Cross Validation & Ensembling¶

Shan-Hung Wu & DataLab
Fall 2016
In [1]:
from IPython.display import Image
from IPython.display import display
import pandas as pd
import numpy as np
import matplotlib.pyplot as plt

# inline plotting instead of popping out
%matplotlib inline

# load utility classes/functions that has been taught in previous labs
# e.g., plot_decision_regions()
import os, sys
module_path = os.path.abspath(os.path.join('.'))
sys.path.append(module_path)
from lib import *


In this lab, we will guide you through the cross validation technique for hyperparameter selection. We will also practice the ensemble learning techniques that combine multiple base-leaners for better performance.

Holdout Method¶

So far, we hold out the validation and testing sets for hyperparameter tuning and performance reporting. Specifically, we partition a dataset $\mathbb{X}$ into the training, validation, and testing sets. We use the training set to fit a model by giving a set of hyperparameters, and then use the validation set to evaluate the performance of the model given the hyperparameters. We repeat these two steps by issuing different sets of hyperparameters and pick the set that leads to the highest validation performance. We then use both the training and validation sets to train our final model, and apply it the testing set to evaluate/report the generalization performance. The following figure illustrates the procedure:

Next, we apply this technique to evaluate the KNeighborsClassifier on the Iris dataset. For simplicity, we consider the sepal width and petal length features only. Let's split the dataset first:

In [2]:
from sklearn.datasets import load_iris
from sklearn.model_selection import train_test_split
from sklearn.preprocessing import StandardScaler
import numpy as np

X, y = iris.data[:,[1,2]], iris.target

# hold out testing set
X_train, X_test, y_train, y_test = train_test_split(X, y, test_size=0.3, random_state=0)
# hold out validation set
X_train, X_val, y_train, y_val = train_test_split(X_train, y_train, test_size=0.3, random_state=0)


Then, we iterate through each value of hyperparameter n_neighbors = 1, 15, 50 to train on the training set and estimate performance on the validation set and record the best:

In [3]:
from sklearn.pipeline import Pipeline
from sklearn.neighbors import KNeighborsClassifier
from sklearn.metrics import accuracy_score

best_k, best_score = -1, -1
clfs = {}

# hyperparameter tuning
for k in [1, 15, 50]:
pipe = Pipeline([['sc', StandardScaler()], ['clf', KNeighborsClassifier(n_neighbors=k)]])
pipe.fit(X_train, y_train)
y_pred = pipe.predict(X_val)
score = accuracy_score(y_val, y_pred)
print('[{}-NN]\nValidation accuracy: {}'.format(k, score))
if score > best_score:
best_k, best_score = k, score
clfs[k] = pipe

# performance reporting
y_pred= clfs[best_k].predict(X_test)
print('\nTest accuracy: %.2f (n_neighbors=%d selected by the holdout method)' %
(accuracy_score(y_test, y_pred), best_k))

[1-NN]
Validation accuracy: 0.9375
[15-NN]
Validation accuracy: 0.90625
[50-NN]
Validation accuracy: 0.4375

Test accuracy: 0.89 (n_neighbors=1 selected by the holdout method)


One major disadvantage of the holdout method is that the validation and testing performance is sensitive to the random splits. If we have a unfortunate split such that the validation (resp. testing) set is unrepresentative, we may end up picking suboptimal hyperparameters (resp. reporting a misleading performance score).

In this case, the hyperparameter n_neighbors = 15 actually leads to better test performance:

In [4]:
y_pred= clfs[15].predict(X_test)
print('Test accuracy: %.2f (n_neighbors=15 selected manually)' %
accuracy_score(y_test, y_pred))

Test accuracy: 0.91 (n_neighbors=15 selected manually)


We can see that the validation set is unrepresentative and leads to indistinguishable validation accuracy scores (1.0) for all values of n_neighbors.

Next, we take a look at a more robust technique called the $K$-Fold Cross-Validation.

$K$-Fold Cross Validation¶

In $K$-fold cross-validation (CV), we randomly split the training dataset into $K$ folds without replacement, where $K−1$ folds are used for the model training and the remaining 1 fold is for testing. This procedure is repeated $K$ times so that we obtain $K$ models and $K$ performance estimates. Then we take their average as the final performance estimate. The following figure illustrate the 10-fold CV:

We can apply $K$-fold CV to either the hyperparameter tuning, performance reporting, or both. The advantage of this approach is that the performance is less sensitive to unfortunate splits of data. In addition, it utilize data better since each example can be used for both training and validation/testing.

Let's use $K$-Fold CV to select the hyperparamter n_neighbors of the KNeighborsClassifier:

In [5]:
from sklearn.model_selection import cross_val_score

X, y = iris.data[:,[1,2]], iris.target

# hold out testing set
X_train, X_test, y_train, y_test = train_test_split(X, y, test_size=0.3, random_state=1)


The dataset is first split into training/testing sets.

In [6]:
best_k, best_score = -1, -1
clfs = {}

for k in [1, 15, 50]: # experiment different hyperparameter
pipe = Pipeline([['sc', StandardScaler()], ['clf', KNeighborsClassifier(n_neighbors=k)]])
pipe.fit(X_train, y_train)
# K-Fold CV
scores = cross_val_score(pipe, X_train, y_train, cv=5)
print('[%d-NN]\nValidation accuracy: %.3f %s' % (k, scores.mean(), scores))
if scores.mean() > best_score:
best_k, best_score = k, scores.mean()
clfs[k] = pipe

[1-NN]
Validation accuracy: 0.894 [ 0.91304348  0.95454545  0.85        0.85        0.9       ]
[15-NN]
Validation accuracy: 0.914 [ 0.95652174  0.86363636  0.85        0.9         1.        ]
[50-NN]
Validation accuracy: 0.801 [ 0.73913043  0.86363636  0.75        0.75        0.9       ]


5-fold CV selects the best n_neighbors = 15 as we expected. Once selecting proper hyperparameter values, we retrain the model on the complete training set and obtain a final performance estimate on the test set:

In [7]:
best_clf = clfs[best_k]
best_clf.fit(X_train, y_train)

# performance reporting
y_pred = best_clf.predict(X_test)
print('Test accuracy: %.2f (n_neighbors=%d selected by 5-fold CV)' %
(accuracy_score(y_test, y_pred), best_k))

Test accuracy: 0.93 (n_neighbors=15 selected by 5-fold CV)


Nested CV¶

We can also apply the $K$-fold CV to both the hyperparameter selection and performance reporting at the same time, this is called the nested CV. Following illustrate the $5\times 2$ nested CV:

where we select the values of hyperparameters by 2-fold CV and estimate the generalized performance by 5-fold CV, respectively. Let's try this ourselves:

In [8]:
from sklearn.model_selection import KFold

outer_cv = KFold(n_splits=5, shuffle=True, random_state=1)
inner_cv = KFold(n_splits=10, shuffle=True, random_state=1)

outer_scores = []
# outer folds
for i, (train_idx, test_idx) in enumerate(outer_cv.split(X, y)):
print('[Outer fold %d/5]' % (i + 1))
X_train, X_test = X[train_idx], X[test_idx]
y_train, y_test = y[train_idx], y[test_idx]

best_k, best_score = -1, -1
clfs = {}
# hyperparameter tuning
for k in [1, 15, 50]:
inner_scores = []
# inner folds
for itrain_idx, val_idx in inner_cv.split(X_train, y_train):
X_itrain, X_val = X_train[itrain_idx], X_train[val_idx]
y_itrain, y_val = y_train[itrain_idx], y_train[val_idx]

pipe = Pipeline([['sc', StandardScaler()],
['clf', KNeighborsClassifier(n_neighbors=k)]])
pipe.fit(X_itrain, y_itrain)

y_pred = pipe.predict(X_val)
inner_scores.append(accuracy_score(y_val, y_pred))
score_mean = np.mean(inner_scores)
if best_score < score_mean:
best_k, best_score = k, score_mean
clfs[k] = pipe

