======= Linear Regression =======


===== Introduction =====





Our emphasis throughout this series of lectures is on understanding
the mathematical aspects of different algorithms used in the fields of
data analysis and machine learning.

However, where possible we will emphasize the importance of using
available software. We start thus with a hands-on and top-down
approach to machine learning. The aim is thus to start with relevant
data or data we have produced and use these to introduce statistical
data analysis concepts and machine learning algorithms before we delve
into the algorithms themselves. The examples we will use in the
beginning, start with simple polynomials with random noise added. We
will use the Python software package
"Scikit-Learn":"http://scikit-learn.org/stable/" and introduce various
machine learning algorithms to make fits of the data and
predictions. We move thereafter to more interesting cases such as data
from say experiments (below we will look at experimental nuclear
binding energies as an example).  These are examples where we can
easily set up the data and then use machine learning algorithms
included in for example _Scikit-Learn_.

These examples will serve us the purpose of getting
started. Furthermore, they allow us to catch more than two birds with
a stone. They will allow us to bring in some programming specific
topics and tools as well as showing the power of various Python
libraries for machine learning and statistical data analysis.

Here, we will mainly focus on two specific Python packages for Machine
Learning, Scikit-Learn and Tensorflow (see below for links etc).
Moreover, the examples we introduce will serve as inputs to many of
our discussions later, as well as allowing you to set up models and
produce your own data and get started with programming.



===== What is Machine Learning?  =====

Statistics, data science and machine learning form important fields of
research in modern science.  They describe how to learn and make
predictions from data, as well as allowing us to extract important
correlations about physical process and the underlying laws of motion
in large data sets. The latter, big data sets, appear frequently in
essentially all disciplines, from the traditional Science, Technology,
Mathematics and Engineering fields to Life Science, Law, education
research, the Humanities and the Social Sciences. 

It has become more
and more common to see research projects on big data in for example
the Social Sciences where extracting patterns from complicated survey
data is one of many research directions.  Having a solid grasp of data
analysis and machine learning is thus becoming central to scientific
computing in many fields, and competences and skills within the fields
of machine learning and scientific computing are nowadays strongly
requested by many potential employers. The latter cannot be
overstated, familiarity with machine learning has almost become a
prerequisite for many of the most exciting employment opportunities,
whether they are in bioinformatics, life science, physics or finance,
in the private or the public sector. This author has had several
students or met students who have been hired recently based on their
skills and competences in scientific computing and data science, often
with marginal knowledge of machine learning.

Machine learning is a subfield of computer science, and is closely
related to computational statistics.  It evolved from the study of
pattern recognition in artificial intelligence (AI) research, and has
made contributions to AI tasks like computer vision, natural language
processing and speech recognition. Many of the methods we will study are also 
strongly rooted in basic mathematics and physics research. 

Ideally, machine learning represents the science of giving computers
the ability to learn without being explicitly programmed.  The idea is
that there exist generic algorithms which can be used to find patterns
in a broad class of data sets without having to write code
specifically for each problem. The algorithm will build its own logic
based on the data.  You should however always keep in mind that
machines and algorithms are to a large extent developed by humans. The
insights and knowledge we have about a specific system, play a central
role when we develop a specific machine learning algorithm. 

Machine learning is an extremely rich field, in spite of its young
age. The increases we have seen during the last three decades in
computational capabilities have been followed by developments of
methods and techniques for analyzing and handling large data sets,
relying heavily on statistics, computer science and mathematics.  The
field is rather new and developing rapidly. Popular software packages
written in Python for machine learning like
"Scikit-learn":"http://scikit-learn.org/stable/",
"Tensorflow":"https://www.tensorflow.org/",
"PyTorch":"http://pytorch.org/" and "Keras":"https://keras.io/", all
freely available at their respective GitHub sites, encompass
communities of developers in the thousands or more. And the number of
code developers and contributors keeps increasing. Not all the
algorithms and methods can be given a rigorous mathematical
justification, opening up thereby large rooms for experimenting and
trial and error and thereby exciting new developments.  However, a
solid command of linear algebra, multivariate theory, probability
theory, statistical data analysis, understanding errors and Monte
Carlo methods are central elements in a proper understanding of many
of algorithms and methods we will discuss.



The approaches to machine learning are many, but are often split into
two main categories.  In *supervised learning* we know the answer to a
problem, and let the computer deduce the logic behind it. On the other
hand, *unsupervised learning* is a method for finding patterns and
relationship in data sets without any prior knowledge of the system.
Some authors also operate with a third category, namely
*reinforcement learning*. This is a paradigm of learning inspired by
behavioral psychology, where learning is achieved by trial-and-error,
solely from rewards and punishment.

Another way to categorize machine learning tasks is to consider the
desired output of a system.  Some of the most common tasks are:

  * Classification: Outputs are divided into two or more classes. The goal is to   produce a model that assigns inputs into one of these classes. An example is to identify  digits based on pictures of hand-written ones. Classification is typically supervised learning.

  * Regression: Finding a functional relationship between an input data set and a reference data set.   The goal is to construct a function that maps input data to continuous output values.

  * Clustering: Data are divided into groups with certain common traits, without knowing the different groups beforehand.  It is thus a form of unsupervised learning.



The methods we cover have three main topics in common, irrespective of
whether we deal with supervised or unsupervised learning.
!bpop
* The first ingredient is normally our data set (which can be subdivided into training, validation  and test data). Many find the most difficult part of using Machine Learning to be the set up of your data in a meaningful way. 

* The second item is a model which is normally a function of some parameters.  The model reflects our knowledge of the system (or lack thereof). As an example, if we know that our data show a behavior similar to what would be predicted by a polynomial, fitting our data to a polynomial of some degree would then determin our model. 

* The last ingredient is a so-called _cost/loss_ function (or error or risk function) which allows us to present an estimate on how good our model is in reproducing the data it is supposed to train.  
!epop


At the heart of basically all Machine Learning algorithms we will encounter so-called minimization or optimization algorithms. A large family of such methods are so-called _gradient methods_.


=== A Frequentist approach to data analysis  ===

When you hear phrases like _predictions and estimations_ and
_correlations and causations_, what do you think of?  May be you think
of the difference between classifying new data points and generating
new data points.
Or perhaps you consider that correlations represent some kind of symmetric statements like
if $A$ is correlated with $B$, then $B$ is correlated with
$A$. Causation on the other hand is directional, that is if $A$ causes $B$, $B$ does not
necessarily cause $A$.

These concepts are in some sense the difference between machine
learning and statistics. In machine learning and prediction based
tasks, we are often interested in developing algorithms that are
capable of learning patterns from given data in an automated fashion,
and then using these learned patterns to make predictions or
assessments of newly given data. In many cases, our primary concern
is the quality of the predictions or assessments, and we are less
concerned about the underlying patterns that were learned in order
to make these predictions.

In machine learning we normally use "a so-called frequentist approach":"https://en.wikipedia.org/wiki/Frequentist_inference",
where the aim is to make predictions and find correlations. We focus
less on for example extracting a probability distribution function (PDF). The PDF can be
used in turn to make estimations and find causations such as given $A$
what is the likelihood of finding $B$.


=== What is a good model? ===

In science and engineering we often end up in situations where we want to infer (or learn) a
quantitative model $M$ for a given set of sample points $\bm{X} \in [x_1, x_2,\dots x_N]$.

As we will see repeatedly in these lectures, we could try to fit these data points to a model given by a
straight line, or if we wish to be more sophisticated to a more complex
function.

The reason for inferring such a model is that it
serves many useful purposes. On the one hand, the model can reveal information
encoded in the data or underlying mechanisms from which the data were generated. For instance, we could discover important
correlations that relate interesting physics interpretations.

In addition, it can simplify the representation of the given data set and help
us in making predictions about  future data samples.

A first important consideration to keep in mind is that inferring the *correct* model
for a given data set is an elusive, if not impossible, task. The fundamental difficulty
is that if we are not specific about what we mean by a *correct* model, there
could easily be many different models that fit the given data set *equally well*.




