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

您可以使用df.as_matrix()函数并创建Numpy-array并传递它。

Y = df.pop()
X = df.as_matrix()
x_train, x_test, y_train, y_test = train_test_split(X, Y, test_size = 0.2)
model.fit(x_train, y_train)
model.test(x_test)

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

可以使用~(波浪符)排除使用df.sample()采样的行，让pandas单独处理索引的采样和过滤，以获得两个集。

train_df = df.sample(frac=0.8, random_state=100)
test_df = df[~df.index.isin(train_df.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:]

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]