# evaluate performance on test fold
best_clf = clfs[best_k]
best_clf.fit(X_train, y_train)
y_pred = best_clf.predict(X_test)
outer_scores.append(accuracy_score(y_test, y_pred))
print('Test accuracy: %.2f (n_neighbors=%d selected by inner 10-fold CV)' %
(outer_scores[i], best_k))

print('\nTest accuracy: %.2f (5x10 nested CV)' % np.mean(outer_scores))

[Outer fold 1/5]
Test accuracy: 0.90 (n_neighbors=1 selected by inner 10-fold CV)
[Outer fold 2/5]
Test accuracy: 0.90 (n_neighbors=15 selected by inner 10-fold CV)
[Outer fold 3/5]
Test accuracy: 0.90 (n_neighbors=15 selected by inner 10-fold CV)
[Outer fold 4/5]
Test accuracy: 0.93 (n_neighbors=15 selected by inner 10-fold CV)
[Outer fold 5/5]
Test accuracy: 1.00 (n_neighbors=15 selected by inner 10-fold CV)

Test accuracy: 0.93 (5x10 nested CV)


As we can see, the 5 inner CVs may select different values for the hyperparameter n_neighbors. In this case, the 1st inner CV selects n_neighbors = 1 due to a unlucky split of the training and testing sets in the outer fold. By doing nested CV, we get a more robust performance estimate.

In fact, we can simplify the above example using the GridSearchCV from Scikit-learn:

In [9]:
from sklearn.model_selection import GridSearchCV

outer_cv = KFold(n_splits=5, shuffle=True, random_state=1)
inner_cv = KFold(n_splits=10, shuffle=True, random_state=1)

outer_scores = []
# outer folds
for i, (train_idx, test_idx) in enumerate(outer_cv.split(X, y)):
print('[Outer fold %d/5]' % (i + 1))
X_train, X_test = X[train_idx], X[test_idx]
y_train, y_test = y[train_idx], y[test_idx]

pipe = Pipeline([['sc', StandardScaler()], ['clf', KNeighborsClassifier()]])
# hyperparameter tuning by grid search CV
param_grid = {'clf__n_neighbors':[1, 15, 50]}
gs = GridSearchCV(estimator=pipe, param_grid=param_grid,
scoring='accuracy', cv=inner_cv)
gs.fit(X_train, y_train)
best_clf = gs.best_estimator_
best_clf.fit(X_train, y_train)
outer_scores.append(best_clf.score(X_test, y_test))
print('Test accuracy: %.2f (n_neighbors=%d selected by inner 10-fold CV)' %
(outer_scores[i], gs.best_params_['clf__n_neighbors']))

print('\nTest accuracy: %.2f (5x10 nested CV)' % np.mean(outer_scores))

[Outer fold 1/5]
Test accuracy: 0.90 (n_neighbors=1 selected by inner 10-fold CV)
[Outer fold 2/5]
Test accuracy: 0.90 (n_neighbors=15 selected by inner 10-fold CV)
[Outer fold 3/5]
Test accuracy: 0.90 (n_neighbors=15 selected by inner 10-fold CV)
[Outer fold 4/5]
Test accuracy: 0.93 (n_neighbors=15 selected by inner 10-fold CV)
[Outer fold 5/5]
Test accuracy: 1.00 (n_neighbors=15 selected by inner 10-fold CV)

Test accuracy: 0.93 (5x10 nested CV)


NOTE: if we have a dataset with imbalance classes, we should use the stratified $K$-fold CV that prepserves the class proportions in each fold to ensure that each fold is representative of the class proportions in the training dataset. To use stratified CV, simply replace

>>> from sklearn.model_selection import KFold
>>> KFold(n_splits=...)

with

>>> from sklearn.model_selection import StratifiedKFold
>>> StratifiedKFold(y=..., n_splits=...)

