问题描述
我正在尝试使用10倍交叉验证来估计逻辑回归.
I am trying to estimate a logistic regression, using the 10-fold cross-validation.
#import libraries
library(car); library(caret); library(e1071); library(verification)
#data import and preparation
data(Chile)
chile <- na.omit(Chile) #remove "na's"
chile <- chile[chile$vote == "Y" | chile$vote == "N" , ] #only "Y" and "N" required
chile$vote <- factor(chile$vote) #required to remove unwanted levels
chile$income <- factor(chile$income) # treat income as a factor
目标是估计glm-模型，该模型根据相关的解释变量，对投票结果"Y"或"N"进行预测，并基于最终模型计算混淆矩阵和ROC曲线，以掌握模型行为不同的阈值水平.
Goal is to estimate a glm - model that predicts to outcome of vote "Y" or "N" depended on relevant explanatory variables and, based on the final model, compute a confusion matrix and ROC curve to grasp the models behaviour for different threshold levels.
模型选择会导致:
res.chileIII <- glm(vote ~
sex +
education +
statusquo ,
family = binomial(),
data = chile)
#prediction
chile.pred <- predict.glm(res.chileIII, type = "response")
生成:
> head(chile.pred)
1 2 3 4 5 6
0.974317861 0.008376988 0.992720134 0.095014139 0.040348115 0.090947144
比较观察值和估计值:
chile.v <- ifelse(chile$vote == "Y", 1, 0) #to compare the two arrays
chile.predt <- function(t) ifelse(chile.pred > t , 1,0) #t is the threshold for which the confusion matrix shall be computed
t = 0.3时的混淆矩阵:
confusion matrix for t = 0.3:
confusionMatrix(chile.predt(0.3), chile.v)
> confusionMatrix(chile.predt(0.3), chile.v)
Confusion Matrix and Statistics
Reference
Prediction 0 1
0 773 44
1 94 792
Accuracy : 0.919
95% CI : (0.905, 0.9315)
No Information Rate : 0.5091
P-Value [Acc > NIR] : < 2.2e-16
和Roc曲线:
roc.plot(chile.v, chile.pred)
这似乎是一个合理的模型.
which seems as a reasonable model.
现在，我要使用10倍的交叉验证估计来测试性能差异，而不是使用正常的" predict.glm()函数.
Now instead of using the "normal" predict.glm() function I want to test out the performance difference to a 10-fold cross-validation estimation.
tc <- trainControl("cv", 10, savePredictions=T) #"cv" = cross-validation, 10-fold
fit <- train(chile$vote ~ chile$sex +
chile$education +
chile$statusquo ,
data = chile ,
method = "glm" ,
family = binomial ,
trControl = tc)
> summary(fit)$coef
Estimate Std. Error z value Pr(>|z|)
(Intercept) 1.0152702 0.1889646 5.372805 7.752101e-08
`chile$sexM` -0.5742442 0.2022308 -2.839549 4.517738e-03
`chile$educationPS` -1.1074079 0.2914253 -3.799971 1.447128e-04
`chile$educationS` -0.6827546 0.2217459 -3.078996 2.076993e-03
`chile$statusquo` 3.1689305 0.1447911 21.886224 3.514468e-106
所有参数均有效.
fitpred <- ifelse(fit$pred$pred == "Y", 1, 0) #to compare with chile.v
> confusionMatrix(fitpred,chile.v)
Confusion Matrix and Statistics
Reference
Prediction 0 1
0 445 429
1 422 407
Accuracy : 0.5003
95% CI : (0.4763, 0.5243)
No Information Rate : 0.5091
P-Value [Acc > NIR] : 0.7738
显然与先前的混淆矩阵有很大不同.我的期望是，交叉验证的结果不会比第一个模型差很多.但是结果显示了其他内容.
which is obviously very different from the previous confusion matrix. My expectation was that the cross validated results should not perform much worse then the first model. However the results show something else.
我的假设是train()参数的设置有误，但我无法弄清楚它是什么.
My assumption is that there is a mistake with the settings of the train() parameters but I can't figure it out what it is.
非常感谢您的帮助.
