如何计算scikit-learn cross_val_predict准确度分数？

Question

使用下面代码中所示的k -fold 方法cross_val_predict(参见doc,v0.18)是否计算每次折叠的准确度并最终平均它们？

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,不是的!

根据交叉验证文档页面,cross_val_predict不返回任何分数,只返回基于此处描述的特定策略的标签:

函数cross_val_predict具有与cross_val_score类似的接口,但是对于输入中的每个元素,返回当它在测试集中时为该元素获得的预测.只能使用将所有元素分配给测试集一次的交叉验证策略(否则会引发异常).

因此,通过调用,accuracy_score(labels, ypred) 您只需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:

然后可以使用这些预测来评估分类器:

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

注意,该计算的结果可能与使用cross_val_score获得的结果略有不同,因为元素以不同方式分组.

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)

编辑:

对于计算Cohen Kappa coefficient和混淆矩阵,我假设你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)

编辑2:

什么cross_val_predict回报？

KFold将数据拆分为k个部分,然后对于i = 1..k迭代执行此操作:将除了部分之外的所有部分作为训练数据,使用它们拟合模型,然后预测第i个部分的标签(测试数据) ).在每次迭代中,预测了第i部分数据的标签.最后,cross_val_predict合并所有部分预测的标签并将它们作为一个整体返回.

此代码逐步显示此过程:

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']

Answer 2

正如你可以从代码中看到cross_val_predict的github上,该函数计算每个倍的预测和连接它们.预测是基于从其他折叠中学习的模型进行的.

以下是代码与代码中提供的示例的组合

from sklearn import datasets, linear_model
from sklearn.model_selection import cross_val_predict, KFold
from sklearn.metrics import accuracy_score

diabetes = datasets.load_diabetes()
X = diabetes.data[:400]
y = diabetes.target[:400]
cv = KFold(n_splits=20)
lasso = linear_model.Lasso()
y_pred = cross_val_predict(lasso, X, y, cv=cv)
accuracy = accuracy_score(y_pred.astype(int), y.astype(int))

print(accuracy)
# >>> 0.0075

最后,回答你的问题:"不,每次折叠的准确性不是平均的"