使用交叉验证评估Logistic回归

Question

我想使用交叉验证来测试/训练我的数据集,并评估逻辑回归模型在整个数据集上的性能,而不仅仅是在测试集上(例如25%).

这些概念对我来说是全新的,我不确定它是否做得对.如果有人能告诉我正确的步骤,我会在错误的地方采取行动,我将不胜感激.我的部分代码如下所示.

另外,如何在当前图形的同一图形上绘制"y2"和"y3"的ROC？

谢谢

import pandas as pd 
Data=pd.read_csv ('C:\\Dataset.csv',index_col='SNo')
feature_cols=['A','B','C','D','E']
X=Data[feature_cols]

Y=Data['Status'] 
Y1=Data['Status1']  # predictions from elsewhere
Y2=Data['Status2'] # predictions from elsewhere

from sklearn.linear_model import LogisticRegression
logreg=LogisticRegression()
logreg.fit(X_train,y_train)

from sklearn.cross_validation import train_test_split
X_train, X_test, y_train, y_test = train_test_split(X, y, test_size=0.2, random_state=0)

from sklearn import metrics, cross_validation
predicted = cross_validation.cross_val_predict(logreg, X, y, cv=10)
metrics.accuracy_score(y, predicted) 

from sklearn.cross_validation import cross_val_score
accuracy = cross_val_score(logreg, X, y, cv=10,scoring='accuracy')
print (accuracy)
print (cross_val_score(logreg, X, y, cv=10,scoring='accuracy').mean())

from nltk import ConfusionMatrix 
print (ConfusionMatrix(list(y), list(predicted)))
#print (ConfusionMatrix(list(y), list(yexpert)))

# sensitivity:
print (metrics.recall_score(y, predicted) )

import matplotlib.pyplot as plt 
probs = logreg.predict_proba(X)[:, 1] 
plt.hist(probs) 
plt.show()

# use 0.5 cutoff for predicting 'default' 
import numpy as np 
preds = np.where(probs > 0.5, 1, 0) 
print (ConfusionMatrix(list(y), list(preds)))

# check accuracy, sensitivity, specificity 
print (metrics.accuracy_score(y, predicted)) 

#ROC CURVES and AUC 
# plot ROC curve 
fpr, tpr, thresholds = metrics.roc_curve(y, probs) 
plt.plot(fpr, tpr) 
plt.xlim([0.0, 1.0]) 
plt.ylim([0.0, 1.0]) 
plt.xlabel('False Positive Rate') 
plt.ylabel('True Positive Rate)') 
plt.show()

# calculate AUC 
print (metrics.roc_auc_score(y, probs))

# use AUC as evaluation metric for cross-validation 
from sklearn.cross_validation import cross_val_score 
logreg = LogisticRegression() 
cross_val_score(logreg, X, y, cv=10, scoring='roc_auc').mean() 

Answer 1

你得到它几乎是正确的.cross_validation.cross_val_predict为您提供整个数据集的预测.您只需要logreg.fit在代码中删除.具体来说,它的作用如下:它将您的数据集划分为n折叠,并在每次迭代中将其中一个折叠作为测试集并在其余折叠(n-1折叠)上训练模型.因此,最终您将获得整个数据的预测.

让我们用sklearn,iris中的一个内置数据集来说明这一点.该数据集包含150个具有4个特征的训练样本.iris['data']是X和iris['target']是y

In [15]: iris['data'].shape
Out[15]: (150, 4)

要通过交叉验证获得整个集合的预测,您可以执行以下操作:

from sklearn.linear_model import LogisticRegression
from sklearn import metrics, cross_validation
from sklearn import datasets
iris = datasets.load_iris()
predicted = cross_validation.cross_val_predict(LogisticRegression(), iris['data'], iris['target'], cv=10)
print metrics.accuracy_score(iris['target'], predicted)

Out [1] : 0.9537

print metrics.classification_report(iris['target'], predicted) 

Out [2] :
                     precision    recall  f1-score   support

                0       1.00      1.00      1.00        50
                1       0.96      0.90      0.93        50
                2       0.91      0.96      0.93        50

      avg / total       0.95      0.95      0.95       150

所以,回到你的代码.你需要的只是这个:

from sklearn import metrics, cross_validation
logreg=LogisticRegression()
predicted = cross_validation.cross_val_predict(logreg, X, y, cv=10)
print metrics.accuracy_score(y, predicted)
print metrics.classification_report(y, predicted) 

要在多类分类中绘制ROC,您可以按照本教程中的内容进行以下操作:

一般来说,sklearn有非常好的教程和文档.我强烈建议阅读他们关于cross_validation的教程.