推荐答案
您正在尝试使用混淆矩阵来了解样本内拟合.使用glm()函数的第一种方法很好.
You are trying to get an idea of the in-sample fit using a confusion matrix. Your first approach using the glm() function is fine.
使用train()的第二种方法的问题在于返回的对象.您正在尝试通过fit$pred$pred从中提取样本内拟合值.但是，fit$pred不包含与或chile$vote对齐的拟合值.它包含不同(10)折的观察值和拟合值:
The problem with the second approach using train() lies in the returned object. You are trying to extract the in-sample fitted values from it by fit$pred$pred. However, fit$pred does not contain the fitted values that are aligned to chile.v or chile$vote. It contains the observations and fitted values of the different (10) folds:
> head(fit$pred)
pred obs rowIndex parameter Resample
1 N N 2 none Fold01
2 Y Y 20 none Fold01
3 Y Y 28 none Fold01
4 N N 38 none Fold01
5 N N 55 none Fold01
6 N N 66 none Fold01
> tail(fit$pred)
pred obs rowIndex parameter Resample
1698 Y Y 1592 none Fold10
1699 Y N 1594 none Fold10
1700 N N 1621 none Fold10
1701 N N 1656 none Fold10
1702 N N 1671 none Fold10
1703 Y Y 1689 none Fold10
因此，由于褶皱的随机性，并且由于您预测的是0或1，因此您可以获得大约50％的准确度.
So, due to the randomness of the folds, and because you are predicting 0 or 1, you get an accuracy of roughly 50%.
您要查找的样本内拟合值在fit$finalModel$fitted.values中.使用这些:
The in-sample fitted values you are looking for are in fit$finalModel$fitted.values. Using those:
fitpred <- fit$finalModel$fitted.values
fitpredt <- function(t) ifelse(fitpred > t , 1,0)
> confusionMatrix(fitpredt(0.3),chile.v)
Confusion Matrix and Statistics
Reference
Prediction 0 1
0 773 44
1 94 792
Accuracy : 0.919
95% CI : (0.905, 0.9315)
No Information Rate : 0.5091
P-Value [Acc > NIR] : < 2.2e-16
Kappa : 0.8381
Mcnemar's Test P-Value : 3.031e-05
Sensitivity : 0.8916
Specificity : 0.9474
Pos Pred Value : 0.9461
Neg Pred Value : 0.8939
Prevalence : 0.5091
Detection Rate : 0.4539
Detection Prevalence : 0.4797
Balanced Accuracy : 0.9195
'Positive' Class : 0
现在精度在预期值附近.将阈值设置为0.5可以产生与10倍交叉验证的估计结果相同的准确性:
Now the accuracy is around the expected value. Setting the threshold to 0.5 yields about the same accuracy as the estimate from the 10-fold cross validation:
> confusionMatrix(fitpredt(0.5),chile.v)
Confusion Matrix and Statistics
Reference
Prediction 0 1
0 809 64
1 58 772
Accuracy : 0.9284
95% CI : (0.9151, 0.9402)
[rest of the output omitted]
> fit
Generalized Linear Model
1703 samples
7 predictors
2 classes: 'N', 'Y'
No pre-processing
Resampling: Cross-Validated (10 fold)
Summary of sample sizes: 1533, 1532, 1532, 1533, 1532, 1533, ...
Resampling results
Accuracy Kappa Accuracy SD Kappa SD
0.927 0.854 0.0134 0.0267
另外，关于您的期望交叉验证的结果应该不会比第一个模型差很多"，请检查summary(res.chileIII)和summary(fit).拟合的模型和系数完全相同，因此它们将给出相同的结果.
Additionally, regarding your expectation "that the cross validated results should not perform much worse than the first model," please check summary(res.chileIII) and summary(fit). The fitted models and coefficients are exactly the same so they will give the same results.
P.S.我知道我对这个问题的答案太晚了-即这是一个很老的问题.反正可以回答这些问题吗?我是新来的，在帮助中未找到有关最新答案"的任何信息.
P.S. I know my answer to this question is late--i.e. this is quite an old question. Is it OK to answer these questions anyway? I am new here and did not find anything about "late answers" in the help.
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