# 2A.ML101.5: Measuring prediction performance¶

**Links:** `notebook`

, html, `PDF`

, `python`

, slides, GitHub

*Source:* Course on machine learning with
scikit-learn
by Gaël Varoquaux

## Using the K-neighbors classifier¶

Here we’ll continue to look at the digits data, but we’ll switch to the
K-Neighbors classifier. The K-neighbors classifier is an instance-based
classifier. The K-neighbors classifier predicts the label of an unknown
point based on the labels of the *K* nearest points in the parameter
space.

```
# Get the data
from sklearn.datasets import load_digits
digits = load_digits()
X = digits.data
y = digits.target
```

```
# Instantiate and train the classifier
from sklearn.neighbors import KNeighborsClassifier
clf = KNeighborsClassifier(n_neighbors=1)
clf.fit(X, y)
```

```
KNeighborsClassifier(n_neighbors=1)
```

```
# Check the results using metrics
from sklearn import metrics
y_pred = clf.predict(X)
```

```
print(metrics.confusion_matrix(y_pred, y))
```

```
[[178 0 0 0 0 0 0 0 0 0]
[ 0 182 0 0 0 0 0 0 0 0]
[ 0 0 177 0 0 0 0 0 0 0]
[ 0 0 0 183 0 0 0 0 0 0]
[ 0 0 0 0 181 0 0 0 0 0]
[ 0 0 0 0 0 182 0 0 0 0]
[ 0 0 0 0 0 0 181 0 0 0]
[ 0 0 0 0 0 0 0 179 0 0]
[ 0 0 0 0 0 0 0 0 174 0]
[ 0 0 0 0 0 0 0 0 0 180]]
```

Apparently, we’ve found a perfect classifier! But this is misleading for
the reasons we saw before: the classifier essentially “memorizes” all
the samples it has already seen. To really test how well this algorithm
does, we need to try some samples it *hasn’t* yet seen.

This problem can also occur with regression models. In the following we fit an other instance-based model named “decision tree” to the Boston Housing price dataset we introduced previously:

```
%matplotlib inline
from matplotlib import pyplot as plt
import numpy as np
```

```
from sklearn.datasets import load_boston
from sklearn.tree import DecisionTreeRegressor
data = load_boston()
clf = DecisionTreeRegressor().fit(data.data, data.target)
predicted = clf.predict(data.data)
expected = data.target
plt.scatter(expected, predicted)
plt.plot([0, 50], [0, 50], '--k')
plt.axis('tight')
plt.xlabel('True price ($1000s)')
plt.ylabel('Predicted price ($1000s)');
```

Here again the predictions are seemingly perfect as the model was able to perfectly memorize the training set.

## A Better Approach: Using a validation set¶

Learning the parameters of a prediction function and testing it on the same data is a methodological mistake: a model that would just repeat the labels of the samples that it has just seen would have a perfect score but would fail to predict anything useful on yet-unseen data.

To avoid over-fitting, we have to define two different sets:

a training set X_train, y_train which is used for learning the parameters of a predictive model

a testing set X_test, y_test which is used for evaluating the fitted predictive model

In scikit-learn such a random split can be quickly computed with the
`train_test_split`

helper function. It can be used this way:

```
from sklearn.model_selection import train_test_split
X = digits.data
y = digits.target
X_train, X_test, y_train, y_test = train_test_split(X, y, test_size=0.25, random_state=0)
print("%r, %r, %r" % (X.shape, X_train.shape, X_test.shape))
```

```
(1797, 64), (1347, 64), (450, 64)
```

Now we train on the training data, and test on the testing data:

```
clf = KNeighborsClassifier(n_neighbors=1).fit(X_train, y_train)
y_pred = clf.predict(X_test)
```

```
print(metrics.confusion_matrix(y_test, y_pred))
```

```
[[37 0 0 0 0 0 0 0 0 0]
[ 0 43 0 0 0 0 0 0 0 0]
[ 0 0 43 1 0 0 0 0 0 0]
[ 0 0 0 45 0 0 0 0 0 0]
[ 0 0 0 0 38 0 0 0 0 0]
[ 0 0 0 0 0 47 0 0 0 1]
[ 0 0 0 0 0 0 52 0 0 0]
[ 0 0 0 0 0 0 0 48 0 0]
[ 0 0 0 0 0 0 0 0 48 0]
[ 0 0 0 1 0 1 0 0 0 45]]
```

```
print(metrics.classification_report(y_test, y_pred))
```

```
precision recall f1-score support
0 1.00 1.00 1.00 37
1 1.00 1.00 1.00 43
2 1.00 0.98 0.99 44
3 0.96 1.00 0.98 45
4 1.00 1.00 1.00 38
5 0.98 0.98 0.98 48
6 1.00 1.00 1.00 52
7 1.00 1.00 1.00 48
8 1.00 1.00 1.00 48
9 0.98 0.96 0.97 47
accuracy 0.99 450
macro avg 0.99 0.99 0.99 450
weighted avg 0.99 0.99 0.99 450
```

