update
This commit is contained in:
Morten Hjorth-Jensen
File diff suppressed because it is too large
Load Diff
@@ -11,7 +11,8 @@ DATE: September 4-8, 2023
|
||||
* Recommended Reading: Hastie et al chapter 3, see URL:"https://link.springer.com/book/10.1007/978-0-387-84858-7"
|
||||
* Presentation and discussion of first project
|
||||
* Material for the lecture on Thursday September 7
|
||||
* Linear Regression and links with Statistics, Resampling methods
|
||||
* Technicalities related to scaling and other issues with data handling
|
||||
* Linear Regression and links with Statistics
|
||||
* Recommended Reading: Goodfellow et al chapter 3 on probability theory, see URL:""
|
||||
* See also Murphy, sections 2.4 (Gaussian distributions) and 3.2 (Bayesian Statistics, basis)
|
||||
|
||||
@@ -923,6 +924,499 @@ plt.show()
|
||||
===== Material for lecture Thursday September 7 =====
|
||||
|
||||
|
||||
===== Important technicalities: More on Rescaling data =====
|
||||
|
||||
|
||||
When you are comparing your own code with for example _Scikit-Learn_'s
|
||||
library, there are some technicalities to keep in mind. The examples
|
||||
here demonstrate some of these aspects with potential pitfalls.
|
||||
|
||||
The discussion here focuses on the role of the intercept, how we can
|
||||
set up the design matrix, what scaling we should use and other topics
|
||||
which tend confuse us.
|
||||
|
||||
The intercept can be interpreted as the expected value of our
|
||||
target/output variables when all other predictors are set to zero.
|
||||
Thus, if we cannot assume that the expected outputs/targets are zero
|
||||
when all predictors are zero (the columns in the design matrix), it
|
||||
may be a bad idea to implement a model which penalizes the intercept.
|
||||
Furthermore, in for example Ridge and Lasso regression, the default solutions
|
||||
from the library _Scikit-Learn_ (when not shrinking $\beta_0$) for the unknown parameters
|
||||
$\bm{\beta}$, are derived under the assumption that both $\bm{y}$ and
|
||||
$\bm{X}$ are zero centered, that is we subtract the mean values.
|
||||
|
||||
|
||||
If our predictors represent different scales, then it is important to
|
||||
standardize the design matrix $\bm{X}$ by subtracting the mean of each
|
||||
column from the corresponding column and dividing the column with its
|
||||
standard deviation. Most machine learning libraries do this as a default. This means that if you compare your code with the results from a given library,
|
||||
the results may differ.
|
||||
|
||||
The
|
||||
"Standardscaler":"https://scikit-learn.org/stable/modules/generated/sklearn.preprocessing.StandardScaler.html"
|
||||
function in _Scikit-Learn_ does this for us. For the data sets we
|
||||
have been studying in our various examples, the data are in many cases
|
||||
already scaled and there is no need to scale them. You as a user of different machine learning algorithms, should always perform a
|
||||
survey of your data, with a critical assessment of them in case you need to scale the data.
|
||||
|
||||
If you need to scale the data, not doing so will give an *unfair*
|
||||
penalization of the parameters since their magnitude depends on the
|
||||
scale of their corresponding predictor.
|
||||
|
||||
Suppose as an example that you
|
||||
you have an input variable given by the heights of different persons.
|
||||
Human height might be measured in inches or meters or
|
||||
kilometers. If measured in kilometers, a standard linear regression
|
||||
model with this predictor would probably give a much bigger
|
||||
coefficient term, than if measured in millimeters.
|
||||
This can clearly lead to problems in evaluating the cost/loss functions.
|
||||
|
||||
|
||||
|
||||
Keep in mind that when you transform your data set before training a model, the same transformation needs to be done
|
||||
on your eventual new data set before making a prediction. If we translate this into a Python code, it would could be implemented as
|
||||
|
||||
!bc pycod
|
||||
"""
|
||||
#Model training, we compute the mean value of y and X
|
||||
y_train_mean = np.mean(y_train)
|
||||
X_train_mean = np.mean(X_train,axis=0)
|
||||
X_train = X_train - X_train_mean
|
||||
y_train = y_train - y_train_mean
|
||||
|
||||
# The we fit our model with the training data
|
||||
trained_model = some_model.fit(X_train,y_train)
|
||||
|
||||
|
||||
#Model prediction, we need also to transform our data set used for the prediction.
