column won't be numeric. Coerce to float and drop these entries.
+Masses['Ebinding'] = pd.to_numeric(Masses['Ebinding'], errors='coerce')
+Masses = Masses.dropna()
+# Convert from keV to MeV.
+Masses['Ebinding'] /= 1000
+
+# Group the DataFrame by nucleon number, A.
+Masses = Masses.groupby('A')
+# Find the rows of the grouped DataFrame with the maximum binding energy.
+Masses = Masses.apply(lambda t: t[t.Ebinding==t.Ebinding.max()])
+A = Masses['A']
+Z = Masses['Z']
+N = Masses['N']
+Element = Masses['Element']
+Energies = Masses['Ebinding']
+
+# Now we set up the design matrix X
+X = np.zeros((len(A),5))
+X[:,0] = 1
+X[:,1] = A
+X[:,2] = A**(2.0/3.0)
+X[:,3] = A**(-1.0/3.0)
+X[:,4] = A**(-1.0)
+# Then nice printout using pandas
+DesignMatrix = pd.DataFrame(X)
+DesignMatrix.index = A
+DesignMatrix.columns = ['1', 'A', 'A^(2/3)', 'A^(-1/3)', '1/A']
+display(DesignMatrix)
+
+
+Optimizing our parameters, more details
+
+
+Interpretations and optimizing our parameters
+
+
+Interpretations and optimizing our parameters
+
+
+Some useful matrix and vector expressions
+
+
+
+Interpretations and optimizing our parameters
+
+
+Own code for Ordinary Least Squares
+
+# matrix inversion to find beta
+beta = np.linalg.inv(X.T.dot(X)).dot(X.T).dot(Energies)
+# and then make the prediction
+ytilde = X @ beta
+
fit = np.linalg.lstsq(X, Energies, rcond =None)[0]
+ytildenp = np.dot(fit,X.T)
+
Masses['Eapprox'] = ytilde
+# Generate a plot comparing the experimental with the fitted values values.
+fig, ax = plt.subplots()
+ax.set_xlabel(r'$A = N + Z$')
+ax.set_ylabel(r'$E_\mathrm{bind}\,/\mathrm{MeV}$')
+ax.plot(Masses['A'], Masses['Ebinding'], alpha=0.7, lw=2,
+ label='Ame2016')
+ax.plot(Masses['A'], Masses['Eapprox'], alpha=0.7, lw=2, c='m',
+ label='Fit')
+ax.legend()
+save_fig("Masses2016OLS")
+plt.show()
+
+
+Adding error analysis and training set up
+
+def R2(y_data, y_model):
+ return 1 - np.sum((y_data - y_model) ** 2) / np.sum((y_data - np.mean(y_data)) ** 2)
+
print(R2(Energies,ytilde))
+def MSE(y_data,y_model):
+ n = np.size(y_model)
+ return np.sum((y_data-y_model)**2)/n
+
+print(MSE(Energies,ytilde))
+
def RelativeError(y_data,y_model):
+ return abs((y_data-y_model)/y_data)
+print(RelativeError(Energies, ytilde))
+
+
+The \( \chi^2 \) function
+
+
+The \( \chi^2 \) function
+
+
+The \( \chi^2 \) function
+
+
+The \( \chi^2 \) function
+
+
+The \( \chi^2 \) function
+
+
+The \( \chi^2 \) function
+
+
+Fitting an Equation of State for Dense Nuclear Matter
+
+
+
+The code
+
+# Common imports
+import os
+import numpy as np
+import pandas as pd
+import matplotlib.pyplot as plt
+import matplotlib.pyplot as plt
+import sklearn.linear_model as skl
+from sklearn.metrics import mean_squared_error, r2_score, mean_absolute_error
+
+# Where to save the figures and data files
+PROJECT_ROOT_DIR = "Results"
+FIGURE_ID = "Results/FigureFiles"
+DATA_ID = "DataFiles/"
+
+if not os.path.exists(PROJECT_ROOT_DIR):
+ os.mkdir(PROJECT_ROOT_DIR)
+
+if not os.path.exists(FIGURE_ID):
+ os.makedirs(FIGURE_ID)
+
+if not os.path.exists(DATA_ID):
+ os.makedirs(DATA_ID)
+
+def image_path(fig_id):
+ return os.path.join(FIGURE_ID, fig_id)
+
+def data_path(dat_id):
+ return os.path.join(DATA_ID, dat_id)
+
+def save_fig(fig_id):
+ plt.savefig(image_path(fig_id) + ".png", format='png')
+
+infile = open(data_path("EoS.csv"),'r')
+
