python - scikit-learn cross_val_predict 准确度分数是如何计算的？

Question

cross_val_predict(参见 doc，v0.18)是否使用如下代码所示的 k-fold 方法计算每个折叠的准确度并最终平均它们或不？

cv = KFold(len(labels), n_folds=20)
clf = SVC()
ypred = cross_val_predict(clf, td, labels, cv=cv)
accuracy = accuracy_score(labels, ypred)
print accuracy

Answer 1

不，它没有!

根据cross validation doc页面，cross_val_predict 不返回任何分数，而只返回基于此处描述的特定策略的标签:

The function cross_val_predict has a similar interface to cross_val_score, but returns, for each element in the input, the prediction that was obtained for that element when it was in the test set. Only cross-validation strategies that assign all elements to a test set exactly once can be used (otherwise, an exception is raised).

因此，通过调用 accuracy_score(labels, ypred) 您只是在计算上述特定策略预测的标签与真实标签相比的准确度分数。这再次在同一文档页面中指定:

These prediction can then be used to evaluate the classifier:

predicted = cross_val_predict(clf, iris.data, iris.target, cv=10) 
metrics.accuracy_score(iris.target, predicted)

Note that the result of this computation may be slightly different from those obtained using cross_val_score as the elements are grouped in different ways.

如果您需要不同折叠的准确度分数，您应该尝试:

>>> scores = cross_val_score(clf, X, y, cv=cv)
>>> scores                                              
array([ 0.96...,  1.  ...,  0.96...,  0.96...,  1.        ])

然后对于所有折叠的平均准确度，使用 scores.mean():

>>> print("Accuracy: %0.2f (+/- %0.2f)" % (scores.mean(), scores.std() * 2))
Accuracy: 0.98 (+/- 0.03)

---

如何计算每一折的 Cohen kappa 系数和混淆矩阵？

为了计算 Cohen Kappa 系数 和混淆矩阵，我假设您的意思是真实标签和每个折叠的预测标签之间的 kappa 系数和混淆矩阵:

from sklearn.model_selection import KFold
from sklearn.svm.classes import SVC
from sklearn.metrics.classification import cohen_kappa_score
from sklearn.metrics import confusion_matrix

cv = KFold(len(labels), n_folds=20)
clf = SVC()
for train_index, test_index in cv.split(X):
    clf.fit(X[train_index], labels[train_index])
    ypred = clf.predict(X[test_index])
    kappa_score = cohen_kappa_score(labels[test_index], ypred)
    confusion_matrix = confusion_matrix(labels[test_index], ypred)

---

cross_val_predict 返回什么？

它使用 KFold 将数据拆分为 k 部分，然后进行 i=1..k 迭代:

• 将第i个部分作为测试数据，其他部分作为训练数据
• 使用训练数据训练模型(除了 i'th 的所有部分)
• 然后通过使用这个经过训练的模型，预测第 i'th 部分(测试数据)的标签

在每次迭代中，预测第 i'th 部分数据的标签。最后 cross_val_predict 合并所有部分预测的标签，并作为最终结果返回。

这段代码一步一步地展示了这个过程:

X = np.array([[0], [1], [2], [3], [4], [5]])
labels = np.array(['a', 'a', 'a', 'b', 'b', 'b'])

cv = KFold(len(labels), n_folds=3)
clf = SVC()
ypred_all = np.chararray((labels.shape))
i = 1
for train_index, test_index in cv.split(X):
    print("iteration", i, ":")
    print("train indices:", train_index)
    print("train data:", X[train_index])
    print("test indices:", test_index)
    print("test data:", X[test_index])
    clf.fit(X[train_index], labels[train_index])
    ypred = clf.predict(X[test_index])
    print("predicted labels for data of indices", test_index, "are:", ypred)
    ypred_all[test_index] = ypred
    print("merged predicted labels:", ypred_all)
    i = i+1
    print("=====================================")
y_cross_val_predict = cross_val_predict(clf, X, labels, cv=cv)
print("predicted labels by cross_val_predict:", y_cross_val_predict)

结果是:

iteration 1 :
train indices: [2 3 4 5]
train data: [[2] [3] [4] [5]]
test indices: [0 1]
test data: [[0] [1]]
predicted labels for data of indices [0 1] are: ['b' 'b']
merged predicted labels: ['b' 'b' '' '' '' '']
=====================================
iteration 2 :
train indices: [0 1 4 5]
train data: [[0] [1] [4] [5]]
test indices: [2 3]
test data: [[2] [3]]
predicted labels for data of indices [2 3] are: ['a' 'b']
merged predicted labels: ['b' 'b' 'a' 'b' '' '']
=====================================
iteration 3 :
train indices: [0 1 2 3]
train data: [[0] [1] [2] [3]]
test indices: [4 5]
test data: [[4] [5]]
predicted labels for data of indices [4 5] are: ['a' 'a']
merged predicted labels: ['b' 'b' 'a' 'b' 'a' 'a']
=====================================
predicted labels by cross_val_predict: ['b' 'b' 'a' 'b' 'a' 'a']