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:]

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

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]

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:]

在我的例子中，我想用特定的数字分割训练、测试和开发中的数据帧。我在这里分享我的解决方案

首先，为数据帧分配一个唯一的id(如果已经不存在的话)

import uuid
df['id'] = [uuid.uuid4() for i in range(len(df))]

以下是我的分割数字:

train = 120765
test  = 4134
dev   = 2816

分裂函数

def df_split(df, n):
    
    first  = df.sample(n)
    second = df[~df.id.isin(list(first['id']))]
    first.reset_index(drop=True, inplace = True)
    second.reset_index(drop=True, inplace = True)
    return first, second

现在分成培训，测试，开发

train, test = df_split(df, 120765)
test, dev   = df_split(test, 4134)