使用sklearn获得精度和召回

Posted 2021-03-29

使用下面的代码，我有Accuracy。现在我想

1）找到每个折叠的precision和recall（总共10倍）

2）获得mean为precision

3）获得mean为recall

这可能类似于下面的print(scores)和print("Accuracy: %0.2f (+/- %0.2f)" % (scores.mean(), scores.std() * 2))。

有什么想法吗？

import numpy as np
from sklearn import cross_validation
from sklearn import datasets
from sklearn import svm
from sklearn.model_selection import StratifiedKFold

iris = datasets.load_iris()
skf = StratifiedKFold(n_splits=10)
clf = svm.SVC(kernel='linear', C=1)
scores = cross_validation.cross_val_score(clf, iris.data, iris.target, cv=10)
print(scores)  #[ 1. 0.93333333   1.  1. 0.86666667  1.  0.93333333   1.  1.  1.]
print("Accuracy: %0.2f (+/- %0.2f)" % (scores.mean(), scores.std() * 2)) # Accuracy: 0.97 (+/- 0.09)

答案

这有点不同，因为cross_val_score无法计算非二进制分类的精度/召回率，因此您需要使用recision_score，recall_score并手动进行交叉验证。参数average ='micro'计算全局精度/召回率。

import numpy as np
from sklearn import cross_validation
from sklearn import datasets
from sklearn import svm
from sklearn.model_selection import StratifiedKFold
from sklearn.metrics import precision_score, recall_score

iris = datasets.load_iris()
skf = StratifiedKFold(n_splits=10)
clf = svm.SVC(kernel='linear', C=1)

X = iris.data
y = iris.target
precision_scores = []
recall_scores = []
for train_index, test_index in skf.split(X, y):
    X_train, X_test = X[train_index], X[test_index]
    y_train, y_test = y[train_index], y[test_index]

    y_pred = clf.fit(X_train, y_train).predict(X_test)
    precision_scores.append(precision_score(y_test, y_pred, average='micro'))
    recall_scores.append(recall_score(y_test, y_pred, average='micro'))

print(precision_scores)
print("Recall: %0.2f (+/- %0.2f)" % (np.mean(precision_scores), np.std(precision_scores) * 2))
print(recall_scores)
print("Recall: %0.2f (+/- %0.2f)" % (np.mean(recall_scores), np.std(recall_scores) * 2))