The central question is this: what leads us to say that a model is correct or
optimal for a given data set? To make the model inference problem well posed, i.e.,
to guarantee that there is a unique optimal model for the given data, we need to
impose additional assumptions or restrictions on the class of models considered. To
this end, we should not be looking for just any model that can describe the data.
Instead, we should look for a _model_ $M$ that is the best among a restricted class
of models. In addition, to make the model inference problem computationally
tractable, we need to specify how restricted the class of models needs to be. A
common strategy is to start 
with the simplest possible class of models that is just necessary to describe the data
or solve the problem at hand. More precisely, the model class should be rich enough
to contain at least one model that can fit the data to a desired accuracy and yet be
restricted enough that it is relatively simple to find the best model for the given data.

Thus, the most popular strategy is to start from the
simplest class of models and increase the complexity of the models only when the
simpler models become inadequate. For instance, if we work with a regression problem to fit a set of sample points, one
may first try the simplest class of models, namely linear models, followed obviously by more complex models.

How to evaluate which model fits best the data is something we will come back to over and over again in these sets of lectures.











===== Simple linear regression model using _scikit-learn_ =====

We start with perhaps our simplest possible example, using
_Scikit-Learn_ to perform linear regression analysis on a data set
produced by us.

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 $\bm{x}$ 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 $x$

!bt
\[
y = 2x+N(0,1),
\]
!et

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 $\tilde{y} =
\alpha + \theta x$ using the function _fit(x,y)_. We call the set of
data $(\bm{x},\bm{y})$ for our training data. The Python package
_scikit-learn_ has also a functionality which extracts the above
fitting parameters $\alpha$ and $\theta$ (see below). Later we will
distinguish between training data and test data.

For plotting we use the Python package
"matplotlib":"https://matplotlib.org/" which produces publication
quality figures. Feel free to explore the extensive
"gallery":"https://matplotlib.org/gallery/index.html" 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.  Note also that _Scikit-Learn_ requires a
matrix as input for the input values $x$ and $y$. In the above code we
have solved this by declaring $x$ and $y$ as arrays of dimension
$n\times 1$.

In the code here we have also made a new array for $x\in [0,1]$. Our
prediction is computed for these values, meaning that they were not
included in the data set used to *train* (or fit) the model.
This is a recurrring theme in machine learning and data analysis. We would like to train a model on a specific given data set.
Thereafter we wish to apply it to data which were not included in the training. Below we will encounter this again in the so-called *train-validate-test* spliting. We will typically split our data into different sets, oen for training, one for validation and finally, our data from the untouched test vault!


The Python code follows here.
!bc pycod
# 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)
# This is our new x-array to which we test our model
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()
!ec

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

!bt
\[
y = 10x+0.01 \times N(0,1),
\]
!et

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 obviously 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))

!bt 
\[ \chi^2 = \frac{1}{n}
\sum_{i=0}^{n-1}\frac{(y_i-\tilde{y}_i)^2}{\sigma_i^2}, 
\] 
!et 

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 $\theta$ 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

!bt
\[
\epsilon_{\mathrm{relative}}= \frac{\vert \bm{y} -\bm{\tilde{y}}\vert}{\vert \bm{y}\vert}.
\]
!et

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
!bc pycod
import numpy as np
import matplotlib.pyplot as plt
from sklearn.linear_model import LinearRegression
# Number of data points
n = 100
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()
!ec

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 $\theta$ 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_.
!bc pycod 
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 theta : \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()

!ec
The function _coef_ gives us the parameter $\theta$ 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 $\theta =5$. 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
!bt 
\[ MSE(\bm{y},\bm{\tilde{y}}) = \frac{1}{n}
\sum_{i=0}^{n-1}(y_i-\tilde{y}_i)^2, 
\] 
!et 

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 $\bm{y}$,
disregarding the input features, would get a $R^2$ score of $0.0$.

If $\tilde{\bm{y}}_i$ 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
!bt
\[
R^2(\bm{y}, \tilde{\bm{y}}) = 1 - \frac{\sum_{i=0}^{n - 1} (y_i - \tilde{y}_i)^2}{\sum_{i=0}^{n - 1} (y_i - \bar{y})^2},
\]
!et
where we have defined the mean value  of $\bm{y}$ as
!bt
\[
\bar{y} =  \frac{1}{n} \sum_{i=0}^{n - 1} y_i.
\]
!et
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
!bt
\[
\text{MAE}(\bm{y}, \bm{\tilde{y}}) = \frac{1}{n} \sum_{i=0}^{n-1} \left| y_i - \tilde{y}_i \right|.
\]
!et
We present the 
squared logarithmic (quadratic) error
!bt
\[
\text{MSLE}(\bm{y}, \bm{\tilde{y}}) = \frac{1}{n} \sum_{i=0}^{n - 1} (\log_e (1 + y_i) - \log_e (1 + \tilde{y}_i) )^2,
\]
!et

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
!bt
\[
H_{\delta}(\bm{a})=\left\{\begin{array}{cc}\frac{1}{2} \bm{a}^{2}& \text{for }|\bm{a}|\leq \delta\\ \delta (|\bm{a}|-\frac{1}{2}\delta ),&\text{otherwise}.\end{array}\right.
\]
!et
Here $\bm{a}=\bm{y} - \bm{\tilde{y}}$.




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. 

!bc pycod
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))
!ec



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
!bt
\[
\Delta M(N, Z) =  M(N, Z) - uA,
\]
!et
where $u$ is the Atomic Mass Unit 
!bt
\[
u = M(^{12}\mathrm{C})/12 = 931.4940954(57) \hspace{0.1cm} \mathrm{MeV}/c^2.
\]
!et
The nucleon masses are
!bt 
\[
m_p =  1.00727646693(9)u,
\] 
!et
and
!bt 
\[
m_n = 939.56536(8)\hspace{0.1cm} \mathrm{MeV}/c^2 = 1.0086649156(6)u.
\]
!et

In the "2016 mass evaluation of by W.J.Huang, G.Audi, M.Wang, F.G.Kondev, S.Naimi and X.Xu":"http://nuclearmasses.org/resources_folder/Wang_2017_Chinese_Phys_C_41_030003.pdf"
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


!bt
\[
BE(N, Z) = ZM_H c^2 + Nm_n c^2 - M(N, Z)c^2 ,
\]
!et
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
!bt
\[
BE(N, Z) = Z\Delta_H c^2 + N\Delta_n c^2 -\Delta(N, Z)c^2 ,
\]
!et
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

!bt
\[ 
BE(N,Z) = a_1A-a_2A^{2/3}-a_3\frac{Z^2}{A^{1/3}}-a_4\frac{(N-Z)^2}{A},
\]
!et

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:

 * There is a volume term $a_1A$ proportional with the number of nucleons (the energy is also an extensive quantity). When an assembly of nucleons of the same size is packed together into the smallest volume, each interior nucleon has a certain number of other nucleons in contact with it. This contribution is proportional to the volume.

 * There is a surface energy term $a_2A^{2/3}$. The assumption here is that a nucleon at the surface of a nucleus interacts with fewer other nucleons than one in the interior of the nucleus and hence its binding energy is less. This surface energy term takes that into account and is therefore negative and is proportional to the surface area.


 * There is a Coulomb energy term $a_3\frac{Z^2}{A^{1/3}}$. The electric repulsion between each pair of protons in a nucleus yields less binding. 

 * There is an asymmetry term $a_4\frac{(N-Z)^2}{A}$. This term is associated with the Pauli exclusion principle and reflects the fact that the proton-neutron interaction is more attractive on the average than the neutron-neutron and proton-proton interactions.

We could also add a so-called pairing term, which is a correction term that
arises from the tendency of proton pairs and neutron pairs to
occur. An even number of particles is more stable than an odd number. 


=== Organizing our data ===

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_.
!bc pycod
# 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')
!ec


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.
!bc pycod 
"""                                                                                                                         
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.                                                          
"""
!ec

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.

!bc pycod 
# 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()])
!ec

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.
!bc pycod 
A = Masses['A']
Z = Masses['Z']
N = Masses['N']
Element = Masses['Element']
Energies = Masses['Ebinding']
print(Masses)
!ec
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 $\bm{X}$.
It has dimensionality $n\times p$, 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. 
!bc pycod
# 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)
!ec

Note well that we have made life simple here. We perform a fit in
terms of the number of nucleons only.  A more sophisticated fit can be
done by including an explicit dependence on the number of protons and
neutrons in the asymmetry and Coulomb terms. We leave this as an exercise to you the reader.

With _Scikit-Learn_ we are now ready to use linear regression and fit our data.
!bc pycod 
clf = skl.LinearRegression().fit(X, Energies)
fity = clf.predict(X)
!ec
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.
!bc pycod
# 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))

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()
!ec


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_!


!bc pycod

#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))
!ec

With a deeper and deeper tree level, we can almost reproduce every
single data point by increasing the max depth of the tree.
We can actually decide to make a decision tree which fits every single point.
As we will
see later, this has the benefit that we can really train a model which
traverses every single data point. However, the price we pay is that
we will easily overfit. That is, if we apply our model to unseen data,
we will most likely fail miserably in our attempt at making
predictions. As an exercise, try to make the tree level larger by adjusting the maximum depth variable. When printing out the predicition, you will note that the binding energy of every nucleus is accurately reproduced.


The _seaborn_ package allows us to visualize data in an efficient way. Note that we use _scikit-learn_'s multi-layer perceptron (or feed forward neural network) 
functionality.
!bc pycod
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()



!ec



=====  Linear Regression, basic elements =====


"Video of Lecture":"https://www.uio.no/studier/emner/matnat/fys/FYS-STK4155/h20/forelesningsvideoer/LectureAug27.mp4?vrtx=view-as-webpage".


Fitting a continuous function with linear parameterization in terms of the parameters  $\bm{\theta}$.
* Method of choice for fitting a continuous function!
* Gives an excellent introduction to central Machine Learning features with _understandable pedagogical_ links to other methods like _Neural Networks_, _Support Vector Machines_ etc
* Analytical expression for the fitting parameters $\bm{\theta}$
* Analytical expressions for statistical propertiers like mean values, variances, confidence intervals and more
* Analytical relation with probabilistic interpretations 
* Easy to introduce basic concepts like bias-variance tradeoff, cross-validation, resampling and regularization techniques and many other ML topics
* Easy to code! And links well with classification problems and logistic regression and neural networks
* Allows for _easy_ hands-on understanding of gradient descent methods
* and many more features

For more discussions of Ridge and Lasso regression, "Wessel van Wieringen's":"https://arxiv.org/abs/1509.09169" article is highly recommended.
Similarly, "Mehta et al's article":"https://arxiv.org/abs/1803.08823" 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 $\bm{x} =[x_0, x_1,\dots, x_{n-1}]^T$. 
The first variable is called the _dependent_, the _outcome_ or the _response_ or just the output variable while the set of variables $\bm{x}$ is called the independent variable, or the predictor variable or the explanatory variable, or just the input variable. We will hereafter call $\bm{y}$ and $\bm{x}x$ for the output and input variables, respectively. 
 
A regression model aims at finding a likelihood function $p(\bm{y}\vert \bm{x})$ (or just a function $f(\bm{x}$) , that is the conditional distribution for $\bm{y}$ with a given $\bm{x}$. The estimation of  $p(\bm{y}\vert \bm{x})$ is made using a data set with 
* $n$ cases $i = 0, 1, 2, \dots, n-1$ 
* Response/output (target, dependent or outcome) variable $y_i$ with $i = 0, 1, 2, \dots, n-1$ 
* $p$ so-called explanatory/input (independent or predictor) variables $\bm{x}_i=[x_{i0}, x_{i1}, \dots, x_{ip-1}]$ with $i = 0, 1, 2, \dots, n-1$ and explanatory variables running from $0$ to $p-1$. See below for more explicit examples.   
 The goal of the regression analysis is to extract/exploit relationship between $\bm{y}$ and $\bm{x}$ 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 $\bm{y}$ (also as above). The variable $\bm{y}$ is
generally referred to as the *response variable*. The aim of
regression analysis is to explain $\bm{y}$ in terms of
$\bm{X}$ 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 $\bm{X}$ and $\bm{y}$. This assumption gives rise to
the *linear regression model* where $\bm{\theta} = [\theta_0, \ldots,
\theta_{p-1}]^{T}$ are the *regression parameters*. 

Linear regression gives us a set of analytical equations for the parameters $\theta_j$.


In order to understand the relation among the predictors $p$, the set of data $n$ and the target (outcome, output etc) $\bm{y}$,
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 
!bt
\[
BE(A) = a_0+a_1A+a_2A^{2/3}+a_3A^{-1/3}+a_4A^{-1},
\]
!et
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 $\bm{X}$.

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":"https://www.sciencedirect.com/science/article/pii/S0957417407006719?via%3Dihub". 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 $\bm{y}=[y_0,y_1,\dots,y_{n-1}]$. 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 $\bm{x}=[x_0,x_1,\dots,x_{n-1}]$, 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
!bt
\[
y=y(x) \rightarrow y(x_i)=\tilde{y}_i+\epsilon_i=\sum_{j=0}^{n-1} \theta_j x_i^j+\epsilon_i,
\]
!et
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
!bt
\begin{align*}
y_0&=\theta_0+\theta_1x_0^1+\theta_2x_0^2+\dots+\theta_{n-1}x_0^{n-1}+\epsilon_0\\
y_1&=\theta_0+\theta_1x_1^1+\theta_2x_1^2+\dots+\theta_{n-1}x_1^{n-1}+\epsilon_1\\
y_2&=\theta_0+\theta_1x_2^1+\theta_2x_2^2+\dots+\theta_{n-1}x_2^{n-1}+\epsilon_2\\
\dots & \dots \\
y_{n-1}&=\theta_0+\theta_1x_{n-1}^1+\theta_2x_{n-1}^2+\dots+\theta_{n-1}x_{n-1}^{n-1}+\epsilon_{n-1}.\\
\end{align*}
!et




Defining the vectors
!bt
\[
\bm{y} = [y_0,y_1, y_2,\dots, y_{n-1}]^T,
\]
!et
and
!bt
\[
\bm{\theta} = [\theta_0,\theta_1, \theta_2,\dots, \theta_{n-1}]^T,
\]
!et
and
!bt
\[
\bm{\epsilon} = [\epsilon_0,\epsilon_1, \epsilon_2,\dots, \epsilon_{n-1}]^T,
\]
!et
and the design matrix
!bt
\[
\bm{X}=
\begin{bmatrix} 
1& x_{0}^1 &x_{0}^2& \dots & \dots &x_{0}^{n-1}\\
1& x_{1}^1 &x_{1}^2& \dots & \dots &x_{1}^{n-1}\\
1& x_{2}^1 &x_{2}^2& \dots & \dots &x_{2}^{n-1}\\                      
\dots& \dots &\dots& \dots & \dots &\dots\\
1& x_{n-1}^1 &x_{n-1}^2& \dots & \dots &x_{n-1}^{n-1}\\
\end{bmatrix} 
\]
!et
we can rewrite our equations as
!bt
\[
\bm{y} = \bm{X}\bm{\theta}+\bm{\epsilon}.
\]
!et
The above design matrix is called a "Vandermonde matrix":"https://en.wikipedia.org/wiki/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

!bt
\begin{align*}
y_0&=\theta_0x_{00}+\theta_1x_{01}+\theta_2x_{02}+\dots+\theta_{n-1}x_{0n-1}+\epsilon_0\\
y_1&=\theta_0x_{10}+\theta_1x_{11}+\theta_2x_{12}+\dots+\theta_{n-1}x_{1n-1}+\epsilon_1\\
y_2&=\theta_0x_{20}+\theta_1x_{21}+\theta_2x_{22}+\dots+\theta_{n-1}x_{2n-1}+\epsilon_2\\
\dots & \dots \\
y_{i}&=\theta_0x_{i0}+\theta_1x_{i1}+\theta_2x_{i2}+\dots+\theta_{n-1}x_{in-1}+\epsilon_i\\
\dots & \dots \\
y_{n-1}&=\theta_0x_{n-1,0}+\theta_1x_{n-1,2}+\theta_2x_{n-1,2}+\dots+\theta_{n-1}x_{n-1,n-1}+\epsilon_{n-1}.\\
\end{align*}
!et

_Note that we have $p=n$ here. The matrix is symmetric. This is generally not the case!_

We redefine in turn the matrix $\bm{X}$ as
!bt
\[
\bm{X}=
\begin{bmatrix} 
x_{00}& x_{01} &x_{02}& \dots & \dots &x_{0,n-1}\\
x_{10}& x_{11} &x_{12}& \dots & \dots &x_{1,n-1}\\
x_{20}& x_{21} &x_{22}& \dots & \dots &x_{2,n-1}\\                      
\dots& \dots &\dots& \dots & \dots &\dots\\
x_{n-1,0}& x_{n-1,1} &x_{n-1,2}& \dots & \dots &x_{n-1,n-1}\\
\end{bmatrix} 
\]
!et
and without loss of generality we rewrite again  our equations as
!bt
\[
\bm{y} = \bm{X}\bm{\theta}+\bm{\epsilon}.
\]
!et
The left-hand side of this equation is kwown. Our error vector $\bm{\epsilon}$ and the parameter vector $\bm{\theta}$ are our unknow quantities. How can we obtain the optimal set of $\theta_i$ values? 

We have defined the matrix $\bm{X}$ via the equations
!bt
\begin{align*}
y_0&=\theta_0x_{00}+\theta_1x_{01}+\theta_2x_{02}+\dots+\theta_{n-1}x_{0n-1}+\epsilon_0\\
y_1&=\theta_0x_{10}+\theta_1x_{11}+\theta_2x_{12}+\dots+\theta_{n-1}x_{1n-1}+\epsilon_1\\
y_2&=\theta_0x_{20}+\theta_1x_{21}+\theta_2x_{22}+\dots+\theta_{n-1}x_{2n-1}+\epsilon_1\\
\dots & \dots \\
y_{i}&=\theta_0x_{i0}+\theta_1x_{i1}+\theta_2x_{i2}+\dots+\theta_{n-1}x_{in-1}+\epsilon_1\\
\dots & \dots \\
y_{n-1}&=\theta_0x_{n-1,0}+\theta_1x_{n-1,2}+\theta_2x_{n-1,2}+\dots+\theta_{n-1}x_{n-1,n-1}+\epsilon_{n-1}.\\
\end{align*}
!et

As we noted above, we stayed with a system with the design matrix 
 $\bm{X}\in {\mathbb{R}}^{n\times n}$, that is we have $p=n$. For reasons to come later (algorithmic arguments) we will hereafter define 
our matrix as $\bm{X}\in {\mathbb{R}}^{n\times p}$, with the predictors refering to the column numbers and the entries $n$ being the row elements.

In our "introductory notes":"https://compphysics.github.io/MachineLearning/doc/pub/How2ReadData/html/How2ReadData.html" we looked at the so-called "liquid drop model":"https://en.wikipedia.org/wiki/Semi-empirical_mass_formula". 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.
!bc pycod
# 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)
!ec

With $\bm{\theta}\in {\mathbb{R}}^{p\times 1}$, it means that we will hereafter write our equations for the approximation as
!bt
\[
\bm{\tilde{y}}= \bm{X}\bm{\theta},
\]
!et
throughout these lectures. 

With the above we use the design matrix to define the approximation $\bm{\tilde{y}}$ via the unknown quantity $\bm{\theta}$ as
!bt
\[
\bm{\tilde{y}}= \bm{X}\bm{\theta},
\]
!et
and in order to find the optimal parameters $\theta_i$ 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
!bt
\[
C(\bm{\theta})=\frac{1}{n}\sum_{i=0}^{n-1}\left(y_i-\tilde{y}_i\right)^2=\frac{1}{n}\left\{\left(\bm{y}-\bm{\tilde{y}}\right)^T\left(\bm{y}-\bm{\tilde{y}}\right)\right\},
\]
!et
or using the matrix $\bm{X}$ and in a more compact matrix-vector notation as
!bt
\[
C(\bm{\theta})=\frac{1}{n}\left\{\left(\bm{y}-\bm{X}\bm{\theta}\right)^T\left(\bm{y}-\bm{X}\bm{\theta}\right)\right\}.
\]
!et
This function is one possible way to define the so-called cost function.



It is also common to define
the function $C$ as

!bt
\[
C(\bm{\theta})=\frac{1}{2n}\sum_{i=0}^{n-1}\left(y_i-\tilde{y}_i\right)^2,
\]
!et
since when taking the first derivative with respect to the unknown parameters $\theta$, the factor of $2$ cancels out. 

The function 
!bt
\[
C(\bm{\theta})=\frac{1}{n}\left\{\left(\bm{y}-\bm{X}\bm{\theta}\right)^T\left(\bm{y}-\bm{X}\bm{\theta}\right)\right\},
\]
!et
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
!bt
\[
y_{i}=\langle y_i \rangle = \theta_0x_{i,0}+\theta_1x_{i,1}+\theta_2x_{i,2}+\dots+\theta_{n-1}x_{i,n-1}+\epsilon_i,
\]
!et

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 $\theta_i$ we will then minimize the spread of $C(\bm{\theta})$, that is we are going to solve the problem
!bt
\[
{\displaystyle \min_{\bm{\theta}\in
{\mathbb{R}}^{p}}}\frac{1}{n}\left\{\left(\bm{y}-\bm{X}\bm{\theta}\right)^T\left(\bm{y}-\bm{X}\bm{\theta}\right)\right\}.
\]
!et
In practical terms it means we will require
!bt
\[
\frac{\partial C(\bm{\theta})}{\partial \theta_j} = \frac{\partial }{\partial \theta_j}\left[ \frac{1}{n}\sum_{i=0}^{n-1}\left(y_i-\theta_0x_{i,0}-\theta_1x_{i,1}-\theta_2x_{i,2}-\dots-\theta_{n-1}x_{i,n-1}\right)^2\right]=0, 
\]
!et
which results in
!bt
\[
\frac{\partial C(\bm{\theta})}{\partial \theta_j} = -\frac{2}{n}\left[ \sum_{i=0}^{n-1}x_{ij}\left(y_i-\theta_0x_{i,0}-\theta_1x_{i,1}-\theta_2x_{i,2}-\dots-\theta_{n-1}x_{i,n-1}\right)\right]=0, 
\]
!et
or in a matrix-vector form as
!bt
\[
\frac{\partial C(\bm{\theta})}{\partial \bm{\theta}} = 0 = \bm{X}^T\left( \bm{y}-\bm{X}\bm{\theta}\right).  
\]
!et



We can rewrite
!bt
\[
\frac{\partial C(\bm{\theta})}{\partial \bm{\theta}} = 0 = \bm{X}^T\left( \bm{y}-\bm{X}\bm{\theta}\right),  
\]
!et
as
!bt
\[
\bm{X}^T\bm{y} = \bm{X}^T\bm{X}\bm{\theta},  
\]
!et
and if the matrix $\bm{X}^T\bm{X}$ is invertible we have the solution
!bt
\[
\bm{\theta} =\left(\bm{X}^T\bm{X}\right)^{-1}\bm{X}^T\bm{y}.
\]
!et

We note also that since our design matrix is defined as $\bm{X}\in
{\mathbb{R}}^{n\times p}$, the product $\bm{X}^T\bm{X} \in
{\mathbb{R}}^{p\times p}$.  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
$\bm{X}^T\bm{X}$. 

_Small question_: Do you think the example we have at hand here (the nuclear binding energies) can lead to problems in inverting the matrix  $\bm{X}^T\bm{X}$? 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. 

!bt
\[
\frac{\partial (\bm{b}^T\bm{a})}{\partial \bm{a}} = \bm{b},
\]
!et
and
!bt
\[
\frac{\partial (\bm{a}^T\bm{A}\bm{a})}{\partial \bm{a}} = \bm{a}^T(\bm{A}+\bm{A}^T),
\]
!et
and
!bt
\[
\frac{\partial \left(\bm{x}-\bm{A}\bm{s}\right)^T\left(\bm{x}-\bm{A}\bm{s}\right)}{\partial \bm{s}} = -2\left(\bm{x}-\bm{A}\bm{s}\right)^T\bm{A},
\]
!et
These and other relations are discussed in the exercises following this chapter (see the end of the chapter).
The latter equation is similar to the equation for the mean-squared error function we have been discussing. 
We can then compute the second derivative of the cost function, which in our case is the second derivative
of the means squared error. This leads to

!bt
\[
\frac{\partial^2 C(\bm{\theta})}{\partial \bm{\theta}^T\partial \bm{\theta}} =\frac{2}{n}\bm{X}^T\bm{X}.
\]
!et
This quantity defines the so- called the Hessian matrix.

The Hessian matrix plays an important role and is defined for the mean squared error  as

!bt
\[
\bm{H}=\bm{X}^T\bm{X}.
\]
!et


The Hessian matrix for ordinary least squares is also proportional to
the covariance matrix. As we will see in the chapter on Ridge and Lasso regression, This means that we can use the Singular Value Decomposition of a matrix  to find
the eigenvalues of the covariance matrix and the Hessian matrix in
terms of the singular values.


The residuals $\bm{\epsilon}$ are in turn given by
!bt
\[
\bm{\epsilon} = \bm{y}-\bm{\tilde{y}} = \bm{y}-\bm{X}\bm{\theta},
\]
!et
and with 
!bt
\[
\bm{X}^T\left( \bm{y}-\bm{X}\bm{\theta}\right)= 0, 
\]
!et
we have
!bt
\[
\bm{X}^T\bm{\epsilon}=\bm{X}^T\left( \bm{y}-\bm{X}\bm{\theta}\right)= 0, 
\]
!et
meaning that the solution for $\bm{\theta}$ 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 $\bm{\theta}$. After having defined the matrix $\bm{X}$ we simply need to 
write
!bc pycod 
# matrix inversion to find theta
theta = np.linalg.inv(X.T.dot(X)).dot(X.T).dot(Energies)
# and then make the prediction
ytilde = X @ theta
!ec
Alternatively, you can use the least squares functionality in _Numpy_ as
!bc pycod 
fit = np.linalg.lstsq(X, Energies, rcond =None)[0]
ytildenp = np.dot(fit,X.T)
!ec

And finally we plot our fit with and compare with data
!bc pycod
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()
!ec

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
!bc pycod 
def R2(y_data, y_model):
    return 1 - np.sum((y_data - y_model) ** 2) / np.sum((y_data - np.mean(y_data)) ** 2)
!ec
and we would be using it as 
!bc pycod 
print(R2(Energies,ytilde))
!ec

We can easily add our _MSE_ score as
!bc pycod
def MSE(y_data,y_model):
    n = np.size(y_model)
    return np.sum((y_data-y_model)**2)/n

print(MSE(Energies,ytilde))
!ec
and finally the relative error as
!bc pycod
def RelativeError(y_data,y_model):
    return abs((y_data-y_model)/y_data)
print(RelativeError(Energies, ytilde))
!ec



=== The $\chi^2$ function  ===

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

!bt
\[
\chi^2(\bm{\theta})=\frac{1}{n}\sum_{i=0}^{n-1}\frac{\left(y_i-\tilde{y}_i\right)^2}{\sigma_i^2}=\frac{1}{n}\left\{\left(\bm{y}-\bm{\tilde{y}}\right)^T\frac{1}{\bm{\Sigma^2}}\left(\bm{y}-\bm{\tilde{y}}\right)\right\},
\]
!et
where the matrix $\bm{\Sigma}$ is a diagonal matrix with $\sigma_i$ as matrix elements. 


In order to find the parameters $\theta_i$ we will then minimize the spread of $\chi^2(\bm{\theta})$ by requiring
!bt
\[
\frac{\partial \chi^2(\bm{\theta})}{\partial \theta_j} = \frac{\partial }{\partial \theta_j}\left[ \frac{1}{n}\sum_{i=0}^{n-1}\left(\frac{y_i-\theta_0x_{i,0}-\theta_1x_{i,1}-\theta_2x_{i,2}-\dots-\theta_{n-1}x_{i,n-1}}{\sigma_i}\right)^2\right]=0, 
\]
!et
which results in
!bt
\[
\frac{\partial \chi^2(\bm{\theta})}{\partial \theta_j} = -\frac{2}{n}\left[ \sum_{i=0}^{n-1}\frac{x_{ij}}{\sigma_i}\left(\frac{y_i-\theta_0x_{i,0}-\theta_1x_{i,1}-\theta_2x_{i,2}-\dots-\theta_{n-1}x_{i,n-1}}{\sigma_i}\right)\right]=0, 
\]
!et
or in a matrix-vector form as
!bt
\[
\frac{\partial \chi^2(\bm{\theta})}{\partial \bm{\theta}} = 0 = \bm{A}^T\left( \bm{b}-\bm{A}\bm{\theta}\right).  
\]
!et
where we have defined the matrix $\bm{A} =\bm{X}/\bm{\Sigma}$ with matrix elements $a_{ij} = x_{ij}/\sigma_i$ and the vector $\bm{b}$ with elements $b_i = y_i/\sigma_i$.   

We can rewrite
!bt
\[
\frac{\partial \chi^2(\bm{\theta})}{\partial \bm{\theta}} = 0 = \bm{A}^T\left( \bm{b}-\bm{A}\bm{\theta}\right),  
\]
!et
as
!bt
\[
\bm{A}^T\bm{b} = \bm{A}^T\bm{A}\bm{\theta},  
\]
!et
and if the matrix $\bm{A}^T\bm{A}$ is invertible we have the solution
!bt
\[
\bm{\theta} =\left(\bm{A}^T\bm{A}\right)^{-1}\bm{A}^T\bm{b}.
\]
!et


If we then introduce the matrix
!bt
\[
\bm{H} =  \left(\bm{A}^T\bm{A}\right)^{-1},
\]
!et
we have then the following expression for the parameters $\theta_j$ (the matrix elements of $\bm{H}$ are $h_{ij}$)
!bt
\[
\theta_j = \sum_{k=0}^{p-1}h_{jk}\sum_{i=0}^{n-1}\frac{y_i}{\sigma_i}\frac{x_{ik}}{\sigma_i} = \sum_{k=0}^{p-1}h_{jk}\sum_{i=0}^{n-1}b_ia_{ik}
\]
!et
We state without proof the expression for the uncertainty  in the parameters $\theta_j$ as (we leave this as an exercise)
!bt
\[
\sigma^2(\theta_j) = \sum_{i=0}^{n-1}\sigma_i^2\left( \frac{\partial \theta_j}{\partial y_i}\right)^2, 
\]
!et
resulting in 
!bt
\[
\sigma^2(\theta_j) = \left(\sum_{k=0}^{p-1}h_{jk}\sum_{i=0}^{n-1}a_{ik}\right)\left(\sum_{l=0}^{p-1}h_{jl}\sum_{m=0}^{n-1}a_{ml}\right) = h_{jj}!
\]
!et

The first step here is to approximate the function $y$ with a first-order polynomial, that is we write
!bt
\[
y=y(x) \rightarrow y(x_i) \approx \theta_0+\theta_1 x_i.
\]
!et
By computing the derivatives of $\chi^2$ with respect to $\theta_0$ and $\theta_1$ show that these are given by
!bt
\[
\frac{\partial \chi^2(\bm{\theta})}{\partial \theta_0} = -2\left[ \frac{1}{n}\sum_{i=0}^{n-1}\left(\frac{y_i-\theta_0-\theta_1x_{i}}{\sigma_i^2}\right)\right]=0,
\]
!et
and
!bt
\[
\frac{\partial \chi^2(\bm{\theta})}{\partial \theta_1} = -\frac{2}{n}\left[ \sum_{i=0}^{n-1}x_i\left(\frac{y_i-\theta_0-\theta_1x_{i}}{\sigma_i^2}\right)\right]=0.
\]
!et

For a linear fit (a first-order polynomial) we don't need to invert a matrix!!  
Defining
!bt
\[
\gamma =  \sum_{i=0}^{n-1}\frac{1}{\sigma_i^2},
\]
!et