How Many Folds?¶

How many folds $K$ do we need? Here are some rules of thumb explained in the lecture:

• For large $K$, the MSE of cross-validation error (to the true expected generalization error of $f_N$) tends to have a small bias but large variance since a classifier in each iteration is trained on more examples but classifiers from different folds are trained on similar examples;
• On the other hand, for small $K$, the cross-validation error tends to have large bias but small variance;
• When dataset is small, the cross-validation error will have both large bias and large variance.

To see these in practice, let's consider the Polynomial regression where the ground truth data generating distribution is known:

$$\mathrm{P}(\mathrm{y}|\mathrm{x}) = \sin(x) + \epsilon, \epsilon\sim\mathcal{N}(0,\sigma^2)$$

We can visualize the bias and variance as follows:

In [28]:
from sklearn.model_selection import KFold
from sklearn.preprocessing import PolynomialFeatures
from sklearn.linear_model import LinearRegression
from sklearn.metrics import mean_squared_error

sigma = 1
n_range = range(10, 50, 2)
k_range = [5, 10]

poly = PolynomialFeatures(degree=2)
X = np.array([])
y = np.array([])
cv5_mean = []
cv5_std = []
cv10_mean = []
cv10_std = []
exp_mean = []
for n in n_range:
# compute the bias and variance of cv5
mse_test = []
for i in range(500):
x, y = gen_data(n, sigma)
X = poly.fit_transform(x[:, np.newaxis])

cv5 = KFold(n_splits=5, random_state=1)
for i, (train, test) in enumerate(cv5.split(X, y)):
lr = LinearRegression()
lr.fit(X[train], y[train])
y_test_pred = lr.predict(X[test])
mse_test.append(mean_squared_error(y[test], y_test_pred))

cv5_mean.append(np.mean(mse_test))
cv5_std.append(np.std(mse_test))

# compute the bias and variance of cv10
mse_test = []
for i in range(500):
x, y = gen_data(n, sigma)
X = poly.fit_transform(x[:, np.newaxis])

cv10 = KFold(n_splits=10, random_state=1)
for i, (train, test) in enumerate(cv10.split(X, y)):
lr = LinearRegression()
lr.fit(X[train], y[train])
y_test_pred = lr.predict(X[test])
mse_test.append(mean_squared_error(y[test], y_test_pred))

cv10_mean.append(np.mean(mse_test))
cv10_std.append(np.std(mse_test))

# compute the expected generalization error of f_N
mse_test = []
for i in range(500):
x, y = gen_data(n, sigma)
X = poly.fit_transform(x[:, np.newaxis])
lr = LinearRegression()
lr.fit(X, y)
x_test, y_test = gen_data(100, sigma)
X_test = poly.transform(x_test[:, np.newaxis])
y_test_pred = lr.predict(X_test)
mse_test.append(mean_squared_error(y_test, y_test_pred))
exp_mean.append(np.mean(mse_test))

In [29]:
plt.plot(n_range, cv5_mean,
markersize=5, label='5-Fold CV', color='blue')
plt.fill_between(n_range,
np.subtract(cv5_mean, cv5_std),
alpha=0.15, color='blue')

plt.plot(n_range, cv10_mean,
markersize=5, label='10-Fold CV', color='green')
plt.fill_between(n_range,
np.subtract(cv10_mean, cv10_std),
alpha=0.15, color='green')

plt.plot(n_range, exp_mean,
markersize=5, label='Exp', color='red')

plt.hlines(y=sigma, xmin=10, xmax=48,
label='Bayes', color='red',
linewidth=2, linestyle='--')

plt.legend(loc='upper right')
plt.xlim([10, 48])
plt.ylim([0, 5])
plt.xlabel('N')
plt.ylabel('MSE')
plt.tight_layout()
plt.savefig('./output/fig-cv-fold.png', dpi=300)
plt.show()


Usually, we set $K=10$ in most applications, $K=5$ for larger datasets, and $K=N$ for very small datasets. The last setting is called the leave-one-out CV.