+# Read the EoS data as csv file and organize the data into two arrays with density and energies
+EoS = pd.read_csv(infile, names=('Density', 'Energy'))
+EoS['Energy'] = pd.to_numeric(EoS['Energy'], errors='coerce')
+EoS = EoS.dropna()
+Energies = EoS['Energy']
+Density = EoS['Density']
+# The design matrix now as function of various polytrops
+X = np.zeros((len(Density),4))
+X[:,3] = Density**(4.0/3.0)
+X[:,2] = Density
+X[:,1] = Density**(2.0/3.0)
+X[:,0] = 1
+
+# We use now Scikit-Learn's linear regressor and ridge regressor
+# OLS part
+clf = skl.LinearRegression().fit(X, Energies)
+ytilde = clf.predict(X)
+EoS['Eols'] = ytilde
+# The mean squared error
+print("Mean squared error: %.2f" % mean_squared_error(Energies, ytilde))
+# Explained variance score: 1 is perfect prediction
+print('Variance score: %.2f' % r2_score(Energies, ytilde))
+# Mean absolute error
+print('Mean absolute error: %.2f' % mean_absolute_error(Energies, ytilde))
+print(clf.coef_, clf.intercept_)
+
+# The Ridge regression with a hyperparameter lambda = 0.1
+_lambda = 0.1
+clf_ridge = skl.Ridge(alpha=_lambda).fit(X, Energies)
+yridge = clf_ridge.predict(X)
+EoS['Eridge'] = yridge
+# The mean squared error
+print("Mean squared error: %.2f" % mean_squared_error(Energies, yridge))
+# Explained variance score: 1 is perfect prediction
+print('Variance score: %.2f' % r2_score(Energies, yridge))
+# Mean absolute error
+print('Mean absolute error: %.2f' % mean_absolute_error(Energies, yridge))
+print(clf_ridge.coef_, clf_ridge.intercept_)
+
+fig, ax = plt.subplots()
+ax.set_xlabel(r'$\rho[\mathrm{fm}^{-3}]$')
+ax.set_ylabel(r'Energy per particle')
+ax.plot(EoS['Density'], EoS['Energy'], alpha=0.7, lw=2,
+ label='Theoretical data')
+ax.plot(EoS['Density'], EoS['Eols'], alpha=0.7, lw=2, c='m',
+ label='OLS')
+ax.plot(EoS['Density'], EoS['Eridge'], alpha=0.7, lw=2, c='g',
+ label='Ridge $\lambda = 0.1$')
+ax.legend()
+save_fig("EoSfitting")
+plt.show()
+
+
+Splitting our Data in Training and Test data
+
+import os
+import numpy as np
+import pandas as pd
+import matplotlib.pyplot as plt
+from sklearn.model_selection import train_test_split
+# Where to save the figures and data files
+PROJECT_ROOT_DIR = "Results"
+FIGURE_ID = "Results/FigureFiles"
+DATA_ID = "DataFiles/"
+
+if not os.path.exists(PROJECT_ROOT_DIR):
+ os.mkdir(PROJECT_ROOT_DIR)
+
+if not os.path.exists(FIGURE_ID):
+ os.makedirs(FIGURE_ID)
+
+if not os.path.exists(DATA_ID):
+ os.makedirs(DATA_ID)
+
+def image_path(fig_id):
+ return os.path.join(FIGURE_ID, fig_id)
+
+def data_path(dat_id):
+ return os.path.join(DATA_ID, dat_id)
+
+def save_fig(fig_id):
+ plt.savefig(image_path(fig_id) + ".png", format='png')
+
+def R2(y_data, y_model):
+ return 1 - np.sum((y_data - y_model) ** 2) / np.sum((y_data - np.mean(y_data)) ** 2)
+def MSE(y_data,y_model):
+ n = np.size(y_model)
+ return np.sum((y_data-y_model)**2)/n
+
+infile = open(data_path("EoS.csv"),'r')
+
+# Read the EoS data as csv file and organized into two arrays with density and energies
+EoS = pd.read_csv(infile, names=('Density', 'Energy'))
+EoS['Energy'] = pd.to_numeric(EoS['Energy'], errors='coerce')
+EoS = EoS.dropna()
+Energies = EoS['Energy']
+Density = EoS['Density']
+# The design matrix now as function of various polytrops
+X = np.zeros((len(Density),5))
+X[:,0] = 1
+X[:,1] = Density**(2.0/3.0)
+X[:,2] = Density
+X[:,3] = Density**(4.0/3.0)
