带有索引的 Scikit-learn train_test_split

Question

使用 train_test_split() 时如何获取数据的原始索引？

我有的是下面的

from sklearn.cross_validation import train_test_split
import numpy as np
data = np.reshape(np.randn(20),(10,2)) # 10 training examples
labels = np.random.randint(2, size=10) # 10 labels
x1, x2, y1, y2 = train_test_split(data, labels, size=0.2)

但这并没有给出原始数据的索引。一种解决方法是将索引添加到数据中（例如 data = [(i, d) for i, d in enumerate(data)]），然后将它们传递到 train_test_split 中，然后再次展开。有没有更清洁的解决方案？

Answer 1

Scikit 学习与 Pandas 配合得很好，所以我建议您使用它。这是一个例子：

In [1]: 
import pandas as pd
import numpy as np
from sklearn.model_selection import train_test_split
data = np.reshape(np.random.randn(20),(10,2)) # 10 training examples
labels = np.random.randint(2, size=10) # 10 labels

In [2]: # Giving columns in X a name
X = pd.DataFrame(data, columns=['Column_1', 'Column_2'])
y = pd.Series(labels)

In [3]:
X_train, X_test, y_train, y_test = train_test_split(X, y, 
                                                    test_size=0.2, 
                                                    random_state=0)

In [4]: X_test
Out[4]:

     Column_1    Column_2
2   -1.39       -1.86
8    0.48       -0.81
4   -0.10       -1.83

In [5]: y_test
Out[5]:

2    1
8    1
4    1
dtype: int32

您可以在 DataFrame/Series 上直接调用任何 scikit 函数，它会起作用。

假设您想进行 LogisticRegression，下面是您如何以一种很好的方式检索系数：

In [6]: 
from sklearn.linear_model import LogisticRegression

model = LogisticRegression()
model = model.fit(X_train, y_train)

# Retrieve coefficients: index is the feature name (['Column_1', 'Column_2'] here)
df_coefs = pd.DataFrame(model.coef_[0], index=X.columns, columns = ['Coefficient'])
df_coefs
Out[6]:
            Coefficient
Column_1    0.076987
Column_2    -0.352463

Answer 2

您可以像 Julien 所说的那样使用 pandas 数据帧或系列，但是如果您想将自己限制为 numpy，您可以传递一个额外的索引数组：

from sklearn.model_selection import train_test_split
import numpy as np
n_samples, n_features, n_classes = 10, 2, 2
data = np.random.randn(n_samples, n_features)  # 10 training examples
labels = np.random.randint(n_classes, size=n_samples)  # 10 labels
indices = np.arange(n_samples)
(
    data_train,
    data_test,
    labels_train,
    labels_test,
    indices_train,
    indices_test,
) = train_test_split(data, labels, indices, test_size=0.2)

Answer 3

docs 提及 train_test_split 只是在 shuffle split 之上的便利功能。

我只是重新整理了他们的一些代码来制作我自己的示例。注意实际的解决方案是中间的代码块。剩下的就是导入，并设置一个可运行的例子。

from sklearn.model_selection import ShuffleSplit
from sklearn.utils import safe_indexing, indexable
from itertools import chain
import numpy as np
X = np.reshape(np.random.randn(20),(10,2)) # 10 training examples
y = np.random.randint(2, size=10) # 10 labels
seed = 1

cv = ShuffleSplit(random_state=seed, test_size=0.25)
arrays = indexable(X, y)
train, test = next(cv.split(X=X))
iterator = list(chain.from_iterable((
    safe_indexing(a, train),
    safe_indexing(a, test),
    train,
    test
    ) for a in arrays)
)
X_train, X_test, train_is, test_is, y_train, y_test, _, _  = iterator

print(X)
print(train_is)
print(X_train)

现在我有了实际的索引：train_is, test_is

Answer 4

这是最简单的解决方案（Jibwa 在另一个答案中让它看起来很复杂），无需自己生成索引 - 只需使用 ShuffleSplit 对象生成 1 个拆分。

import numpy as np 
from sklearn.model_selection import ShuffleSplit # or StratifiedShuffleSplit
sss = ShuffleSplit(n_splits=1, test_size=0.1)

data_size = 100
X = np.reshape(np.random.rand(data_size*2),(data_size,2))
y = np.random.randint(2, size=data_size)

sss.get_n_splits(X, y)
train_index, test_index = next(sss.split(X, y)) 

X_train, X_test = X[train_index], X[test_index] 
y_train, y_test = y[train_index], y[test_index]

Answer 5

如果您使用的是 pandas，您可以通过调用您希望模拟的任何数组的 .index 来访问索引。 train_test_split 将 pandas 索引转移到新数据帧。

在您的代码中，您只需使用 x1.index 返回的数组是与 x.

中原始位置相关的索引

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