我能想到的最简单的方法是将分割分数映射到数组下标，如下所示:

train_set = data[:int((len(data)+1)*train_fraction)]
test_set = data[int((len(data)+1)*train_fraction):int((len(data)+1)*(train_fraction+test_fraction))]
val_set = data[int((len(data)+1)*(train_fraction+test_fraction)):]

其中data = random.shuffle(data)

注意:

函数被编写来处理随机集创建的播种。你不应该依赖集分割，它不会随机化集合。

import numpy as np
import pandas as pd

def train_validate_test_split(df, train_percent=.6, validate_percent=.2, seed=None):
    np.random.seed(seed)
    perm = np.random.permutation(df.index)
    m = len(df.index)
    train_end = int(train_percent * m)
    validate_end = int(validate_percent * m) + train_end
    train = df.iloc[perm[:train_end]]
    validate = df.iloc[perm[train_end:validate_end]]
    test = df.iloc[perm[validate_end:]]
    return train, validate, test

示范

np.random.seed([3,1415])
df = pd.DataFrame(np.random.rand(10, 5), columns=list('ABCDE'))
df

train, validate, test = train_validate_test_split(df)

train

validate

test

回答任意数量的子集:

def _separate_dataset(patches, label_patches, percentage, shuffle: bool = True):
    """
    :param patches: data patches
    :param label_patches: label patches
    :param percentage: list of percentages for each value, example [0.9, 0.02, 0.08] to get 90% train, 2% val and 8% test.
    :param shuffle: Shuffle dataset before split.
    :return: tuple of two lists of size = len(percentage), one with data x and other with labels y.
    """
    x_test = patches
    y_test = label_patches
    percentage = list(percentage)       # need it to be mutable
    assert sum(percentage) == 1., f"percentage must add to 1, but it adds to sum{percentage} = {sum(percentage)}"
    x = []
    y = []
    for i, per in enumerate(percentage[:-1]):
        x_train, x_test, y_train, y_test = train_test_split(x_test, y_test, test_size=1-per, shuffle=shuffle)
        percentage[i+1:] = [value / (1-percentage[i]) for value in percentage[i+1:]]
        x.append(x_train)
        y.append(y_train)
    x.append(x_test)
    y.append(y_test)
    return x, y

这适用于任何比例。在本例中，您应该执行percentage = [train_percentage, val_percentage, test_percentage]。

我能想到的最简单的方法是将分割分数映射到数组下标，如下所示:

train_set = data[:int((len(data)+1)*train_fraction)]
test_set = data[int((len(data)+1)*train_fraction):int((len(data)+1)*(train_fraction+test_fraction))]
val_set = data[int((len(data)+1)*(train_fraction+test_fraction)):]

其中data = random.shuffle(data)

在监督学习的情况下，你可能想拆分X和y(其中X是你的输入，y是基本真理输出)。 你只需要注意在分割之前以同样的方式洗牌X和y。

在这里，X和y在同一个数据帧中，所以我们对它们进行洗牌，将它们分开，并对每个数据帧应用拆分(就像在选择的答案中一样)，或者X和y在两个不同的数据帧中，所以我们洗牌X，将y按洗牌X的方式重新排序，并对每个数据帧应用拆分。

# 1st case: df contains X and y (where y is the "target" column of df)
df_shuffled = df.sample(frac=1)
X_shuffled = df_shuffled.drop("target", axis = 1)
y_shuffled = df_shuffled["target"]

# 2nd case: X and y are two separated dataframes
X_shuffled = X.sample(frac=1)
y_shuffled = y[X_shuffled.index]

# We do the split as in the chosen answer
X_train, X_validation, X_test = np.split(X_shuffled, [int(0.6*len(X)),int(0.8*len(X))])
y_train, y_validation, y_test = np.split(y_shuffled, [int(0.6*len(X)),int(0.8*len(X))])

使用train_test_split非常方便，不需要在划分到几个集后执行重新索引，也不需要编写一些额外的代码。上面的最佳答案没有提到使用train_test_split分隔两次而不改变分区大小将不会给出最初预期的分区:

x_train, x_remain = train_test_split(x, test_size=(val_size + test_size))

那么x_remain中的验证集和测试集的部分就会发生变化，可以算作

new_test_size = np.around(test_size / (val_size + test_size), 2)
# To preserve (new_test_size + new_val_size) = 1.0 
new_val_size = 1.0 - new_test_size

x_val, x_test = train_test_split(x_remain, test_size=new_test_size)

在这种情况下，将保存所有初始分区。