KFold.split()等类似解释

 

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1	class sklearn.model_selection.KFold(n_splits=5, *, shuffle=False, random_state=None

　　

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>>> import numpy as np >>> from sklearn.model_selection import KFold >>> X = np.array([[1, 2], [3, 4], [1, 2], [3, 4]]) >>> y = np.array([1, 2, 3, 4]) >>> kf = KFold(n_splits=2) >>> kf.get_n_splits(X) 2 >>> print(kf) KFold(n_splits=2, random_state=None, shuffle=False) >>> for train_index, test_index in kf.split(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] TRAIN: [2 3] TEST: [0 1] TRAIN: [0 1] TEST: [2 3]

　　

Methods

get_n_splits([X, y, groups])	Returns the number of splitting iterations in the cross-validator
split(X[, y, groups])	Generate indices to split data into training and test set.

 

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1 2	for flod_idx, (train_idx, val_idx) in enumerate(skf.split(train_jpg, train_jpg)):     ...

　　

 

话不多说，用例子说话：

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from sklearn.model_selection import KFold    kf = KFold(n_splits=5, random_state=43, shuffle=True) a=[[1,2],[3,4],[5,6],[7,8],[9,10]] b=[1,2,3,4,5] for i,j in kf.split(a,b):     print(i,j)   #输出： [0 1 2 4] [3] [0 1 3 4] [2] [0 2 3 4] [1] [1 2 3 4] [0] [0 1 2 3] [4]

不用多说了吧，上面的数都是索引，其实就是从0-4索引里，选一个作为输出，其他都是输入。