!bt
\[
\gamma_x =  \sum_{i=0}^{n-1}\frac{x_{i}}{\sigma_i^2},
\]
!et

!bt
\[
\gamma_y = \sum_{i=0}^{n-1}\left(\frac{y_i}{\sigma_i^2}\right),
\]
!et

!bt
\[
\gamma_{xx} =  \sum_{i=0}^{n-1}\frac{x_ix_{i}}{\sigma_i^2},
\]
!et

!bt
\[
\gamma_{xy} = \sum_{i=0}^{n-1}\frac{y_ix_{i}}{\sigma_i^2},
\]
!et

we obtain

!bt
\[
\theta_0 = \frac{\gamma_{xx}\gamma_y-\gamma_x\gamma_y}{\gamma\gamma_{xx}-\gamma_x^2},
\]
!et

!bt
\[
\theta_1 = \frac{\gamma_{xy}\gamma-\gamma_x\gamma_y}{\gamma\gamma_{xx}-\gamma_x^2}.
\]
!et

This approach (different linear and non-linear regression) suffers
often from both being underdetermined and overdetermined in the
unknown coefficients $\theta_i$.  A better approach is to use the
Singular Value Decomposition (SVD) method discussed below. Or using
Lasso and Ridge regression. See below.


=== Fitting an Equation of State for Dense Nuclear Matter ===

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":"https://www.sciencedirect.com/science/article/pii/S0370157399001106". 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.



!bc pycod 
# 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_)


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.legend()
save_fig("EoSfitting")
plt.show()
!ec

The above simple polynomial in density $\rho$ gives an excellent fit
to the data. 



===== Splitting our Data in Training and Test data =====


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.

Let us study some examples. The first code here takes a simple
one-dimensional second-order polynomial and we fit it to a
second-order polynomial. Depending on the strength of the added noise,
the various measures like the $R2$ score or the mean-squared error,
the fit becomes better or worse.

!bc pycod
import os
import numpy as np
import pandas as pd
import matplotlib.pyplot as plt
from sklearn.model_selection import train_test_split


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 theta
theta = np.linalg.inv(X_train.T @ X_train) @ X_train.T @ y_train
print(theta)
# and then make the prediction
ytilde = X_train @ theta
print("Training R2")
print(R2(y_train,ytilde))
print("Training MSE")
print(MSE(y_train,ytilde))
ypredict = X_test @ theta
print("Test R2")
print(R2(y_test,ypredict))
print("Test MSE")
print(MSE(y_test,ypredict))
!ec

Alternatively, you could write your own test-train splitting function as shown here.

!bc pycod
# equivalently in numpy
def train_test_split_numpy(inputs, labels, train_size, test_size):
    n_inputs = len(inputs)
    inputs_shuffled = inputs.copy()
    labels_shuffled = labels.copy()

    np.random.shuffle(inputs_shuffled)
    np.random.shuffle(labels_shuffled)

    train_end = int(n_inputs*train_size)
    X_train, X_test = inputs_shuffled[:train_end], inputs_shuffled[train_end:]
    Y_train, Y_test = labels_shuffled[:train_end], labels_shuffled[train_end:]

    return X_train, X_test, Y_train, Y_test
!ec

But since _scikit-learn_ has its own function for doing this and since
it interfaces easily with _tensorflow_ and other libraries, we
normally recommend using the latter functionality.


As another example, we apply the training and testing split to 
to the above equation of state fitting example
but now splitting the data into a training set and a test set.

!bc pycod 
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 theta
theta = np.linalg.inv(X_train.T.dot(X_train)).dot(X_train.T).dot(y_train)
# and then make the prediction
ytilde = X_train @ theta
print("Training R2")
print(R2(y_train,ytilde))
print("Training MSE")
print(MSE(y_train,ytilde))
ypredict = X_test @ theta
print("Test R2")
print(R2(y_test,ypredict))
print("Test MSE")
print(MSE(y_test,ypredict))
!ec



===== Reducing the number of degrees of freedom, overarching view  =====

Many Machine Learning problems involve thousands or even millions of
features for each training instance. Not only does this make training
extremely slow, it can also make it much harder to find a good
solution, as we will see. This problem is often referred to as the
curse of dimensionality.  Fortunately, in real-world problems, it is
often possible to reduce the number of features considerably, turning
an intractable problem into a tractable one.

Later  we will discuss some of the most popular dimensionality reduction
techniques: the principal component analysis (PCA), Kernel PCA, and
Locally Linear Embedding (LLE).  


Principal component analysis and its various variants deal with the
problem of fitting a low-dimensional "affine
subspace":"https://en.wikipedia.org/wiki/Affine_space" to a set of of
data points in a high-dimensional space. With its family of methods it
is one of the most used tools in data modeling, compression and
visualization.



Before we proceed however, we will discuss how to preprocess our
data. Till now and in connection with our previous examples we have
not met so many cases where we are too sensitive to the scaling of 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.

For data sets gathered for real world applications, it is rather normal that
different features have very different units and
numerical scales. For example, a data set detailing health habits may include
features such as _age_ in the range $0-80$, and _caloric intake_ of order $2000$.
Many machine learning methods sensitive to the scales of the features and may perform poorly if they
are very different scales. Therefore, it is typical to scale
the features in a way to avoid such outlier values.


_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.


Many features are often scaled using standardization to improve
performance. In _Scikit-Learn_ this is given by the _StandardScaler_
function as discussed above. It is easy however to write your own.
Mathematically, this involves subtracting the mean and divide by the
standard deviation over the data set, for each feature:

!bt
\[
    x_j^{(i)} \rightarrow \frac{x_j^{(i)} - \overline{x}_j}{\sigma(x_j)},
\]
!et

where $\overline{x}_j$ and $\sigma(x_j)$ are the mean and standard
deviation, respectively, of the feature $x_j$.  This ensures that each
feature has zero mean and unit standard deviation.  For data sets
where we do not have the standard deviation or don't wish to calculate
it, it is then common to simply set it to one.



Let us consider the following vanilla example where we use both
_Scikit-Learn_ and write our own function as well.  We produce a
simple test design matrix with random numbers. Each column could then
represent a specific feature whose mean value is subracted.


!bc pycod
import sklearn.linear_model as skl
from sklearn.metrics import mean_squared_error
from sklearn.model_selection import  train_test_split
from sklearn.preprocessing import MinMaxScaler, StandardScaler, Normalizer
import numpy as np
import pandas as pd
from IPython.display import display
np.random.seed(100)
# setting up a 10 x 5 matrix
rows = 10
cols = 5
X = np.random.randn(rows,cols)
XPandas = pd.DataFrame(X)
display(XPandas)
print(XPandas.mean())
print(XPandas.std())
XPandas = (XPandas -XPandas.mean())
display(XPandas)
#  This option does not include the standard deviation
scaler = StandardScaler(with_std=False)
scaler.fit(X)
Xscaled = scaler.transform(X)
display(XPandas-Xscaled)
!ec

Small exercise: perform the standard scaling by including the standard deviation and compare with what Scikit-Learn gives.



Another commonly used scaling method is min-max scaling. This is very
useful for when we want the features to lie in a certain interval. To
scale the feature $x_j$ to the interval $[a, b]$, we can apply the
transformation

!bt
\[
x_j^{(i)} \rightarrow (b-a)\frac{x_j^{(i)} - \min(x_j)}{\max(x_j) - \min(x_j)} - a
\]
!et
where $\min(x_j)$ and $\max(x_j)$ return the minimum and maximum value of $x_j$ over the data set, respectively.




===== Testing the Means Squared Error as function of Complexity =====


Before we proceed with a more detailed analysis of the so-called
Bias-Variance tradeoff, we present here an example of the relation
between model complexity and the mean squared error for the triaining
data and the test data.

The results here tell us clearly that for the data not included in the
training, there is an optimal model as function of the complexity of
ourmodel (here in terms of the polynomial degree of the model).

The results here will vary as function of model complexity and the amount od data used for training. 


Our data is defined by $x\in [-3,3]$ with a total of for example $100$ data points.

