Python 带有索引的 Scikit-learn train_test_split
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Scikit-learn train_test_split with indices
提问by CentAu
How do I get the original indices of the data when using train_test_split()?
使用 train_test_split() 时如何获取数据的原始索引？
What I have is the following
我所拥有的是以下内容
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)
But this does not give the indices of the original data.
One workaround is to add the indices to data (e.g. data = [(i, d) for i, d in enumerate(data)]) and then pass them inside train_test_splitand then expand again.
Are there any cleaner solutions?
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)]），然后将它们传递到内部train_test_split，然后再次展开。有没有更清洁的解决方案？
采纳答案by Julien Marrec
Scikit learn plays really well with Pandas, so I suggest you use it. Here's an example:
Scikit learn 与 Pandas 配合得很好，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
You can directly call any scikit functions on DataFrame/Series and it will work.
您可以直接调用 DataFrame/Series 上的任何 scikit 函数，它会起作用。
Let's say you wanted to do a LogisticRegression, here's how you could retrieve the coefficients in a nice way:
假设你想做一个 LogisticRegression，下面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
回答by ogrisel
You can use pandas dataframes or series as Julien said but if you want to restrict your-self to numpy you can pass an additional array of indices:
您可以使用 Julien 所说的 Pandas 数据帧或系列，但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)
x1, x2, y1, y2, idx1, idx2 = train_test_split(
data, labels, indices, test_size=0.2)
回答by Jibwa
The docsmention train_test_split is just a convenience function on top of shuffle split.
该文档提到train_test_split只是在洗牌分裂的顶部方便的功能。
I just rearranged some of their code to make my own example. Note the actual solution is the middle block of code. The rest is imports, and setup for a runnable example.
我只是重新排列了他们的一些代码来制作我自己的例子。请注意，实际的解决方案是中间的代码块。其余的是导入和可运行示例的设置。
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)
Now I have the actual indexes: train_is, test_is
现在我有了实际的索引： train_is, test_is
回答by m3h0w
Here's the simplest solution (Jibwa made it seem complicated in another answer), without having to generate indices yourself - just using the ShuffleSplit object to generate 1 split.
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]
