Question

我知道sklearn有很好的方法来获得交叉验证分数：

 from sklearn.model_selection import cross_val_score
 clf = svm.SVC(kernel='linear', C=1)
 scores = cross_val_score(clf, iris.data, iris.target, cv=5)
 scores

我想知道具有特定训练和测试集的分数：

train_list = [train1, train2, train3] # train1,2,3 is the training data sets
test_list = [test1, test2, test3] # # test1,2,3 is the test data sets
clf = svm.SVC(kernel='linear', C=1)
scores = some_nice_method(clf, train_list, test_list)

是否有这样的方法在python中提供了大量特定的分离数据集？

Answer 1

这正是两行代码：

for tr, te in zip(train_list, test_list):
    svm.SVC(kernel='linear', C=1).train(X[tr, :], y[tr]).score(X[te, :], y[te])

请参阅sklearn.svn.SVC.score：

score(X, y, sample_weight=None)

返回给定测试数据和标签的平均准确度。

Answer 2

我的建议是使用下面的kfold cross validation。在这种情况下，您将获得特定拆分的列车，测试指数以及准确度得分。在Sklearn的新版本中，有一些变化。

from sklearn import svm
from sklearn import datasets
from sklearn.model_selection import KFold
from sklearn.metrics import accuracy_score

iris = datasets.load_iris()
X = iris.data
y = iris.target

clf = svm.SVC(kernel='linear', C=1)
kf = KFold(n_splits=5)

for train_index, test_index in kf.split(range(len(X))):
    print("TRAIN:", train_index, "TEST:", test_index)
    X_train, X_test = X[train_index], X[test_index]
    y_train, y_test = y[train_index], y[test_index]
    clf.fit(X_train, y_train)
    y_pred = clf.predict(X_test)
    score = accuracy_score(y_test, y_pred)
    print score

使用Python与特定数据集列表进行交叉验证

2 个答案: