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)

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)

def train_val_test_split(X, y, train_size, val_size, test_size):
    X_train_val, X_test, y_train_val, y_test = train_test_split(X, y, test_size = test_size)
    relative_train_size = train_size / (val_size + train_size)
    X_train, X_val, y_train, y_val = train_test_split(X_train_val, y_train_val,
                                                      train_size = relative_train_size, test_size = 1-relative_train_size)
    return X_train, X_val, X_test, y_train, y_val, y_test

在这里，我们使用sklearn的train_test_split将数据分割2次

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))])

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

Numpy解决方案。我们将首先洗牌整个数据集(df。Sample (frac=1, random_state=42))，然后将我们的数据集分成以下部分:

60% -列车集， 20% -验证集， 20% -测试装置

In [305]: train, validate, test = \
              np.split(df.sample(frac=1, random_state=42), 
                       [int(.6*len(df)), int(.8*len(df))])

In [306]: train
Out[306]:
          A         B         C         D         E
0  0.046919  0.792216  0.206294  0.440346  0.038960
2  0.301010  0.625697  0.604724  0.936968  0.870064
1  0.642237  0.690403  0.813658  0.525379  0.396053
9  0.488484  0.389640  0.599637  0.122919  0.106505
8  0.842717  0.793315  0.554084  0.100361  0.367465
7  0.185214  0.603661  0.217677  0.281780  0.938540

In [307]: validate
Out[307]:
          A         B         C         D         E
5  0.806176  0.008896  0.362878  0.058903  0.026328
6  0.145777  0.485765  0.589272  0.806329  0.703479

In [308]: test
Out[308]:
          A         B         C         D         E
4  0.521640  0.332210  0.370177  0.859169  0.401087
3  0.333348  0.964011  0.083498  0.670386  0.169619

[int(.6*len(df))， int(.8*len(df))] -是numpy.split()的indices_or_sections数组。

下面是一个np.split()使用的小演示-让我们把20个元素的数组分成以下部分:80%，10%，10%:

In [45]: a = np.arange(1, 21)

In [46]: a
Out[46]: array([ 1,  2,  3,  4,  5,  6,  7,  8,  9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20])

In [47]: np.split(a, [int(.8 * len(a)), int(.9 * len(a))])
Out[47]:
[array([ 1,  2,  3,  4,  5,  6,  7,  8,  9, 10, 11, 12, 13, 14, 15, 16]),
 array([17, 18]),
 array([19, 20])]