Ensemble Methods¶

No free lunch theorem states that no machine learning algorithm is universally better than the others in all domains. To goal of ensembling is to combine multiple learner to improve the applicability and get better performance.

NOTE: it is possible that the final model performs no better than the most accurate learners in the ensemble. But it at least reduces the probability of selecting a poor one and increases the applicability.

Voting¶

Voting is arguably the most straightforward way to combine multiple learners $d^{(j)}(\cdot)$. The idea is to to taking a linear combination of the predictions made by the learners. For example, in multiclass classification, we have

$$\tilde{y}_k =\sum_j^L w_j d^{(j)}_k(\boldsymbol{x}), \text{ where }w_j\geq 0\text{ and }\sum_j w_j=1,$$

for any class $k$, where $L$ is the number of voters. This can be simplified to the plurarity vote where each voter has the same weight:

$$\tilde{y}_k =\sum_j \frac{1}{L} d^{(j)}_k(\boldsymbol{x}).$$

Let's use the VotingClassifier from Scikit-learn to combine KNeighborsClassifer, LogisticRegression, and DecisionTreeClassifier together and train on the synthetic two-moon dataset:

In [50]:
#import pandas as pd
from sklearn.neighbors import KNeighborsClassifier
from sklearn.tree import DecisionTreeClassifier
from sklearn.linear_model import LogisticRegression
from sklearn.datasets import make_moons

X, y = make_moons(n_samples=500, noise=0.3, random_state=0)
X_train, X_test, y_train, y_test = train_test_split(X, y, test_size=0.2, random_state=3)

plt.scatter(X[y == 0, 0], X[y == 0, 1], label='Class 0',
c='r', marker='s', alpha=0.5)
plt.scatter(X[y == 1, 0], X[y == 1, 1], label='Class 1',
c='b', marker='x', alpha=0.5)
plt.scatter(X_test[:, 0], X_test[:, 1],
c='', marker='o', label='Class 1')
plt.show()

pipe1 = Pipeline([['sc', StandardScaler()], ['clf', LogisticRegression(C=10, random_state=0)]])
pipe2 = Pipeline([['clf', DecisionTreeClassifier(max_depth=3, random_state=0)]])
pipe3 = Pipeline([['sc', StandardScaler()], ['clf', KNeighborsClassifier(n_neighbors=5)]])


We can estimate the performance of individual classifiers via the 10-fold CV:

In [51]:
from sklearn.model_selection import cross_val_score

clf_labels = ['LogisticRegression', 'DecisionTree', 'KNN']
print('[Individual]')
for pipe, label in zip([pipe1, pipe2, pipe3], clf_labels):
scores = cross_val_score(estimator=pipe, X=X_train, y=y_train, cv=10, scoring='roc_auc')
print('%s: %.3f (+/- %.3f)' % (label, scores.mean(), scores.std()))

[Individual]
LogisticRegression: 0.929 (+/- 0.029)
DecisionTree: 0.941 (+/- 0.023)
KNN: 0.948 (+/- 0.026)


Let's combined the classifiers by VotingClassifer from Scikit-learn and experiment some weight combinations:

In [52]:
from sklearn.ensemble import VotingClassifier
import itertools

print('[Voting]')
best_vt, best_w, best_score = None, (), -1
for a, b, c in list(itertools.permutations(range(0,3))): # try some weight combination
clf = VotingClassifier(estimators=[('lr', pipe1), ('dt', pipe2), ('knn', pipe3)],
voting='soft', weights=[a,b,c])
scores = cross_val_score(estimator=clf, X=X_train, y=y_train, cv=10, scoring='roc_auc')
print('%s: %.3f (+/- %.3f)' % ((a,b,c), scores.mean(), scores.std()))
if best_score < scores.mean():
best_vt, best_w, best_score = clf, (a, b, c), scores.mean()

print('\nBest %s: %.3f' % (best_w, best_score))

