有很多有效的答案。又多了一个。 从sklearn。交叉验证导入train_test_split

#gets a random 80% of the entire set
X_train = X.sample(frac=0.8, random_state=1)
#gets the left out portion of the dataset
X_test = X.loc[~df_model.index.isin(X_train.index)]

像这样从df中选择range row

row_count = df.shape[0]
split_point = int(row_count*1/5)
test_data, train_data = df[:split_point], df[split_point:]

shuffle = np.random.permutation(len(df))
test_size = int(len(df) * 0.2)
test_aux = shuffle[:test_size]
train_aux = shuffle[test_size:]
TRAIN_DF =df.iloc[train_aux]
TEST_DF = df.iloc[test_aux]

你也可以考虑分层划分为训练集和测试集。设定划分也随机生成训练集和测试集，但保留了原始的类比例。这使得训练集和测试集更好地反映原始数据集的属性。

import numpy as np  

def get_train_test_inds(y,train_proportion=0.7):
    '''Generates indices, making random stratified split into training set and testing sets
    with proportions train_proportion and (1-train_proportion) of initial sample.
    y is any iterable indicating classes of each observation in the sample.
    Initial proportions of classes inside training and 
    testing sets are preserved (stratified sampling).
    '''

    y=np.array(y)
    train_inds = np.zeros(len(y),dtype=bool)
    test_inds = np.zeros(len(y),dtype=bool)
    values = np.unique(y)
    for value in values:
        value_inds = np.nonzero(y==value)[0]
        np.random.shuffle(value_inds)
        n = int(train_proportion*len(value_inds))

        train_inds[value_inds[:n]]=True
        test_inds[value_inds[n:]]=True

    return train_inds,test_inds

df[train_inds]和df[test_inds]为您提供原始DataFrame df的训练和测试集。

如果你希望有一个数据帧和两个数据帧(不是numpy数组)，这应该可以做到:

def split_data(df, train_perc = 0.8):

   df['train'] = np.random.rand(len(df)) < train_perc

   train = df[df.train == 1]

   test = df[df.train == 0]

   split_data ={'train': train, 'test': test}

   return split_data

不需要转换为numpy。只要用pandas df来做拆分，它就会返回一个pandas df。

from sklearn.model_selection import train_test_split

train, test = train_test_split(df, test_size=0.2)

如果你想把x和y分开

X_train, X_test, y_train, y_test = train_test_split(df[list_of_x_cols], df[y_col],test_size=0.2)

如果要分割整个df

X, y = df[list_of_x_cols], df[y_col]