scikit-learn的LogisticRegression()可以自动将输入数据标准化为z分数吗?

时间:2015-06-24 23:15:05

标签: python scikit-learn logistic-regression

有没有办法让LogisticRegression()的实例自动规范化为适合/培训提供的数据z-scores来构建模型? LinearRegression()有一个normalize=True参数,但这可能对LogisticRegression()没有意义吗?

如果是这样,在调用predict_proba()之前,我是否必须手动标记未标记的输入向量(即重新计算每列的均值,标准偏差)?如果模型已经执行了可能代价高昂的计算,那就太奇怪了。

由于

1 个答案:

答案 0 :(得分:6)

这是你在找什么?

from sklearn.datasets import make_classification
from sklearn.preprocessing import StandardScaler
from sklearn.pipeline import make_pipeline
from sklearn.linear_model import LogisticRegression


X, y = make_classification(n_samples=1000, n_features=100, weights=[0.1, 0.9], random_state=0)
X.shape

# build pipe: first standardize by substracting mean and dividing std
# next do classificaiton
pipe = make_pipeline(StandardScaler(), LogisticRegression(class_weight='auto'))

# fit
pipe.fit(X, y)
# predict
pipe.predict_proba(X)

# to get back mean/std
scaler = pipe.steps[0][1]
scaler.mean_
Out[12]: array([ 0.0313, -0.0334,  0.0145, ..., -0.0247,  0.0191,  0.0439])

scaler.std_
Out[13]: array([ 1.    ,  1.0553,  0.9805, ...,  1.0033,  1.0097,  0.9884])