update
This commit is contained in:
Morten Hjorth-Jensen
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-101
@@ -253,56 +253,149 @@ which may confuse us.
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Yes, it could be a bad idea to include the intercept column for the
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exact reason you stated. If no transformation is applied to your data,
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the intercept can be interpreted as the expected value of your target
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variable when all your predictors are put to zero. Therefore, whenever
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you cannot assume that the expected target variable is zero when all
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your predictors are zero, it could be a bad idea to apply a model
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which penalizes the intercept. Also, the analytical solution to the
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ridge regression coefficients (when not shrinking $$\beta_0$$) is
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derived under the assumption that both y and X are zero centered (mean
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subtracted). What you are doing is correct, but you should also zero
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center X (subtracting the mean of each column from the corresponding
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column).
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The intercept can be interpreted as the expected value of our
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target/output variables when all other predictors are set to zero.
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Thus, if we cannot assume that the expected outputs/targets are zero
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when all predictors are zero (the columns in the design matrix), it
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may be a bad idea to implement a model which penalizes the intercept.
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Furthermore, in for example Ridge and Lasso regression, the solutions
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(when not shrinking $$\beta_0$$) for the unknown parameters
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$\bm{\beta}$ are derived under the assumption that both $\bm{y}$ and
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$\bm{X}$ are zero centered, that is we subtract the mean values.
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If your predictors are of different scales, I would advice you to
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standardize X by subtracting the mean of each column from the
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corresponding column and dividing the column with its standard
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deviation. If you dont do this, you will give an "unfair" penalization
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of the parameters since their magnitude depends on the scale of their
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corresponding predictor. Suppose that you have an input variable
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"height". Human height might be measured in inches or meters or
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!split
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===== More thinking =====
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If our predictors represent different scales, then it is important to
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standardize the design matrix $\bm{X}$ by subtracting the mean of each
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column from the corresponding column and dividing the column with its
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standard deviation.
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The
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"Standadscaler":"https://scikit-learn.org/stable/modules/generated/sklearn.preprocessing.StandardScaler.html"
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function in _Scikit-Learn_ does this for us. For the data sets we
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have been studying in our various examples, the data are in many cases
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already scaled and there is no need to scale them.
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If you need to scale the data, not doing so will give an *unfair*
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penalization of the parameters since their magnitude depends on the
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scale of their corresponding predictor.
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Suppose as an example that you
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you have an input variable given by the heights of different persons.
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Human height might be measured in inches or meters or
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kilometers. If measured in kilometers, a standard linear regression
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model with this predictor would probably give a much bigger
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coefficient term, than if measured in millimeters. You may see how
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this could become a problem when considering the loss function for
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ridge regression.
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coefficient term, than if measured in millimeters.
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This can clearly lead to problems in evaluating the cost/loss functions.
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!split
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===== Still thinking =====
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Remember that when you do any transformation to your dataset before
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training, the exact same transformation has to be applied to new data
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before making a prediction. In your case, this means:
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Keep in mind that when you transform your data set before training a model, the same transformation needs to be done
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on your eventual new data set before making a prediction. If we translate this into a Python code, it would could be implemented as follows
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!bc pycod
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#Model training:
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#Model training, we compute the mean value of y and X
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y_train_mean = np.mean(y_train)
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X_train_mean = np.mean(X_train,axis=0)
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X_train = X_train - X_train_mean
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y_train = y_train - y_train_mean
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# The we fit our model with the training data
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trained_model = some_model.fit(X_train,y_train)
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#Model prediction:
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#Model prediction, here we need also to transform our data set used for the prediction.
