我有一个数据框架形式的相当大的数据集,我想知道我如何能够将数据框架分成两个随机样本(80%和20%)进行训练和测试。

谢谢!


当前回答

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)]

其他回答

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]

熊猫随机抽样也可以

train=df.sample(frac=0.8,random_state=200)
test=df.drop(train.index)

对于相同的random_state值,您将始终在训练集和测试集中获得相同的确切数据。这带来了一定程度的可重复性,同时还随机分离训练和测试数据。

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
import pandas as pd

from sklearn.model_selection import train_test_split

datafile_name = 'path_to_data_file'

data = pd.read_csv(datafile_name)

target_attribute = data['column_name']

X_train, X_test, y_train, y_test = train_test_split(data, target_attribute, test_size=0.8)

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)]