|
||||
X_test = X_test - X_train_mean #Use mean from training data
|
||||
y_pred = trained_model(X_test)
|
||||
y_pred = y_pred + y_train_mean
|
||||
"""
|
||||
!ec
|
||||
|
||||
|
||||
|
||||
Let us try to understand what this may imply mathematically when we
|
||||
subtract the mean values, also known as *zero centering*. For
|
||||
simplicity, we will focus on ordinary regression, as done in the above example.
|
||||
|
||||
The cost/loss function for regression is
|
||||
!bt
|
||||
\[
|
||||
C(\beta_0, \beta_1, ... , \beta_{p-1}) = \frac{1}{n}\sum_{i=0}^{n} \left(y_i - \beta_0 - \sum_{j=1}^{p-1} X_{ij}\beta_j\right)^2,.
|
||||
\]
|
||||
!et
|
||||
|
||||
Recall also that we use the squared value. This expression can lead to an
|
||||
increased penalty for higher differences between predicted and
|
||||
output/target values.
|
||||
|
||||
What we have done is to single out the $\beta_0$ term in the
|
||||
definition of the mean squared error (MSE). The design matrix $X$
|
||||
does in this case not contain any intercept column. When we take the
|
||||
derivative with respect to $\beta_0$, we want the derivative to obey
|
||||
|
||||
!bt
|
||||
\[
|
||||
\frac{\partial C}{\partial \beta_j} = 0,
|
||||
\]
|
||||
!et
|
||||
|
||||
for all $j$. For $\beta_0$ we have
|
||||
|
||||
!bt
|
||||
\[
|
||||
\frac{\partial C}{\partial \beta_0} = -\frac{2}{n}\sum_{i=0}^{n-1} \left(y_i - \beta_0 - \sum_{j=1}^{p-1} X_{ij} \beta_j\right).
|
||||
\]
|
||||
!et
|
||||
Multiplying away the constant $2/n$, we obtain
|
||||
!bt
|
||||
\[
|
||||
\sum_{i=0}^{n-1} \beta_0 = \sum_{i=0}^{n-1}y_i - \sum_{i=0}^{n-1} \sum_{j=1}^{p-1} X_{ij} \beta_j.
|
||||
\]
|
||||
!et
|
||||
|
||||
Let us specialize first to the case where we have only two parameters $\beta_0$ and $\beta_1$.
|
||||
Our result for $\beta_0$ simplifies then to
|
||||
!bt
|
||||
\[
|
||||
n\beta_0 = \sum_{i=0}^{n-1}y_i - \sum_{i=0}^{n-1} X_{i1} \beta_1.
|
||||
\]
|
||||
!et
|
||||
We obtain then
|
||||
!bt
|
||||
\[
|
||||
\beta_0 = \frac{1}{n}\sum_{i=0}^{n-1}y_i - \beta_1\frac{1}{n}\sum_{i=0}^{n-1} X_{i1}.
|
||||
\]
|
||||
!et
|
||||
If we define
|
||||
!bt
|
||||
\[
|
||||
\mu_{\bm{x}_1}=\frac{1}{n}\sum_{i=0}^{n-1} X_{i1},
|
||||
\]
|
||||
!et
|
||||
and the mean value of the outputs as
|
||||
!bt
|
||||
\[
|
||||
\mu_y=\frac{1}{n}\sum_{i=0}^{n-1}y_i,
|
||||
\]
|
||||
!et
|
||||
we have
|
||||
!bt
|
||||
\[
|
||||
\beta_0 = \mu_y - \beta_1\mu_{\bm{x}_1}.
|
||||
\]
|
||||
!et
|
||||
In the general case with more parameters than $\beta_0$ and $\beta_1$, we have
|
||||
!bt
|
||||
\[
|
||||
\beta_0 = \frac{1}{n}\sum_{i=0}^{n-1}y_i - \frac{1}{n}\sum_{i=0}^{n-1}\sum_{j=1}^{p-1} X_{ij}\beta_j.