+X[:,4] = Density**(5.0/3.0)
+# We split the data in test and training data
+X_train, X_test, y_train, y_test = train_test_split(X, Energies, test_size=0.2)
+# matrix inversion to find beta
+beta = np.linalg.inv(X_train.T.dot(X_train)).dot(X_train.T).dot(y_train)
+# and then make the prediction
+ytilde = X_train @ beta
+print("Training R2")
+print(R2(y_train,ytilde))
+print("Training MSE")
+print(MSE(y_train,ytilde))
+ypredict = X_test @ beta
+print("Test R2")
+print(R2(y_test,ypredict))
+print("Test MSE")
+print(MSE(y_test,ypredict))
+
+
Exercises for week 36
Here are three possible exercises for week 36 and the lab sessions of Wednesday September 1..
diff --git a/doc/pub/week34/html/week34.html b/doc/pub/week34/html/week34.html
index ad6e49168..c2f7fa037 100644
--- a/doc/pub/week34/html/week34.html
+++ b/doc/pub/week34/html/week34.html
@@ -150,8 +150,6 @@ div { text-align: justify; text-justify: inter-word; }
('Matrices in Python', 2, None, 'matrices-in-python'),
('Meet the Pandas', 2, None, 'meet-the-pandas'),
('Friday August 27', 2, None, 'friday-august-27'),
- ('Reading Data and fitting', 2, None, 'reading-data-and-fitting'),
- ('Friday August 27', 2, None, 'friday-august-27'),
('Simple linear regression model using _scikit-learn_',
3,
None,
@@ -171,6 +169,88 @@ div { text-align: justify; text-justify: inter-word; }
None,
'and-what-about-using-neural-networks'),
('A first summary', 2, None, 'a-first-summary'),
+ ('Why Linear Regression (aka Ordinary Least Squares and family)',
+ 2,
+ None,
+ 'why-linear-regression-aka-ordinary-least-squares-and-family'),
+ ('Regression analysis, overarching aims',
+ 2,
+ None,
+ 'regression-analysis-overarching-aims'),
+ ('Regression analysis, overarching aims II',
+ 2,
+ None,
+ 'regression-analysis-overarching-aims-ii'),
+ ('Examples', 2, None, 'examples'),
+ ('General linear models', 2, None, 'general-linear-models'),
+ ('Rewriting the fitting procedure as a linear algebra problem',
+ 2,
+ None,
+ 'rewriting-the-fitting-procedure-as-a-linear-algebra-problem'),
+ ('Rewriting the fitting procedure as a linear algebra problem, '
+ 'more details',
+ 2,
+ None,
+ 'rewriting-the-fitting-procedure-as-a-linear-algebra-problem-more-details'),
+ ('Generalizing the fitting procedure as a linear algebra problem',
+ 2,
+ None,
+ 'generalizing-the-fitting-procedure-as-a-linear-algebra-problem'),
+ ('Generalizing the fitting procedure as a linear algebra problem',
+ 2,
+ None,
+ 'generalizing-the-fitting-procedure-as-a-linear-algebra-problem'),
+ ('Optimizing our parameters',
+ 2,
+ None,
+ 'optimizing-our-parameters'),
+ ('Our model for the nuclear binding energies',
+ 2,
+ None,
+ 'our-model-for-the-nuclear-binding-energies'),
+ ('Optimizing our parameters, more details',
+ 2,
+ None,
+ 'optimizing-our-parameters-more-details'),
+ ('Interpretations and optimizing our parameters',
+ 2,
+ None,
+ 'interpretations-and-optimizing-our-parameters'),
+ ('Interpretations and optimizing our parameters',
+ 2,
+ None,
+ 'interpretations-and-optimizing-our-parameters'),
+ ('Some useful matrix and vector expressions',
+ 2,
+ None,
+ 'some-useful-matrix-and-vector-expressions'),
+ ('Interpretations and optimizing our parameters',
+ 2,
+ None,
+ 'interpretations-and-optimizing-our-parameters'),
+ ('Own code for Ordinary Least Squares',
+ 2,
+ None,
+ 'own-code-for-ordinary-least-squares'),