The averaged f1-score is often used as a convenient measure of the overall performance of an algorithm. It appears in the bottom row of the classification report; it can also be accessed directly:

```
metrics.f1_score(y_test, y_pred, average="macro")
```

```
0.9913675218842191
```

The over-fitting we saw previously can be quantified by computing the f1-score on the training data itself:

```
metrics.f1_score(y_train, clf.predict(X_train), average="macro")
```

```
1.0
```

**Regression metrics** In the case of regression models, we need to use
different metrics, such as explained variance.

## Application: Model Selection via Validation¶

In the previous notebook, we saw Gaussian Naive Bayes classification of the digits. Here we saw K-neighbors classification of the digits. We’ve also seen support vector machine classification of digits. Now that we have these validation tools in place, we can ask quantitatively which of the three estimators works best for the digits dataset.

With the default hyper-parameters for each estimator, which gives the best f1 score on the

**validation set**? Recall that hyperparameters are the parameters set when you instantiate the classifier: for example, the`n_neighbors`

inclf = KNeighborsClassifier(n_neighbors=1)

For each classifier, which value for the hyperparameters gives the best results for the digits data? For

`LinearSVC`

, use`loss='l2'`

and`loss='l1'`

. For`KNeighborsClassifier`

we use`n_neighbors`

between 1 and 10. Note that`GaussianNB`

does not have any adjustable hyperparameters.

```
from sklearn.svm import LinearSVC
from sklearn.naive_bayes import GaussianNB
from sklearn.neighbors import KNeighborsClassifier
import warnings # suppress warnings from older versions of KNeighbors
warnings.filterwarnings('ignore', message='kneighbors*')
X = digits.data
y = digits.target
X_train, X_test, y_train, y_test = train_test_split(X, y, test_size=0.25, random_state=0)
for Model in [LinearSVC, GaussianNB, KNeighborsClassifier]:
clf = Model().fit(X_train, y_train)
y_pred = clf.predict(X_test)
print(Model.__name__,
metrics.f1_score(y_test, y_pred, average="macro"))
print('------------------')
# test SVC loss
for loss, p, dual in [('squared_hinge', 'l1', False), ('squared_hinge', 'l2', True)]:
clf = LinearSVC(penalty=p, loss=loss, dual=dual)
clf.fit(X_train, y_train)
y_pred = clf.predict(X_test)
print("LinearSVC(penalty='{0}', loss='{1}')".format(p, loss),
metrics.f1_score(y_test, y_pred, average="macro"))
print('-------------------')
# test K-neighbors
for n_neighbors in range(1, 11):
clf = KNeighborsClassifier(n_neighbors=n_neighbors).fit(X_train, y_train)
y_pred = clf.predict(X_test)
print("KNeighbors(n_neighbors={0})".format(n_neighbors),
metrics.f1_score(y_test, y_pred, average="macro"))
```

C:xavierdupre__home_github_forkscikit-learnsklearnsvm_base.py:975: ConvergenceWarning: Liblinear failed to converge, increase the number of iterations. "the number of iterations.", ConvergenceWarning)

```
LinearSVC 0.9293524810149194
GaussianNB 0.8332741681010101
KNeighborsClassifier 0.9804562804949924
------------------
```

C:xavierdupre__home_github_forkscikit-learnsklearnsvm_base.py:975: ConvergenceWarning: Liblinear failed to converge, increase the number of iterations. "the number of iterations.", ConvergenceWarning)

```
LinearSVC(penalty='l1', loss='squared_hinge') 0.9446323900921667
```

C:xavierdupre__home_github_forkscikit-learnsklearnsvm_base.py:975: ConvergenceWarning: Liblinear failed to converge, increase the number of iterations. "the number of iterations.", ConvergenceWarning)

```
LinearSVC(penalty='l2', loss='squared_hinge') 0.9319089652789225
-------------------
KNeighbors(n_neighbors=1) 0.9913675218842191
KNeighbors(n_neighbors=2) 0.9848442068835102
KNeighbors(n_neighbors=3) 0.9867753449543099
KNeighbors(n_neighbors=4) 0.9803719053818863
KNeighbors(n_neighbors=5) 0.9804562804949924
KNeighbors(n_neighbors=6) 0.9757924194139573
KNeighbors(n_neighbors=7) 0.9780645792142071
KNeighbors(n_neighbors=8) 0.9780645792142071
KNeighbors(n_neighbors=9) 0.9780645792142071
KNeighbors(n_neighbors=10) 0.9755550897728812
```