|
||||
\]
|
||||
!et
|
||||
|
||||
We can rewrite the latter equation as
|
||||
!bt
|
||||
\[
|
||||
\beta_0 = \frac{1}{n}\sum_{i=0}^{n-1}y_i - \sum_{j=1}^{p-1} \mu_{\bm{x}_j}\beta_j,
|
||||
\]
|
||||
!et
|
||||
where we have defined
|
||||
!bt
|
||||
\[
|
||||
\mu_{\bm{x}_j}=\frac{1}{n}\sum_{i=0}^{n-1} X_{ij},
|
||||
\]
|
||||
!et
|
||||
the mean value for all elements of the column vector $\bm{x}_j$.
|
||||
|
||||
|
||||
|
||||
Replacing $y_i$ with $y_i - y_i - \overline{\bm{y}}$ and centering also our design matrix results in a cost function (in vector-matrix disguise)
|
||||
!bt
|
||||
\[
|
||||
C(\boldsymbol{\beta}) = (\boldsymbol{\tilde{y}} - \tilde{X}\boldsymbol{\beta})^T(\boldsymbol{\tilde{y}} - \tilde{X}\boldsymbol{\beta}).
|
||||
\]
|
||||
!et
|
||||
|
||||
|
||||
|
||||
If we minimize with respect to $\bm{\beta}$ we have then
|
||||
|
||||
!bt
|
||||
\[
|
||||
\hat{\bm{\beta}} = (\tilde{X}^T\tilde{X})^{-1}\tilde{X}^T\boldsymbol{\tilde{y}},
|
||||
\]
|
||||
!et
|
||||
|
||||
where $\boldsymbol{\tilde{y}} = \boldsymbol{y} - \overline{\bm{y}}$
|
||||
and $\tilde{X}_{ij} = X_{ij} - \frac{1}{n}\sum_{k=0}^{n-1}X_{kj}$.
|
||||
|
||||
For Ridge regression we need to add $\lambda \boldsymbol{\beta}^T\boldsymbol{\beta}$ to the cost function and get then
|
||||
!bt
|
||||
\[
|
||||
\hat{\bm{\beta}} = (\tilde{X}^T\tilde{X} + \lambda I)^{-1}\tilde{X}^T\boldsymbol{\tilde{y}}.
|
||||
\]
|
||||
!et
|
||||
|
||||
What does this mean? And why do we insist on all this? Let us look at some examples.
|
||||
|
||||
|
||||
This code shows a simple first-order fit to a data set using the above transformed data, where we consider the role of the intercept first, by either excluding it or including it (*code example thanks to Øyvind Sigmundson Schøyen*). Here our scaling of the data is done by subtracting the mean values only.
|
||||
Note also that we do not split the data into training and test.
|
||||
|
||||
!bc pycod
|
||||
import numpy as np
|
||||
import matplotlib.pyplot as plt
|
||||
|
||||
from sklearn.linear_model import LinearRegression
|
||||
|
||||
|
||||
np.random.seed(2021)
|
||||
|
||||
def MSE(y_data,y_model):
|
||||
n = np.size(y_model)
|
||||
return np.sum((y_data-y_model)**2)/n
|
||||
|
||||
|
||||
def fit_beta(X, y):
|
||||
return np.linalg.pinv(X.T @ X) @ X.T @ y
|
||||
|
||||
|
||||
true_beta = [2, 0.5, 3.7]
|
||||
|
||||
x = np.linspace(0, 1, 11)
|
||||
y = np.sum(
|
||||
np.asarray([x ** p * b for p, b in enumerate(true_beta)]), axis=0
|
||||
) + 0.1 * np.random.normal(size=len(x))
|
||||
|
||||
degree = 3
|
||||
X = np.zeros((len(x), degree))
|
||||
|
||||
# Include the intercept in the design matrix
|
||||
for p in range(degree):
|
||||
X[:, p] = x ** p
|
||||
|
||||
beta = fit_beta(X, y)
|
||||
|
||||
# Intercept is included in the design matrix
|
||||
skl = LinearRegression(fit_intercept=False).fit(X, y)
|
||||
|
||||
print(f"True beta: {true_beta}")
|
||||
print(f"Fitted beta: {beta}")
|
||||
print(f"Sklearn fitted beta: {skl.coef_}")
|
||||
ypredictOwn = X @ beta
|
||||
ypredictSKL = skl.predict(X)
|
||||
print(f"MSE with intercept column")
|
||||
print(MSE(y,ypredictOwn))
|
||||
print(f"MSE with intercept column from SKL")
|
||||