+ ('Adding error analysis and training set up',
+ 2,
+ None,
+ 'adding-error-analysis-and-training-set-up'),
+ ('The $\\chi^2$ function', 2, None, 'the-chi-2-function'),
+ ('The $\\chi^2$ function', 2, None, 'the-chi-2-function'),
+ ('The $\\chi^2$ function', 2, None, 'the-chi-2-function'),
+ ('The $\\chi^2$ function', 2, None, 'the-chi-2-function'),
+ ('The $\\chi^2$ function', 2, None, 'the-chi-2-function'),
+ ('The $\\chi^2$ function', 2, None, 'the-chi-2-function'),
+ ('Fitting an Equation of State for Dense Nuclear Matter',
+ 2,
+ None,
+ 'fitting-an-equation-of-state-for-dense-nuclear-matter'),
+ ('The code', 2, None, 'the-code'),
+ ('Splitting our Data in Training and Test data',
+ 2,
+ None,
+ 'splitting-our-data-in-training-and-test-data'),
('Exercises for week 36', 2, None, 'exercises-for-week-36'),
('Exercise 1: Setting up various Python environments',
2,
@@ -275,10 +355,10 @@ Reading recommendations this week: Refresh linear algebra, GBC chapters 1 and 2.
Thursday August 26
Lectures and ComputerLab
@@ -1473,50 +1556,6 @@ For multidimensional arrays, we recommend strongly Reading Data and fitting
-
-
-
-Friday August 27
-
-
-
Simple linear regression model using scikit-learn
+Why Linear Regression (aka Ordinary Least Squares and family)
+
+
+
+
+For more discussions of Ridge and Lasso regression, Wessel van Wieringen's article is highly recommended.
+Similarly, Mehta et al's article is also recommended.
+
+
+
+Regression analysis, overarching aims
+
+
+
+ The goal of the regression analysis is to extract/exploit relationship between \( \boldsymbol{y} \) and \( \boldsymbol{x} \) in or to infer causal dependencies, approximations to the likelihood functions, functional relationships and to make predictions, making fits and many other things.
+
+
+Regression analysis, overarching aims II
+
+
+Examples
+
+
+General linear models
+
+
+Rewriting the fitting procedure as a linear algebra problem
+
+
+Rewriting the fitting procedure as a linear algebra problem, more details
+
+
+Generalizing the fitting procedure as a linear algebra problem
+
+
+Generalizing the fitting procedure as a linear algebra problem
+
+
+Optimizing our parameters
+
+
+Our model for the nuclear binding energies
+
+# Common imports
+import numpy as np
+import pandas as pd
+import matplotlib.pyplot as plt
+from IPython.display import display
+import os
+
+# Where to save the figures and data files
+PROJECT_ROOT_DIR = "Results"
+FIGURE_ID = "Results/FigureFiles"
+DATA_ID = "DataFiles/"
+
+if not os.path.exists(PROJECT_ROOT_DIR):
+ os.mkdir(PROJECT_ROOT_DIR)
+
+if not os.path.exists(FIGURE_ID):
+ os.makedirs(FIGURE_ID)
+
+if not os.path.exists(DATA_ID):
+ os.makedirs(DATA_ID)
+
+def image_path(fig_id):
+ return os.path.join(FIGURE_ID, fig_id)
+
+def data_path(dat_id):
+ return os.path.join(DATA_ID, dat_id)
+
+def save_fig(fig_id):
+ plt.savefig(image_path(fig_id) + ".png", format='png')
+
+infile = open(data_path("MassEval2016.dat"),'r')
+
+
+# Read the experimental data with Pandas
+Masses = pd.read_fwf(infile, usecols=(2,3,4,6,11),
+ names=('N', 'Z', 'A', 'Element', 'Ebinding'),
+ widths=(1,3,5,5,5,1,3,4,1,13,11,11,9,1,2,11,9,1,3,1,12,11,1),
+ header=39,
+ index_col=False)
+
+# Extrapolated values are indicated by '#' in place of the decimal place, so
+# the Ebinding column won't be numeric. Coerce to float and drop these entries.