!bc pycod
import matplotlib.pyplot as plt
import numpy as np
from sklearn.linear_model import LinearRegression, Ridge, Lasso
from sklearn.preprocessing import PolynomialFeatures
from sklearn.model_selection import train_test_split
from sklearn.pipeline import make_pipeline


np.random.seed(2018)
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)
TestError = np.zeros(maxdegree)
TrainError = np.zeros(maxdegree)
polydegree = np.zeros(maxdegree)
x_train, x_test, y_train, y_test = train_test_split(x, y, test_size=0.2)


for degree in range(maxdegree):
    model = make_pipeline(PolynomialFeatures(degree=degree), LinearRegression(fit_intercept=False))
    clf = model.fit(x_train,y_train)
    y_fit = clf.predict(x_train)
    y_pred = clf.predict(x_test) 
    polydegree[degree] = degree
    TestError[degree] = np.mean( np.mean((y_test - y_pred)**2) )
    TrainError[degree] = np.mean( np.mean((y_train - y_fit)**2) )

plt.plot(polydegree, TestError, label='Test Error')
plt.plot(polydegree, TrainError, label='Train Error')
plt.legend()
plt.show()
!ec






===== Exercises =====


===== Exercise: Setting up various Python environments  =====

The first exercise here is of a mere technical art. We want you to have 
* git as a version control software and to establish a user account on a provider like GitHub. Other providers like GitLab etc are equally fine. You can also use the University of Oslo "GitHub facilities":"https://www.uio.no/tjenester/it/maskin/filer/versjonskontroll/github.html". 
* Install various Python packages

We will make extensive use of Python as programming language and its
myriad of available libraries.  You will find
IPython/Jupyter notebooks invaluable in your work.  You can run _R_
codes in the Jupyter/IPython notebooks, with the immediate benefit of
visualizing your data. You can also use compiled languages like C++,
Rust, Fortran etc if you prefer. The focus in these lectures will be
on Python.

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 

o pip install numpy scipy matplotlib ipython scikit-learn sympy pandas pillow 

For _Tensorflow_, we recommend following the instructions in the text of 
"Aurelien Geron, Hands‑On Machine Learning with Scikit‑Learn and TensorFlow, O'Reilly":"http://shop.oreilly.com/product/0636920052289.do"

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 

o brew install python3

For Linux users, with its variety of distributions like for example the widely popular Ubuntu distribution,
you can use _pip_ as well and simply install Python as 

o sudo apt-get install python3  (or python for Python2.7)

If you don't want to perform these operations separately and venture
into the hassle of exploring how to set up dependencies and paths, we
recommend two widely used distrubutions which set up all relevant
dependencies for Python, namely 

* "Anaconda":"https://docs.anaconda.com/", 

which is an open source
distribution of the Python and R programming languages for large-scale
data processing, predictive analytics, and scientific computing, that
aims to simplify package management and deployment. Package versions
are managed by the package management system _conda_. 

* "Enthought canopy":"https://www.enthought.com/product/canopy/" 

is a Python
distribution for scientific and analytic computing distribution and
analysis environment, available for free and under a commercial
license.

We recommend using _Anaconda_ if you are not too familiar with setting paths in a terminal environment.




===== Exercise: making your own data and exploring scikit-learn =====


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).
!bc pycod
x = np.random.rand(100,1)
y = 2.0+5*x*x+0.1*np.random.randn(100,1)
!ec

o Write your own code (following the examples under the "regression notes":"https://compphysics.github.io/MachineLearning/doc/LectureNotes/_build/html/chapter1.html") for computing the parametrization of the data set fitting a second-order polynomial. 
o Use thereafter _scikit-learn_ (see again the examples in the regression slides) and compare with your own code.   When compairing with _scikit_learn_, make sure you set the option for the intercept to _FALSE_, see URL:"https://scikit-learn.org/stable/modules/generated/sklearn.linear_model.LinearRegression.html". This feature will be explained in more detail during the lectures of week 35 and week 36. You can find more in URL:"https://compphysics.github.io/MachineLearning/doc/LectureNotes/_build/html/chapter3.html#more-on-rescaling-data".
o Using scikit-learn, compute also the mean square error, a risk metric corresponding to the expected value of the squared (quadratic) error defined as
!bt 
\[ MSE(\bm{y},\bm{\tilde{y}}) = \frac{1}{n}
\sum_{i=0}^{n-1}(y_i-\tilde{y}_i)^2, 
\] 
!et 
and the $R^2$ score function.
If $\tilde{\bm{y}}_i$ 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
!bt
\[
R^2(\bm{y}, \tilde{\bm{y}}) = 1 - \frac{\sum_{i=0}^{n - 1} (y_i - \tilde{y}_i)^2}{\sum_{i=0}^{n - 1} (y_i - \bar{y})^2},
\]
!et
where we have defined the mean value  of $\bm{y}$ as
!bt
\[
\bar{y} =  \frac{1}{n} \sum_{i=0}^{n - 1} y_i.
\]
!et
You can use the functionality included in scikit-learn. If you feel for it, you can use your own program and define functions which compute the above two functions. 
Discuss the meaning of these results. Try also to vary the coefficient in front of the added stochastic noise term and discuss the quality of the fits.

!bsol
The code here is an example of where we define our own design matrix and fit parameters $\theta$.
!bc pycod
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 theta
theta = np.linalg.inv(X_train.T @ X_train) @ X_train.T @ y_train
print(theta)
# and then make the prediction
ytilde = X_train @ theta
print("Training R2")
print(R2(y_train,ytilde))
print("Training MSE")
print(MSE(y_train,ytilde))
ypredict = X_test @ theta
print("Test R2")
print(R2(y_test,ypredict))
print("Test MSE")
print(MSE(y_test,ypredict))
!ec
!esol



===== Exercise: Normalizing our data  =====


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 $\bm{X}$ and data $\bm{y}$ (remember to import _scikit-learn_)
!bc pycod
# split in training and test data
X_train, X_test, y_train, y_test = train_test_split(X,y,test_size=0.2)
!ec
Then we can use the standard scaler to scale our data as
!bc pycod
scaler = StandardScaler()
scaler.fit(X_train)
X_train_scaled = scaler.transform(X_train)
X_test_scaled = scaler.transform(X_test)
!ec


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":"https://github.com/CompPhysics/MLErasmus/blob/master/doc/Textbooks/elementsstat.pdf".



Our data is defined by $x\in [-3,3]$ with a total of for example $100$ data points.
!bc pycod
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)
!ec
where $y$ is the function we want to fit with a given polynomial.
!bsubex
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. 
!esubex

!bsubex
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.
!esubex

!bsubex
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)?

!esubex


!bsol
We present here the solution for the last exercise. All elements here can be used to solve exercises a) and b) as well.
Note that in this example we have used the polynomial fitting functions of _scikit-learn_.
!bc pycod
import matplotlib.pyplot as plt
import numpy as np
from sklearn.linear_model import LinearRegression, Ridge, Lasso
from sklearn.preprocessing import PolynomialFeatures
from sklearn.model_selection import train_test_split
from sklearn.pipeline import make_pipeline


np.random.seed(2018)
n = 30
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)
TestError = np.zeros(maxdegree)
TrainError = np.zeros(maxdegree)
polydegree = np.zeros(maxdegree)
x_train, x_test, y_train, y_test = train_test_split(x, y, test_size=0.2)


for degree in range(maxdegree):
    model = make_pipeline(PolynomialFeatures(degree=degree), LinearRegression(fit_intercept=False))
    clf = model.fit(x_train,y_train)
    y_fit = clf.predict(x_train)
    y_pred = clf.predict(x_test) 
    polydegree[degree] = degree
    TestError[degree] = np.mean( np.mean((y_test - y_pred)**2) )
    TrainError[degree] = np.mean( np.mean((y_train - y_fit)**2) )

plt.plot(polydegree, TestError, label='Test Error')
plt.plot(polydegree, TrainError, label='Train Error')
plt.legend()
plt.show()
!ec
!esol




===== Exercise: Adding Ridge Regression  =====


This exercise is a continuation of exercise 2. We will use the same function to
generate our data set, still staying with a simple function $y(x)$
which we want to fit using linear regression, but now extending the
analysis to include the Ridge regression method.

