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]

上面有很多很好的答案，所以我只想再加一个例子，在这种情况下，你想通过使用numpy库来指定火车和测试集的确切样本数量。

# set the random seed for the reproducibility
np.random.seed(17)

# e.g. number of samples for the training set is 1000
n_train = 1000

# shuffle the indexes
shuffled_indexes = np.arange(len(data_df))
np.random.shuffle(shuffled_indexes)

# use 'n_train' samples for training and the rest for testing
train_ids = shuffled_indexes[:n_train]
test_ids = shuffled_indexes[n_train:]

train_data = data_df.iloc[train_ids]
train_labels = labels_df.iloc[train_ids]

test_data = data_df.iloc[test_ids]
test_labels = data_df.iloc[test_ids]

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)]

如果你想把它分成训练集、测试集和验证集，你可以使用这个函数:

from sklearn.model_selection import train_test_split
import pandas as pd

def train_test_val_split(df, test_size=0.15, val_size=0.45):
    temp, test = train_test_split(df, test_size=test_size)
    total_items_count = len(df.index)
    val_length = total_items_count * val_size
    new_val_propotion = val_length / len(temp.index) 
    train, val = train_test_split(temp, test_size=new_val_propotion)
    return train, test, val

我认为你还需要一个副本，而不是一个切片的数据框架，如果你想以后添加列。

msk = np.random.rand(len(df)) < 0.8
train, test = df[msk].copy(deep = True), df[~msk].copy(deep = True)

我会使用numpy的randn:

In [11]: df = pd.DataFrame(np.random.randn(100, 2))

In [12]: msk = np.random.rand(len(df)) < 0.8

In [13]: train = df[msk]

In [14]: test = df[~msk]

为了证明这是有效的:

In [15]: len(test)
Out[15]: 21

In [16]: len(train)
Out[16]: 79