MNT Improve wording in Exercise M4.03

python_scripts/linear_models_sol_03.py

@@ -8,8 +8,8 @@
 # %% [markdown]
 # # 📃 Solution for Exercise M4.03
 #
-# Now, we tackle a more realistic classification problem instead of making a
-# synthetic dataset. We start by loading the Adult Census dataset with the
+# Now, we tackle a (relatively) realistic classification problem instead of making
+# a synthetic dataset. We start by loading the Adult Census dataset with the
 # following snippet. For the moment we retain only the **numerical features**.
 
 # %%
@@ -24,10 +24,12 @@
 # %% [markdown]
 # We confirm that all the selected features are numerical.
 #
-# Compute the generalization performance in terms of accuracy of a linear model
-# composed of a `StandardScaler` and a `LogisticRegression`. Use a 10-fold
-# cross-validation with `return_estimator=True` to be able to inspect the
-# trained estimators.
+# Define a linear model composed of a `StandardScaler` followed by a
+# `LogisticRegression` with default parameters.
+#
+# Then use a 10-fold cross-validation to estimate its generalization performance
+# in terms of accuracy. Also set `return_estimator=True` to be able to inspect
+# the trained estimators.
 
 # %%
 # solution
@@ -84,11 +86,11 @@
 # - The numerical data must be scaled.
 # - The categorical data must be one-hot encoded, set `min_frequency=0.01` to
 #   group categories concerning less than 1% of the total samples.
-# - The predictor is a `LogisticRegression`. You may need to increase the number
-#   of `max_iter`, which is 100 by default.
+# - The predictor is a `LogisticRegression` with default parameters, except that
+#   you may need to increase the number of `max_iter`, which is 100 by default.
 #
 # Use the same 10-fold cross-validation strategy with `return_estimator=True` as
-# above to evaluate this complex pipeline.
+# above to evaluate this scaled and encoded pipeline.
 
 # %%
 # solution
@@ -195,23 +197,29 @@
 
 # %% [markdown]
 # Now create a similar pipeline consisting of the same preprocessor as above,
-# followed by a `PolynomialFeatures` and a logistic regression with `C=0.01`.
-# Set `degree=2` and `interaction_only=True` to the feature engineering step.
-# Remember not to include a "bias" feature to avoid introducing a redundancy
-# with the intercept of the subsequent logistic regression.
+# followed by a `PolynomialFeatures` and a logistic regression with `C=0.01` and
+# enough `max_iter`. Set `degree=2` and `interaction_only=True` to the feature
+# engineering step. Remember not to include a "bias" feature to avoid
+# introducing a redundancy with the intercept of the subsequent logistic
+# regression.
 
 # %%
 # solution
 from sklearn.preprocessing import PolynomialFeatures
 
-model_with_interaction = make_pipeline(
+model_with_interactions = make_pipeline(
     preprocessor,
     PolynomialFeatures(degree=2, include_bias=False, interaction_only=True),
     LogisticRegression(C=0.01, max_iter=5_000),
 )
-model_with_interaction
+model_with_interactions
 
 # %% [markdown]
+# Use the same 10-fold cross-validation strategy as above to evaluate this
+# pipeline with interactions. In this case there is no need to return the
+# estimator, as the number of features is much larger to visually explore the
+# learned coefficients.
+#
 # By comparing the cross-validation test scores of both models fold-to-fold,
 # count the number of times the model using multiplicative interactions and both
 # numerical and categorical features has a better test score than the model
@@ -220,11 +228,10 @@
 # %%
 # solution
 cv_results_interactions = cross_validate(
-    model_with_interaction,
+    model_with_interactions,
     data,
     target,
     cv=10,
-    return_estimator=True,
     n_jobs=2,
 )
 test_score_interactions = cv_results_interactions["test_score"]