+Masses['Ebinding'] = pd.to_numeric(Masses['Ebinding'], errors='coerce')
+Masses = Masses.dropna()
+# Convert from keV to MeV.
+Masses['Ebinding'] /= 1000
+
+# Group the DataFrame by nucleon number, A.
+Masses = Masses.groupby('A')
+# Find the rows of the grouped DataFrame with the maximum binding energy.
+Masses = Masses.apply(lambda t: t[t.Ebinding==t.Ebinding.max()])
+A = Masses['A']
+Z = Masses['Z']
+N = Masses['N']
+Element = Masses['Element']
+Energies = Masses['Ebinding']
+
+# Now we set up the design matrix X
+X = np.zeros((len(A),5))
+X[:,0] = 1
+X[:,1] = A
+X[:,2] = A**(2.0/3.0)
+X[:,3] = A**(-1.0/3.0)
+X[:,4] = A**(-1.0)
+# Then nice printout using pandas
+DesignMatrix = pd.DataFrame(X)
+DesignMatrix.index = A
+DesignMatrix.columns = ['1', 'A', 'A^(2/3)', 'A^(-1/3)', '1/A']
+display(DesignMatrix)
+
+
+Optimizing our parameters, more details
+
+
+Interpretations and optimizing our parameters
+
+
+Interpretations and optimizing our parameters
+
+
+Some useful matrix and vector expressions
+
+
+
+Interpretations and optimizing our parameters
+
+
+Own code for Ordinary Least Squares
+
+# matrix inversion to find beta
+beta = np.linalg.inv(X.T.dot(X)).dot(X.T).dot(Energies)
+# and then make the prediction
+ytilde = X @ beta
+
fit = np.linalg.lstsq(X, Energies, rcond =None)[0]
+ytildenp = np.dot(fit,X.T)
+
Masses['Eapprox'] = ytilde
+# Generate a plot comparing the experimental with the fitted values values.
