如何在scikit-learn中使用k折交叉验证来获得每折的精确召回率？

Question

Let's say I have this scenario:

from sklearn import model_selection
from sklearn.linear_model import LogisticRegression

kfold = model_selection.KFold(n_splits=5, random_state=7)
acc_per_fold = model_selection.cross_val_score(LogisticRegression(),
               x_inputs, np.ravel(y_response), cv=kfold, scoring='accuracy')

What else can I get from model_selection.cross_val_score() ? Is there a way to see what happens inside every actual fold? Can I get precision-recall per fold? Predicted values? How about using a trained model from a fold to make predictions on unseen data?

Answer 1

You can use the cross_validate function to see what happens in each fold.

import numpy as np
from sklearn.datasets import make_classification
from sklearn.model_selection import cross_validate
from sklearn.linear_model import LogisticRegression
from sklearn.metrics import accuracy_score, confusion_matrix, recall_score, roc_auc_score, precision_score

X, y = make_classification(
    n_classes=2, class_sep=1.5, weights=[0.9, 0.1],
    n_features=20, n_samples=1000, random_state=10
)
clf = LogisticRegression(class_weight="balanced")
scoring = {'accuracy': 'accuracy',
           'recall': 'recall',
           'precision': 'precision',
           'roc_auc': 'roc_auc'}
cross_val_scores = cross_validate(clf, X, y, cv=3, scoring=scoring)

The output is the following,

{'fit_time': array([ 0.        ,  0.        ,  0.01559997]),
 'score_time': array([ 0.01559997,  0.        ,  0.        ]),
 'test_accuracy': array([ 0.9251497 ,  0.95808383,  0.93674699]),
 'test_precision': array([ 0.59183673,  0.70833333,  0.63636364]),
 'test_recall': array([ 0.85294118,  1.        ,  0.84848485]),
 'test_roc_auc': array([ 0.96401961,  0.99343137,  0.96787271]),
 'train_accuracy': array([ 0.96096096,  0.93693694,  0.95209581]),
 'train_precision': array([ 0.73033708,  0.62376238,  0.69148936]),
 'train_recall': array([ 0.97014925,  0.94029851,  0.95588235]),
 'train_roc_auc': array([ 0.99426906,  0.98509954,  0.99223039])}

So what happened in the first fold ?

FOLD, METRIC = (0, 'test_precision')
cross_val_scores[METRIC][FOLD]

And is the precision score stable ?

np.std(cross_val_scores[METRIC])