print(MSE(y,ypredictSKL))
|
||||
|
||||
|
||||
plt.figure()
|
||||
plt.scatter(x, y, label="Data")
|
||||
plt.plot(x, X @ beta, label="Fit")
|
||||
plt.plot(x, skl.predict(X), label="Sklearn (fit_intercept=False)")
|
||||
|
||||
|
||||
# Do not include the intercept in the design matrix
|
||||
X = np.zeros((len(x), degree - 1))
|
||||
|
||||
for p in range(degree - 1):
|
||||
X[:, p] = x ** (p + 1)
|
||||
|
||||
# Intercept is not included in the design matrix
|
||||
skl = LinearRegression(fit_intercept=True).fit(X, y)
|
||||
|
||||
# Use centered values for X and y when computing coefficients
|
||||
y_offset = np.average(y, axis=0)
|
||||
X_offset = np.average(X, axis=0)
|
||||
|
||||
beta = fit_beta(X - X_offset, y - y_offset)
|
||||
intercept = np.mean(y_offset - X_offset @ beta)
|
||||
|
||||
print(f"Manual intercept: {intercept}")
|
||||
print(f"Fitted beta (wiothout intercept): {beta}")
|
||||
print(f"Sklearn intercept: {skl.intercept_}")
|
||||
print(f"Sklearn fitted beta (without intercept): {skl.coef_}")
|
||||
ypredictOwn = X @ beta
|
||||
ypredictSKL = skl.predict(X)
|
||||
print(f"MSE with Manual intercept")
|
||||
print(MSE(y,ypredictOwn+intercept))
|
||||
print(f"MSE with Sklearn intercept")
|
||||
print(MSE(y,ypredictSKL))
|
||||
|
||||
plt.plot(x, X @ beta + intercept, "--", label="Fit (manual intercept)")
|
||||
plt.plot(x, skl.predict(X), "--", label="Sklearn (fit_intercept=True)")
|
||||
plt.grid()
|
||||
plt.legend()
|
||||
|
||||
plt.show()
|
||||
|
||||
!ec
|
||||
|
||||
The intercept is the value of our output/target variable
|
||||
when all our features are zero and our function crosses the $y$-axis (for a one-dimensional case).
|
||||
|
||||
Printing the MSE, we see first that both methods give the same MSE, as
|
||||
they should. However, when we move to for example Ridge regression,
|
||||
the way we treat the intercept may give a larger or smaller MSE,
|
||||
meaning that the MSE can be penalized by the value of the
|
||||
intercept. Not including the intercept in the fit, means that the
|
||||
regularization term does not include $\beta_0$. For different values
|
||||
of $\lambda$, this may lead to different MSE values.
|
||||
|
||||
To remind the reader, the regularization term, with the intercept in Ridge regression, is given by
|
||||
!bt
|
||||
\[
|
||||
\lambda \vert\vert \bm{\beta} \vert\vert_2^2 = \lambda \sum_{j=0}^{p-1}\beta_j^2,
|
||||
\]
|
||||
!et
|
||||
but when we take out the intercept, this equation becomes
|
||||
!bt
|
||||
\[
|
||||
\lambda \vert\vert \bm{\beta} \vert\vert_2^2 = \lambda \sum_{j=1}^{p-1}\beta_j^2.
|
||||
\]
|
||||
!et
|
||||
|
||||
For Lasso regression we have
|
||||
!bt
|
||||
\[
|
||||
\lambda \vert\vert \bm{\beta} \vert\vert_1 = \lambda \sum_{j=1}^{p-1}\vert\beta_j\vert.
|
||||
\]
|
||||
!et
|
||||
|
||||
It means that, when scaling the design matrix and the outputs/targets,
|
||||
by subtracting the mean values, we have an optimization problem which
|
||||
is not penalized by the intercept. The MSE value can then be smaller
|
||||
since it focuses only on the remaining quantities. If we however bring
|
||||
back the intercept, we will get a MSE which then contains the
|
||||
intercept.
|
||||
|
||||
|
||||
Armed with this wisdom, we attempt first to simply set the intercept equal to _False_ in our implementation of Ridge regression for our well-known vanilla data set.
|
||||
|
||||
!bc pycod
|
||||
import numpy as np
|
||||
import pandas as pd
|
||||
import matplotlib.pyplot as plt
|
||||
from sklearn.model_selection import train_test_split
|
||||
from sklearn import linear_model
|
||||
|
||||
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)
|
||||
|
||||
n = 100
|
||||
x = np.random.rand(n)
|
||||
y = np.exp(-x**2) + 1.5 * np.exp(-(x-2)**2)
|
||||
|
||||
Maxpolydegree = 20
|
||||
X = np.zeros((n,Maxpolydegree))
|
||||
#We include explicitely the intercept column
|
||||
for degree in range(Maxpolydegree):
|
||||
X[:,degree] = x**degree
|
||||
# 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)
|
||||
|
||||
p = Maxpolydegree
|
||||
I = np.eye(p,p)
|
||||
# Decide which values of lambda to use
|
||||
nlambdas = 6
|
||||
MSEOwnRidgePredict = np.zeros(nlambdas)
|
||||
MSERidgePredict = np.zeros(nlambdas)
|
||||
lambdas = np.logspace(-4, 2, nlambdas)
|
||||
for i in range(nlambdas):
|
||||
lmb = lambdas[i]
|
||||
OwnRidgeBeta = np.linalg.pinv(X_train.T @ X_train+lmb*I) @ X_train.T @ y_train
|
||||
# Note: we include the intercept column and no scaling
|
||||
RegRidge = linear_model.Ridge(lmb,fit_intercept=False)
|
||||
RegRidge.fit(X_train,y_train)
|
||||
# and then make the prediction
|
||||
ytildeOwnRidge = X_train @ OwnRidgeBeta
|
||||
ypredictOwnRidge = X_test @ OwnRidgeBeta
|
||||
ytildeRidge = RegRidge.predict(X_train)
|
||||
ypredictRidge = RegRidge.predict(X_test)
|
||||
MSEOwnRidgePredict[i] = MSE(y_test,ypredictOwnRidge)
|
||||
MSERidgePredict[i] = MSE(y_test,ypredictRidge)
|
||||
print("Beta values for own Ridge implementation")
|
||||
print(OwnRidgeBeta)
|
||||
print("Beta values for Scikit-Learn Ridge implementation")
|
||||
print(RegRidge.coef_)
|
||||
print("MSE values for own Ridge implementation")
|
||||
print(MSEOwnRidgePredict[i])
|
||||
print("MSE values for Scikit-Learn Ridge implementation")
|
||||
print(MSERidgePredict[i])
|
||||
|
||||
# Now plot the results
|
||||
plt.figure()
|
||||
plt.plot(np.log10(lambdas), MSEOwnRidgePredict, 'r', label = 'MSE own Ridge Test')
|
||||
plt.plot(np.log10(lambdas), MSERidgePredict, 'g', label = 'MSE Ridge Test')
|
||||
|
||||
plt.xlabel('log10(lambda)')
|
||||
plt.ylabel('MSE')
|
||||
plt.legend()
|
||||
plt.show()
|
||||
|
||||
!ec
|
||||
|
||||
The results here agree when we force _Scikit-Learn_'s Ridge function to include the first column in our design matrix.
|
||||
We see that the results agree very well. Here we have thus explicitely included the intercept column in the design matrix.
|
||||
What happens if we do not include the intercept in our fit?
|
||||
Let us see how we can change this code by zero centering.