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X_test = X_test - X_train_mean #Use mean from training data
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y_pred = trained_model(X_test)
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y_pred = y_pred + y_train_mean
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!ec
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!split
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===== Linear Regression code, Intercept handling first =====
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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*)
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!bc pycod
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import numpy as np
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import matplotlib.pyplot as plt
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from sklearn.linear_model import LinearRegression
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np.random.seed(2021)
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def fit_beta(X, y):
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return np.linalg.pinv(X.T @ X) @ X.T @ y
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true_beta = [2, 0.5, 3.7]
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x = np.linspace(0, 1, 11)
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y = np.sum(
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np.asarray([x ** p * b for p, b in enumerate(true_beta)]), axis=0
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) + 0.1 * np.random.normal(size=len(x))
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degree = 3
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X = np.zeros((len(x), degree))
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# Include the intercept in the design matrix
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for p in range(degree):
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X[:, p] = x ** p
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beta = fit_beta(X, y)
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# Intercept is included in the design matrix
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clf = LinearRegression(fit_intercept=False).fit(X, y)
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print(f"True beta: {true_beta}")
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print(f"Fitted beta: {beta}")
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print(f"Sklearn fitted beta: {clf.coef_}")
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plt.figure()
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plt.scatter(x, y, label="Data")
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plt.plot(x, X @ beta, label="Fit")
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plt.plot(x, clf.predict(X), label="Sklearn (fit_intercept=False)")
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# Do not include the intercept in the design matrix
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X = np.zeros((len(x), degree - 1))
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for p in range(degree - 1):
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X[:, p] = x ** (p + 1)
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# Intercept is not included in the design matrix
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clf = LinearRegression(fit_intercept=True).fit(X, y)
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# Use centered values for X and y when computing coefficients
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y_offset = np.average(y, axis=0)
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X_offset = np.average(X, axis=0)
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beta = fit_beta(X - X_offset, y - y_offset)
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intercept = np.mean(y_offset - X_offset @ beta)
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print(f"Manual intercept: {intercept}")
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print(f"Fitted beta (sans intercept): {beta}")
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print(f"Sklearn intercept: {clf.intercept_}")
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print(f"Sklearn fitted beta (sans intercept): {clf.coef_}")
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plt.plot(x, X @ beta + intercept, "--", label="Fit (manual intercept)")
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plt.plot(x, clf.predict(X), "--", label="Sklearn (fit_intercept=True)")
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plt.grid()
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plt.legend()
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plt.show()
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!ec
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!split
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===== What does centering mean mathematically? =====
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Here is a mathematical explanation of the zero centering:
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@@ -602,79 +695,6 @@ plt.show()
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!ec
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!bc pycod
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import numpy as np
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import matplotlib.pyplot as plt
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from sklearn.linear_model import LinearRegression
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np.random.seed(2021)
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def fit_beta(X, y):
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return np.linalg.pinv(X.T @ X) @ X.T @ y
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true_beta = [2, 0.5, 3.7]
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x = np.linspace(0, 1, 11)
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y = np.sum(
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np.asarray([x ** p * b for p, b in enumerate(true_beta)]), axis=0
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) + 0.1 * np.random.normal(size=len(x))
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degree = 3
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X = np.zeros((len(x), degree))
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# Include the intercept in the design matrix
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for p in range(degree):
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X[:, p] = x ** p
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beta = fit_beta(X, y)
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# Intercept is included in the design matrix
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clf = LinearRegression(fit_intercept=False).fit(X, y)
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print(f"True beta: {true_beta}")
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print(f"Fitted beta: {beta}")
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print(f"Sklearn fitted beta: {clf.coef_}")
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plt.figure()
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plt.scatter(x, y, label="Data")
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plt.plot(x, X @ beta, label="Fit")
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plt.plot(x, clf.predict(X), label="Sklearn (fit_intercept=False)")
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# Do not include the intercept in the design matrix
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X = np.zeros((len(x), degree - 1))
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for p in range(degree - 1):
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X[:, p] = x ** (p + 1)
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# Intercept is not included in the design matrix
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clf = LinearRegression(fit_intercept=True).fit(X, y)
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# Use centered values for X and y when computing coefficients
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y_offset = np.average(y, axis=0)
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X_offset = np.average(X, axis=0)
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beta = fit_beta(X - X_offset, y - y_offset)
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intercept = np.mean(y_offset - X_offset @ beta)
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print(f"Manual intercept: {intercept}")
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print(f"Fitted beta (sans intercept): {beta}")
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print(f"Sklearn intercept: {clf.intercept_}")
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print(f"Sklearn fitted beta (sans intercept): {clf.coef_}")
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plt.plot(x, X @ beta + intercept, "--", label="Fit (manual intercept)")
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plt.plot(x, clf.predict(X), "--", label="Sklearn (fit_intercept=True)")
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plt.grid()
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plt.legend()
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plt.show()
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!ec
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!split
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