## Cross-validation¶

Cross-validation consists in repetively splitting the data in pairs of train and test sets, called ‘folds’. Scikit-learn comes with a function to automatically compute score on all these folds. Here we do ‘K-fold’ with k=5.

```
clf = KNeighborsClassifier()
from sklearn.model_selection import cross_val_score
cross_val_score(clf, X, y, cv=5)
```

```
array([0.94722222, 0.95555556, 0.96657382, 0.98050139, 0.9637883 ])
```

We can use different splitting strategies, such as random splitting

```
from sklearn.model_selection import ShuffleSplit
cv = ShuffleSplit(n_splits=5)
cross_val_score(clf, X, y, cv=cv)
```

```
array([0.97777778, 0.99444444, 0.99444444, 0.97777778, 0.98333333])
```

There exists many different cross-validation strategies in scikit-learn. They are often useful to take in account non iid datasets.

## Hyperparameter optimization with cross-validation¶

Consider regularized linear models, such as *Ridge Regression*, which
uses regularlization, and *Lasso Regression*, which uses
regularization. Choosing their regularization parameter
is important.

Let us set these paramaters on the Diabetes dataset, a simple regression problem. The diabetes data consists of 10 physiological variables (age, sex, weight, blood pressure) measure on 442 patients, and an indication of disease progression after one year:

```
from sklearn.datasets import load_diabetes
data = load_diabetes()
X, y = data.data, data.target
print(X.shape)
```

```
(442, 10)
```

With the default hyper-parameters: we use the cross-validation score to determine goodness-of-fit:

```
from sklearn.linear_model import Ridge, Lasso
for Model in [Ridge, Lasso]:
model = Model()
print(Model.__name__, cross_val_score(model, X, y).mean())
```

```
Ridge 0.4101758336587286
Lasso 0.3375597834274946
```

### Basic Hyperparameter Optimization¶

We compute the cross-validation score as a function of alpha, the strength of the regularization for Lasso and Ridge. We choose 20 values of alpha between 0.0001 and 1:

```
alphas = np.logspace(-3, -1, 30)
for Model in [Lasso, Ridge]:
scores = [cross_val_score(Model(alpha), X, y, cv=3).mean()
for alpha in alphas]
plt.plot(alphas, scores, label=Model.__name__)
plt.legend(loc='lower left');
```

Can we trust our results to be actually useful?

### Automatically Performing Grid Search¶

```
from sklearn.model_selection import GridSearchCV
```

`GridSearchCV`

is constructed with an estimator, as well as a
dictionary of parameter values to be searched. We can find the optimal
parameters this way:

```
for Model in [Ridge, Lasso]:
gscv = GridSearchCV(Model(), dict(alpha=alphas), cv=3).fit(X, y)
print(Model.__name__, gscv.best_params_)
```

```
Ridge {'alpha': 0.06210169418915616}
Lasso {'alpha': 0.01268961003167922}
```

### Built-in Hyperparameter Search¶

For some models within scikit-learn, cross-validation can be performed
more efficiently on large datasets. In this case, a cross-validated
version of the particular model is included. The cross-validated
versions of `Ridge`

and `Lasso`

are `RidgeCV`

and `LassoCV`

,
respectively. The grid search on these estimators can be performed as
follows:

```
from sklearn.linear_model import RidgeCV, LassoCV
for Model in [RidgeCV, LassoCV]:
model = Model(alphas=alphas, cv=3).fit(X, y)
print(Model.__name__, model.alpha_)
```

```
RidgeCV 0.06210169418915616
LassoCV 0.01268961003167922
```

We see that the results match those returned by GridSearchCV

### Nested cross-validation¶

How do we measure the performance of these estimators? We have used data
to set the hyperparameters, so we need to test on actually new data. We
can do this by running `cross_val_score`

on our CV objects. Here there
are 2 cross-validation loops going on, this is called ‘nested cross
validation’:

```
for Model in [RidgeCV, LassoCV]:
scores = cross_val_score(Model(alphas=alphas, cv=3), X, y, cv=3)
print(Model.__name__, np.mean(scores))
```

```
RidgeCV 0.4891614245496554
LassoCV 0.4854913966864775
```

Note that these results do not match the best results of our curves
above, and `LassoCV`

seems to under-perform `RidgeCV`

. The reason is
that setting the hyper-parameter is harder for Lasso, thus the
estimation error on this hyper-parameter is larger.