[Voting]
(0, 1, 2): 0.959 (+/- 0.021)
(0, 2, 1): 0.960 (+/- 0.019)
(1, 0, 2): 0.958 (+/- 0.023)
(1, 2, 0): 0.944 (+/- 0.025)
(2, 0, 1): 0.948 (+/- 0.023)
(2, 1, 0): 0.939 (+/- 0.025)

Best (0, 2, 1): 0.960


The best ensemble combines the DecisionTreeClassifier and KNeighborsClassifier. This is a reasonable choice because these two models "complement" each other in design: DecisionTreeClassifier makes predictions based on informative features; while KNeighborsClassifier makes predictions based on representative examples.

To compare the VotingClassifer with individual classifiers on the testing set, we can plot the ROC curves:

In [56]:
import matplotlib.pyplot as plt
from sklearn.metrics import roc_curve
from sklearn.metrics import auc

clf_labels =['LogisticRegression', 'DecisionTree', 'KNN', 'Voting']
colors = ['black', 'orange', 'blue', 'green']
linestyles = ['-', '-', '-', '--']
for clf, label, clr, ls in zip([pipe1, pipe2, pipe3, best_vt], clf_labels, colors, linestyles):
# assume positive class is at dimension 2
clf.fit(X_train, y_train)
y_pred = clf.predict_proba(X_test)[:, 1]

fpr, tpr, thresholds = roc_curve(y_true=y_test, y_score=y_pred)
roc_auc = auc(x=fpr, y=tpr)

plt.plot(fpr, tpr, color=clr, linestyle=ls, label='%s (auc=%0.2f)' % (label, roc_auc))
plt.legend(loc='lower right')
plt.plot([0, 1], [0, 1], linestyle='--', color='gray')
plt.xlim([-0.02, 1])
plt.ylim([-0.1, 1.1])
plt.grid()
plt.xlabel('FPR')
plt.ylabel('TPR')

plt.grid()
plt.tight_layout()
plt.savefig('./output/fig-vote-roc.png', dpi=300)
plt.show()


As we can see, the VotingClassifer can successfully combine the base-learners to give a higher true-positive rate at a low false-positive rate. Let's see the decision boundaries:

In [35]:
X_combined = np.vstack((X_train, X_test))
y_combined = np.hstack((y_train, y_test))

plot_decision_regions(X=X_combined, y=y_combined,
classifier=pipe1,
test_idx=range(len(y_train),
len(y_train) + len(y_test)))
plt.title('Logistic regression')
plt.tight_layout()
plt.savefig('./output/fig-vote-logistic-regressio-boundary.png', dpi=300)
plt.show()

plot_decision_regions(X=X_combined, y=y_combined,
classifier=pipe2,
test_idx=range(len(y_train),
len(y_train) + len(y_test)))
plt.title('Decision tree')
plt.tight_layout()
plt.savefig('./output/fig-vote-decision-tree-boundary.png', dpi=300)
plt.show()

plot_decision_regions(X=X_combined, y=y_combined,
classifier=pipe3,
test_idx=range(len(y_train),
len(y_train) + len(y_test)))
plt.title('KNN')
plt.tight_layout()
plt.savefig('./output/fig-voting-knn-boundary.png', dpi=300)
plt.show()

plot_decision_regions(X=X_combined, y=y_combined,
classifier=best_vt,
test_idx=range(len(y_train),
len(y_train) + len(y_test)))
plt.title('Voting')
plt.tight_layout()
plt.savefig('./output/fig-voting-boundary.png', dpi=300)
plt.show()


The decision boundaries of DecisionTreeClassifier and VotingClassifier looks similar. But they have different soft decision boundaries that take into account the probability/confidence of predictions:

In [36]:
plot_decision_regions(X=X_combined, y=y_combined,
classifier=pipe2, soft=True,
test_idx=range(len(y_train),
len(y_train) + len(y_test)))
plt.title('Decision tree')
plt.tight_layout()
plt.savefig('./output/fig-vote-decision-tree-boundary-soft.png', dpi=300)
plt.show()

plot_decision_regions(X=X_combined, y=y_combined,
classifier=best_vt, soft=True,
test_idx=range(len(y_train),
len(y_train) + len(y_test)))
plt.title('Voting')
plt.tight_layout()
plt.savefig('./output/fig-voting-boundary-soft.png', dpi=300)
plt.show()


The different soft decision boundaries result in different ROC curves.

