Question

作为交叉验证的一部分，需要将火车阵列分成N折。然后，每折叠进行一次实验。后者意味着我需要将N-1折合并为一个阵列，然后将其余折用于验证。

假设我将binary_train_X作为初始数组，并希望将其拆分为5折。我有一些有效的代码：

num_folds = 5
train_folds_X = []

# Split the training data in folds
step = int(binary_train_X.shape[0] / num_folds)
for i in range(num_folds):
    train_folds_X.append(binary_train_X[i*step:(i+1)*step])

# Prepare train and test arrays
for i in range(num_folds):
    if i == 0:
        train_temp_X = np.concatenate((train_folds_X[1:]))
    elif i == num_folds - 1:
        train_temp_X = np.concatenate((train_folds_X[0:(num_folds - 1)]))
    else:
        train_temp_X1 = np.concatenate((train_folds_X[0:i]))
        train_temp_X2 = np.concatenate((train_folds_X[(i+1):(num_folds)]))
        train_temp_X = np.concatenate((train_temp_X1, train_temp_X2))
    test_temp_X = train_folds_X[i]

    # Run classifier based on train_temp_X and test_temp_X
    ...
    pass

问题-如何以更优雅的方式做到这一点？

Answer 1

为什么不这样做：

splits = np.array_split(binary_train_X, num_folds)

for i in range(num_folds):
    fold_train_X = np.concatenate([*splits[:i], *splits[i + 1:]])
    fold_test_X = splits[i]
    # use your folds here

如果要使用预构建的解决方案，则可以使用sklearn.model_selection.KFold：

kf = KFold(num_folds)

for train_index, test_index in kf.split(binary_train_X):
    fold_train_X = binary_train_X[train_index]
    fold_test_X = binary_test_X[train_index]
    # use your folds here

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