We will thus again 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).
!bc pycod
x = np.random.rand(100)
y = 2.0+5*x*x+0.1*np.random.randn(100)
!ec


Write your own code for the Ridge method (see chapter 3.4 of Hastie *et al.*, equations (3.43) and (3.44)) and compute the parametrization for different values of $\lambda$. Compare and analyze your results with those from exercise 3. Study the dependence on $\lambda$ while also varying the strength of the noise in your expression for $y(x)$. 


Repeat the above but using the functionality of
_Scikit-Learn_. Compare your code with the results from
_Scikit-Learn_. Remember to run with the same random numbers for
generating $x$ and $y$.  Observe also that when you compare with _Scikit-Learn_, you need to pay attention to how the intercept is dealt with.



Finally, using _Scikit-Learn_ or your own code, compute also the mean square error, a risk metric corresponding to the expected value of the squared (quadratic) error defined as
!bt 
\[ MSE(\hat{y},\hat{\tilde{y}}) = \frac{1}{n}
\sum_{i=0}^{n-1}(y_i-\tilde{y}_i)^2, 
\] 
!et 
and the $R^2$ score function.
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
!bt
\[
R^2(\hat{y}, \tilde{\hat{y}}) = 1 - \frac{\sum_{i=0}^{n - 1} (y_i - \tilde{y}_i)^2}{\sum_{i=0}^{n - 1} (y_i - \bar{y})^2},
\]
!et
where we have defined the mean value  of $\hat{y}$ as
!bt
\[
\bar{y} =  \frac{1}{n} \sum_{i=0}^{n - 1} y_i.
\]
!et
Discuss these quantities as functions of the variable $\lambda$ in Ridge regression.


!bsol
The code here allows you to perform your own Ridge calculation and
perform calculations for various values of the regularization
parameter $\lambda$. This program can easily be extended upon.

!bc pycod
import os
import numpy as np
import pandas as pd
import matplotlib.pyplot as plt
from sklearn.model_selection import train_test_split
from sklearn.preprocessing import StandardScaler
from sklearn import linear_model

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


# A seed just to ensure that the random numbers are the same for every run.
# Useful for eventual debugging.
np.random.seed(3155)

x = np.random.rand(100)
y = 2.0+5*x*x+0.1*np.random.randn(100)

# number of features p (here degree of polynomial
p = 3
#  The design matrix now as function of a given polynomial
X = np.zeros((len(x),p))
X[:,0] = 1.0
X[:,1] = x
X[:,2] = x*x
# 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 theta
OLStheta = np.linalg.inv(X_train.T @ X_train) @ X_train.T @ y_train
print(OLStheta)
# and then make the prediction
ytildeOLS = X_train @ OLStheta
print("Training R2 for OLS")
print(R2(y_train,ytildeOLS))
print("Training MSE for OLS")
print(MSE(y_train,ytildeOLS))
ypredictOLS = X_test @ OLStheta
print("Test R2 for OLS")
print(R2(y_test,ypredictOLS))
print("Test MSE OLS")
print(MSE(y_test,ypredictOLS))


# Repeat now for Ridge regression and various values of the regularization parameter
I = np.eye(p,p)
# Decide which values of lambda to use
nlambdas = 20
OwnMSEPredict = np.zeros(nlambdas)
OwnMSETrain = np.zeros(nlambdas)
MSERidgePredict =  np.zeros(nlambdas)
lambdas = np.logspace(-4, 1, nlambdas)
for i in range(nlambdas):
    lmb = lambdas[i]
    OwnRidgetheta = np.linalg.inv(X_train.T @ X_train+lmb*I) @ X_train.T @ y_train
    # and then make the prediction
    OwnytildeRidge = X_train @ OwnRidgetheta
    OwnypredictRidge = X_test @ OwnRidgetheta
    OwnMSEPredict[i] = MSE(y_test,OwnypredictRidge)
    OwnMSETrain[i] = MSE(y_train,OwnytildeRidge)
    # Make the fit using Ridge from Sklearn
    RegRidge = linear_model.Ridge(lmb,fit_intercept=False)
    RegRidge.fit(X_train,y_train)
    # and then make the prediction
    ypredictRidge = RegRidge.predict(X_test)
    # Compute the MSE and print it
    MSERidgePredict[i] = MSE(y_test,ypredictRidge)

# Now plot the results
plt.figure()
plt.plot(np.log10(lambdas), OwnMSETrain, label = 'MSE Ridge train, Own code')
plt.plot(np.log10(lambdas), OwnMSEPredict, 'r--', label = 'MSE Ridge Test, Own code')
plt.plot(np.log10(lambdas), MSERidgePredict, 'g--', label = 'MSE Ridge Test, Sklearn code')
plt.xlabel('log10(lambda)')
plt.ylabel('MSE')
plt.legend()
plt.show()
!ec
!esol


===== Exercise: Analytical exercises  =====

In this exercise we derive the expressions for various derivatives of
products of vectors and matrices. Such derivatives are central to the
optimization of various cost functions. Although we will often use
automatic differentiation in actual calculations, to be able to have
analytical expressions is extremely helpful in case we have simpler
derivatives as well as when we analyze various properties (like second
derivatives) of the chosen cost functions.  Vectors are always written
as boldfaced lower case letters and matrices as upper case boldfaced
letters.

Show that
!bt
\[
\frac{\partial (\bm{b}^T\bm{a})}{\partial \bm{a}} = \bm{b},
\]
!et
and
!bt
\[
\frac{\partial (\bm{a}^T\bm{A}\bm{a})}{\partial \bm{a}} = \bm{a}^T(\bm{A}+\bm{A}^T),
\]
!et
and
!bt
\[
\frac{\partial \left(\bm{x}-\bm{A}\bm{s}\right)^T\left(\bm{x}-\bm{A}\bm{s}\right)}{\partial \bm{s}} = -2\left(\bm{x}-\bm{A}\bm{s}\right)^T\bm{A},
\]
!et
and finally find the second derivative of this function with respect to the vector $\bm{s}$.

!bsol
In these exercises it is always useful to write out with summation indices the various quantities.
As an example, consider the function

!bt
\[
f(\bm{x}) =\bm{A}\bm{x},
\]
!et
which reads for a specific component $f_i$ (we define the matrix $\bm{A}$ to have dimension $n\times n$ and the vector $\bm{x}$ to have length $n$)

!bt
\[
f_i =\sum_{j=0}^{n-1}a_{ij}x_j, 
\]
!et
which leads to
!bt
\[
\frac{\partial f_i}{\partial x_j}= a_{ij},
\]
!et
and written out in terms of the vector $\bm{x}$ we have
!bt
\[
\frac{\partial f(\bm{x})}{\partial \bm{x}}= \bm{A}.
\]
!et

For the first derivative
!bt
\[
\frac{\partial (\bm{b}^T\bm{a})}{\partial \bm{a}} = \bm{b},
\]
!et
we can write out the inner product as (assuming all elements are real)
!bt
\[
\bm{b}^T\bm{a}=\sum_i b_ia_i, 
\]
!et
taking the derivative
!bt
\[
\frac{\partial \left( \sum_i b_ia_i\right)}{\partial a_k}= b_k, 
\]
!et
leading to 
!bt
\[
\frac{\partial \bm{b}^T\bm{a}}{\partial \bm{a}}= \begin{bmatrix} b_0 \\ b_1 \\ b_2 \\ \dots \\ \dots \\ b_{n-1}\end{bmatrix} = \bm{b}. 
\]
!et

For the second exercise we have 
!bt
\[
\frac{\partial (\bm{a}^T\bm{A}\bm{a})}{\partial \bm{a}}.
\]
!et
Defining a vector $\bm{f}=\bm{A}\bm{a}$ with components $f_i=\sum_ja_{ij}a_i$  we have
!bt
\[
\frac{\partial (\bm{a}^T\bm{f})}{\partial \bm{a}}=\bm{a}^T\bm{A}+\bm{f}^T=\bm{a}^T\left(\bm{A}+\bm{A}^T\right),
\]
!et
since $f$ depends on $a$ and we have used the chain rule for derivatives on the derivative of $f$ with respect to $a$. 


!esol





