python – 使用交叉验证评估Logistic回归
作者:互联网
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()
解决方法:
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.您只需要在代码中删除logreg.fit即可.具体来说,它的作用如下:
它将您的数据集划分为n个折叠,并且在每次迭代中,它将其中一个折叠作为测试集并在其余折叠上训练模型(n-1倍).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,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)
要在多类别分类中绘制ROC,您可以按照this tutorial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的tutorial on cross_validation.
标签:python,scikit-learn,cross-validation,logistic-regression 来源: https://codeday.me/bug/20190519/1134596.html