|
||||
|
||||
!bc pycod
|
||||
import numpy as np
|
||||
import pandas as pd
|
||||
import matplotlib.pyplot as plt
|
||||
from sklearn.model_selection import train_test_split
|
||||
from sklearn import linear_model
|
||||
from sklearn.preprocessing import StandardScaler
|
||||
|
||||
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(315)
|
||||
|
||||
n = 100
|
||||
x = np.random.rand(n)
|
||||
y = np.exp(-x**2) + 1.5 * np.exp(-(x-2)**2)
|
||||
|
||||
Maxpolydegree = 20
|
||||
X = np.zeros((n,Maxpolydegree-1))
|
||||
|
||||
for degree in range(1,Maxpolydegree): #No intercept column
|
||||
X[:,degree-1] = x**(degree)
|
||||
|
||||
# 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)
|
||||
|
||||
#For our own implementation, we will need to deal with the intercept by centering the design matrix and the target variable
|
||||
X_train_mean = np.mean(X_train,axis=0)
|
||||
#Center by removing mean from each feature
|
||||
X_train_scaled = X_train - X_train_mean
|
||||
X_test_scaled = X_test - X_train_mean
|
||||
#The model intercept (called y_scaler) is given by the mean of the target variable (IF X is centered)
|
||||
#Remove the intercept from the training data.
|
||||
y_scaler = np.mean(y_train)
|
||||
y_train_scaled = y_train - y_scaler
|
||||
|
||||
p = Maxpolydegree-1
|
||||
I = np.eye(p,p)
|
||||
# Decide which values of lambda to use
|
||||
nlambdas = 6
|
||||
MSEOwnRidgePredict = np.zeros(nlambdas)
|
||||
MSERidgePredict = np.zeros(nlambdas)
|
||||
|
||||
lambdas = np.logspace(-4, 2, nlambdas)
|
||||
for i in range(nlambdas):
|
||||
lmb = lambdas[i]
|
||||
OwnRidgeBeta = np.linalg.pinv(X_train_scaled.T @ X_train_scaled+lmb*I) @ X_train_scaled.T @ (y_train_scaled)
|
||||
intercept_ = y_scaler - X_train_mean@OwnRidgeBeta #The intercept can be shifted so the model can predict on uncentered data
|
||||
#Add intercept to prediction
|
||||
ypredictOwnRidge = X_test_scaled @ OwnRidgeBeta + y_scaler
|
||||
RegRidge = linear_model.Ridge(lmb)
|
||||
RegRidge.fit(X_train,y_train)
|
||||
ypredictRidge = RegRidge.predict(X_test)
|
||||
MSEOwnRidgePredict[i] = MSE(y_test,ypredictOwnRidge)
|
||||
MSERidgePredict[i] = MSE(y_test,ypredictRidge)
|
||||
print("Beta values for own Ridge implementation")
|
||||
print(OwnRidgeBeta) #Intercept is given by mean of target variable
|
||||
print("Beta values for Scikit-Learn Ridge implementation")
|
||||
print(RegRidge.coef_)
|
||||
print('Intercept from own implementation:')
|
||||
print(intercept_)
|
||||
print('Intercept from Scikit-Learn Ridge implementation')
|
||||
print(RegRidge.intercept_)
|
||||
print("MSE values for own Ridge implementation")
|
||||
print(MSEOwnRidgePredict[i])
|
||||
print("MSE values for Scikit-Learn Ridge implementation")
|
||||
print(MSERidgePredict[i])
|
||||
|
||||
|
||||
# Now plot the results
|
||||
plt.figure()
|
||||
plt.plot(np.log10(lambdas), MSEOwnRidgePredict, 'b--', label = 'MSE own Ridge Test')
|
||||
plt.plot(np.log10(lambdas), MSERidgePredict, 'g--', label = 'MSE SL Ridge Test')
|
||||
plt.xlabel('log10(lambda)')
|
||||
plt.ylabel('MSE')
|
||||
plt.legend()
|
||||
plt.show()
|
||||
!ec
|
||||
We see here, when compared to the code which includes explicitely the
|
||||
intercept column, that our MSE value is actually smaller. This is
|
||||
because the regularization term does not include the intercept value
|
||||
$\beta_0$ in the fitting. This applies to Lasso regularization as
|
||||
well. It means that our optimization is now done only with the
|
||||
centered matrix and/or vector that enter the fitting procedure.
|
||||
|
||||
|
||||
!split
|
||||
===== Linking the regression analysis with a statistical interpretation =====
|
||||
|
||||
@@ -1567,3 +2061,5 @@ which is our Lasso cost function!
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
Reference in New Issue
Block a user