+fig, ax = plt.subplots()
+ax.set_xlabel(r'$A = N + Z$')
+ax.set_ylabel(r'$E_\mathrm{bind}\,/\mathrm{MeV}$')
+ax.plot(Masses['A'], Masses['Ebinding'], alpha=0.7, lw=2,
+ label='Ame2016')
+ax.plot(Masses['A'], Masses['Eapprox'], alpha=0.7, lw=2, c='m',
+ label='Fit')
+ax.legend()
+save_fig("Masses2016OLS")
+plt.show()
+
+
+Adding error analysis and training set up
+
+def R2(y_data, y_model):
+ return 1 - np.sum((y_data - y_model) ** 2) / np.sum((y_data - np.mean(y_data)) ** 2)
+
print(R2(Energies,ytilde))
+def MSE(y_data,y_model):
+ n = np.size(y_model)
+ return np.sum((y_data-y_model)**2)/n
+
+print(MSE(Energies,ytilde))
+
def RelativeError(y_data,y_model):
+ return abs((y_data-y_model)/y_data)
+print(RelativeError(Energies, ytilde))
+
+
+The \( \chi^2 \) function
+
+
+The \( \chi^2 \) function
+
+
+The \( \chi^2 \) function
+
+
+The \( \chi^2 \) function
+
+
+The \( \chi^2 \) function
+
+
+The \( \chi^2 \) function
+
+
+Fitting an Equation of State for Dense Nuclear Matter
+
+
+
+The code
+
+# Common imports
+import os
+import numpy as np
+import pandas as pd
+import matplotlib.pyplot as plt
+import matplotlib.pyplot as plt
+import sklearn.linear_model as skl
+from sklearn.metrics import mean_squared_error, r2_score, mean_absolute_error
+
+# Where to save the figures and data files
+PROJECT_ROOT_DIR = "Results"
+FIGURE_ID = "Results/FigureFiles"
+DATA_ID = "DataFiles/"
+
+if not os.path.exists(PROJECT_ROOT_DIR):
+ os.mkdir(PROJECT_ROOT_DIR)
+
+if not os.path.exists(FIGURE_ID):
+ os.makedirs(FIGURE_ID)
+
+if not os.path.exists(DATA_ID):
+ os.makedirs(DATA_ID)
+
+def image_path(fig_id):
+ return os.path.join(FIGURE_ID, fig_id)
+
+def data_path(dat_id):
+ return os.path.join(DATA_ID, dat_id)
+
+def save_fig(fig_id):
+ plt.savefig(image_path(fig_id) + ".png", format='png')
+
+infile = open(data_path("EoS.csv"),'r')
+
+# Read the EoS data as csv file and organize the data into two arrays with density and energies
+EoS = pd.read_csv(infile, names=('Density', 'Energy'))
+EoS['Energy'] = pd.to_numeric(EoS['Energy'], errors='coerce')
+EoS = EoS.dropna()
+Energies = EoS['Energy']
+Density = EoS['Density']
+# The design matrix now as function of various polytrops
+X = np.zeros((len(Density),4))
+X[:,3] = Density**(4.0/3.0)
+X[:,2] = Density
+X[:,1] = Density**(2.0/3.0)
+X[:,0] = 1
+
+# We use now Scikit-Learn's linear regressor and ridge regressor
+# OLS part
+clf = skl.LinearRegression().fit(X, Energies)
+ytilde = clf.predict(X)
+EoS['Eols'] = ytilde
+# The mean squared error
+print("Mean squared error: %.2f" % mean_squared_error(Energies, ytilde))
+# Explained variance score: 1 is perfect prediction
+print('Variance score: %.2f' % r2_score(Energies, ytilde))
+# Mean absolute error
+print('Mean absolute error: %.2f' % mean_absolute_error(Energies, ytilde))
+print(clf.coef_, clf.intercept_)
+
+# The Ridge regression with a hyperparameter lambda = 0.1
+_lambda = 0.1
+clf_ridge = skl.Ridge(alpha=_lambda).fit(X, Energies)
+yridge = clf_ridge.predict(X)
+EoS['Eridge'] = yridge
+# The mean squared error
+print("Mean squared error: %.2f" % mean_squared_error(Energies, yridge))
+# Explained variance score: 1 is perfect prediction
+print('Variance score: %.2f' % r2_score(Energies, yridge))
+# Mean absolute error
+print('Mean absolute error: %.2f' % mean_absolute_error(Energies, yridge))
+print(clf_ridge.coef_, clf_ridge.intercept_)
+
+fig, ax = plt.subplots()
+ax.set_xlabel(r'$\rho[\mathrm{fm}^{-3}]$')
