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)]

对我来说，更优雅一点的方法是创建一个随机列，然后按它进行分割，这样我们就可以得到一个符合我们需求的随机分割。

def split_df(df, p=[0.8, 0.2]):
import numpy as np
df["rand"]=np.random.choice(len(p), len(df), p=p)
r = [df[df["rand"]==val] for val in df["rand"].unique()]
return r

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]

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)

我将使用scikit-learn自己的training_test_split，并从索引生成它

from sklearn.model_selection import train_test_split


y = df.pop('output')
X = df

X_train,X_test,y_train,y_test = train_test_split(X.index,y,test_size=0.2)
X.iloc[X_train] # return dataframe train

这是我在需要分割数据帧时所写的。我考虑过使用上面安迪的方法，但不喜欢我不能精确地控制数据集的大小(例如，有时是79，有时是81，等等)。

def make_sets(data_df, test_portion):
    import random as rnd

    tot_ix = range(len(data_df))
    test_ix = sort(rnd.sample(tot_ix, int(test_portion * len(data_df))))
    train_ix = list(set(tot_ix) ^ set(test_ix))

    test_df = data_df.ix[test_ix]
    train_df = data_df.ix[train_ix]

    return train_df, test_df


train_df, test_df = make_sets(data_df, 0.2)
test_df.head()