NOTE: here we extend the plot_decision_regions() function such that it draws a "soft" decision boundary of a binary classifier (using the predict_proba() method, if existing) when fed by a parameter soft=True. Please refer to the lib.py for more details.

Bagging¶

Bagging (Bootstrap AGgragating) is a voting method where each base-learner are trained over a slightly different training set. The procedure of bagging is summarized below:

1. Train $L$ classifiers, each on a dataset generated by bootstrapping (draw with replacement);
2. Predict by voting (aggregating all predictions of the $L$ classifiers).

Bagging can reduce the variance since voters now only see different training sets and become less positively correlated with each other. Also, bagging is more robust to noise and outliers since we do the resampling on dataset. However, the model bias cannot be reduced, and this is why we usually use classifiers with low bias, for example, decision trees or nonlinear SVMs, as the base-learners in bagging.

NOTE: when the amount of data is large enough, bagging doesn't help since each classifier will have low variance. We can introduce additional diversity to bagging by randomly selecting features of training examples. The random forest model is this kind of ensembling of decision trees.

The BaggingClassifier is provided by Scikit-learn. Let's use the unpruned DecisionTreeClassifier as the base-learner and create an ensemble of 500 decision trees fitted on different bootstrap examples of the training set:

In [37]:
from sklearn.ensemble import BaggingClassifier

tree = DecisionTreeClassifier(criterion='entropy', max_depth=None, random_state=0)
bag = BaggingClassifier(base_estimator=tree, n_estimators=500,
max_samples=0.7, bootstrap=True,
max_features=1.0, bootstrap_features=False,
n_jobs=1, random_state=1)


The parameter max_samples controls the number of bootstrapped examples and max_feature controls the proportion of features from the feature set that will be sampled to train the base classifiers. We disable feature bootstrapping here.

Next, we compare the performance of the trained BaggingClassifier to a single unpruned DecisionTreeClassifier:

In [38]:
from sklearn.metrics import accuracy_score

# single DecisionTree
tree = tree.fit(X_train, y_train)
y_train_pred = tree.predict(X_train)
y_test_pred = tree.predict(X_test)

tree_train = accuracy_score(y_train, y_train_pred)
tree_test = accuracy_score(y_test, y_test_pred)
print('[DecisionTree] accuracy-train = %.3f, accuracy-test = %.3f' % (tree_train, tree_test))
# Bagging
bag = bag.fit(X_train, y_train)
y_train_pred = bag.predict(X_train)
y_test_pred = bag.predict(X_test)

bag_train = accuracy_score(y_train, y_train_pred)
bag_test = accuracy_score(y_test, y_test_pred)
print('[Bagging] auc-train = %.3f, auc-test = %.3f' % (bag_train, bag_test))

[DecisionTree] accuracy-train = 1.000, accuracy-test = 0.840
[Bagging] auc-train = 0.993, auc-test = 0.860


We sample $0.7N$ examples in each bootstrap to make the base-learners more uncorrelated. The BaggingClassifer successfully mitigates the overfitting behavior of the unpruned DecisionTreeClassifier and gives better generalization performance. We can see this more clearly by comparing the decision boundaries of the two models:

In [39]:
import numpy as np
import matplotlib.pyplot as plt

X_combined = np.vstack((X_train, X_test))
y_combined = np.hstack((y_train, y_test))

plot_decision_regions(X=X_combined, y=y_combined,
classifier=tree,
test_idx=range(len(y_train),
len(y_train) + len(y_test)))
plt.title('Decision tree')
plt.tight_layout()
plt.savefig('./output/fig-bagging-decision-tree-boundary.png', dpi=300)
plt.show()

plot_decision_regions(X=X_combined, y=y_combined,
classifier=bag,
test_idx=range(len(y_train),
len(y_train) + len(y_test)))
plt.title('Bagging')
plt.tight_layout()
plt.savefig('./output/fig-bagging-boundary.png', dpi=300)
plt.show()

plot_decision_regions(X=X_combined, y=y_combined,
classifier=bag, soft=True,
test_idx=range(len(y_train),
len(y_train) + len(y_test)))
plt.title('Bagging (soft)')
plt.tight_layout()
plt.savefig('./output/fig-bagging-boundary-soft.png', dpi=300)
plt.show()


The BaggingClassifer give a smoother decision boundary that less overfits the training data.

Boosting¶

The key idea of boosting is to create complementary base-leanrers by training the new learner using the examples that the previous leaners do not agree. A common implementation is AdaBoost (Adaptive Boosting), which can be summarized as followings:

1. Initialize $\Pr^{(i,1)}=\frac{1}{N}$ for all $i$;
2. Start from $j=1$:
1. Randomly draw $\mathbb{X}^{(j)}$ from $\mathbb{X}$ with probabilities $\Pr^{(i,j)}$ and train $d^{(j)}$;
2. Stop adding new base-learners if $\epsilon^{(j)}=\sum_{i}\Pr^{(i,j)}1(y^{(i)}\neq d^{(j)}(\boldsymbol{x}^{(i)}))\geq\frac{1}{2}$;
3. Define $\alpha_{j}=\frac{1}{2}\log\left(\frac{1-\varepsilon^{(j)}}{\varepsilon^{(j)}}\right)>0$ and set $\Pr^{(i,j+1)}=\Pr^{(i,j)}\cdot\exp(-\alpha_{j}y^{(i)}d^{(j)}(\boldsymbol{x}^{(i)}))$ for all $i$;

After training, the (soft) prediction $\tilde{y}$ is made by voting: $\tilde{y}=\sum_{j}\alpha_{j}d^{(j)}(\boldsymbol{x})$.

Let's train an AdaBoostClassifier from Scikit-learn with 500 decision trees of depth 1:

In [40]:
from sklearn.ensemble import AdaBoostClassifier
from sklearn import datasets

tree = DecisionTreeClassifier(criterion='entropy', max_depth=1)
# single decision tree
tree = tree.fit(X_train, y_train)
y_train_pred = tree.predict(X_train)
y_test_pred = tree.predict(X_test)

tree_train = accuracy_score(y_train, y_train_pred)
tree_test = accuracy_score(y_test, y_test_pred)
print('[DecisionTree] accuracy-train = %.3f, accuracy-test = %.3f' %
(tree_train, tree_test))

ada_train = accuracy_score(y_train, y_train_pred)
ada_test = accuracy_score(y_test, y_test_pred)
print('[AdaBoost] accuracy-train = %.3f, accuracy-test = %.3f' %

[DecisionTree] accuracy-train = 0.838, accuracy-test = 0.710
[AdaBoost] accuracy-train = 1.000, accuracy-test = 0.870


The AdaBoostClassifier predicts all training examples correctly but not so well on testing set, which is a sign of overfitting. However, since AdaBoostClassifier increases the margins of training examples (as discussed in the lecture), the variance can be controlled. Overall, AdaBoostClassifier gives better generalization performance due to a smaller bias.

Let's check the decision boundaries:

In [41]:
plot_decision_regions(X=X_combined, y=y_combined,
classifier=tree,
test_idx=range(len(y_train),
len(y_train) + len(y_test)))
plt.title('Decision tree')
plt.tight_layout()
plt.show()

plot_decision_regions(X=X_combined, y=y_combined,
test_idx=range(len(y_train),
len(y_train) + len(y_test)))