+ax.set_ylabel(r'Energy per particle')
+ax.plot(EoS['Density'], EoS['Energy'], alpha=0.7, lw=2,
+ label='Theoretical data')
+ax.plot(EoS['Density'], EoS['Eols'], alpha=0.7, lw=2, c='m',
+ label='OLS')
+ax.plot(EoS['Density'], EoS['Eridge'], alpha=0.7, lw=2, c='g',
+ label='Ridge $\lambda = 0.1$')
+ax.legend()
+save_fig("EoSfitting")
+plt.show()
+
+
+Splitting our Data in Training and Test data
+
+import os
+import numpy as np
+import pandas as pd
+import matplotlib.pyplot as plt
+from sklearn.model_selection import train_test_split
+# Where to save the figures and data files
+PROJECT_ROOT_DIR = "Results"
+FIGURE_ID = "Results/FigureFiles"
+DATA_ID = "DataFiles/"
+
+if not os.path.exists(PROJECT_ROOT_DIR):
+ os.mkdir(PROJECT_ROOT_DIR)
+
+if not os.path.exists(FIGURE_ID):
+ os.makedirs(FIGURE_ID)
+
+if not os.path.exists(DATA_ID):
+ os.makedirs(DATA_ID)
+
+def image_path(fig_id):
+ return os.path.join(FIGURE_ID, fig_id)
+
+def data_path(dat_id):
+ return os.path.join(DATA_ID, dat_id)
+
+def save_fig(fig_id):
+ plt.savefig(image_path(fig_id) + ".png", format='png')
+
+def R2(y_data, y_model):
+ return 1 - np.sum((y_data - y_model) ** 2) / np.sum((y_data - np.mean(y_data)) ** 2)
+def MSE(y_data,y_model):
+ n = np.size(y_model)
+ return np.sum((y_data-y_model)**2)/n
+
+infile = open(data_path("EoS.csv"),'r')
+
+# Read the EoS data as csv file and organized into two arrays with density and energies
+EoS = pd.read_csv(infile, names=('Density', 'Energy'))
+EoS['Energy'] = pd.to_numeric(EoS['Energy'], errors='coerce')
+EoS = EoS.dropna()
+Energies = EoS['Energy']
+Density = EoS['Density']
+# The design matrix now as function of various polytrops
+X = np.zeros((len(Density),5))
+X[:,0] = 1
+X[:,1] = Density**(2.0/3.0)
+X[:,2] = Density
+X[:,3] = Density**(4.0/3.0)
+X[:,4] = Density**(5.0/3.0)
+# We split the data in test and training data
+X_train, X_test, y_train, y_test = train_test_split(X, Energies, test_size=0.2)
+# matrix inversion to find beta
+beta = np.linalg.inv(X_train.T.dot(X_train)).dot(X_train.T).dot(y_train)
+# and then make the prediction
+ytilde = X_train @ beta
+print("Training R2")
+print(R2(y_train,ytilde))
+print("Training MSE")
+print(MSE(y_train,ytilde))
+ypredict = X_test @ beta
+print("Test R2")
+print(R2(y_test,ypredict))
+print("Test MSE")
+print(MSE(y_test,ypredict))
+
+
Exercises for week 36
Here are three possible exercises for week 36 and the lab sessions of Wednesday September 1..
diff --git a/doc/pub/week34/ipynb/ipynb-week34-src.tar.gz b/doc/pub/week34/ipynb/ipynb-week34-src.tar.gz
index f82161be2..849ae396e 100644
Binary files a/doc/pub/week34/ipynb/ipynb-week34-src.tar.gz and b/doc/pub/week34/ipynb/ipynb-week34-src.tar.gz differ
diff --git a/doc/pub/week34/ipynb/week34.ipynb b/doc/pub/week34/ipynb/week34.ipynb
index ba76954c7..a78310b0a 100644
--- a/doc/pub/week34/ipynb/week34.ipynb
+++ b/doc/pub/week34/ipynb/week34.ipynb
@@ -54,16 +54,24 @@
"## Thursday August 26\n",
"\n",
"\n",
- "[Video of Lecture from Fall Semester 2020](https://www.uio.no/studier/emner/matnat/fys/FYS-STK4155/h20/forelesningsvideoer/zoom_0.mp4?vrtx=view-as-webpage).\n",
+ "\n",
"\n",
"The lectures will be recorded and updated videos will be posted after the lectures. \n",
"\n",
+ "\"Video of Lecture August 26, 2021\":\"https://www.uio.no/studier/emner/matnat/fys/FYS-STK4155/h21/forelesningsvideoer/LectureThursdayAugust26.mp4?vrtx=view-as-webpage\n",
+ "\n",
+ "\n",
"**Zoom link for lectures**: