Scikit-learn带索引的train_test_split

Question

使用train_testrongplit()时如何获得数据的原始索引?

我有以下内容

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(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

你可以像Julien说的那样使用pandas dataframes或series，但如果你想限制自己使用numpy，你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 2

Scikit learn和Pandas玩得很好，所以我建议你使用它。下面是一个例子:

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

你可以直接调用DataFrame/Series上的任何scikit函数，它将会工作。

让我们假设你想做一个逻辑回归，这里是你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 3

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

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 4

文档中提到train_testrongplit只是在shuffle split之上的一个方便函数。

我只是重新安排了他们的一些代码，使我自己的例子。注意，实际的解决方案是中间的代码块。剩下的是导入，以及可运行示例的设置。

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 5

如果使用的是pandas，则可以通过调用想要模拟的任何数组的.index来访问索引。train_testrongplit将pandas索引传递到新的数据帧。

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。