R Caret软件包中的Logistic回归调整参数网格? [英] Logistic Regression Tuning Parameter Grid in R Caret Package?
问题描述
我正在尝试使用caret package在R中拟合逻辑回归模型.我已经做了以下事情:
I am trying to fit a logistic regression model in R using the caret package. I have done the following:
model <- train(dec_var ~., data=vars, method="glm", family="binomial",
trControl = ctrl, tuneGrid=expand.grid(C=c(0.001, 0.01, 0.1, 1,10,100, 1000)))
但是，我不确定该模型的调整参数应该是什么，并且我很难找到它.我假设它是C，因为C是sklearn中使用的参数.目前，我收到以下错误-
However, I am unsure what the tuning parameter should be for this model and I am having a difficult time finding it. I assumed it is C because C is the parameter used in sklearn. Currently, I am getting the following error -
错误:调整参数网格应具有列参数
Error: The tuning parameter grid should have columns parameter
您对如何解决此问题有任何建议吗?
Do you have any suggestions on how to fix this?
推荐答案
Per Max Kuhn's web-book - search for method = 'glm' here ,there is no tuning parameter glm within caret.
通过测试一些基本的train调用，我们可以轻松地验证这种情况.首先，让我们从一个方法(rpart)开始，该方法的确对每个网络书都有一个调整参数(cp).
We can easily verify this is the case by testing out a few basic train calls. First off, let's start with a method (rpart) that does have a tuning parameter (cp) per the web book.
library(caret)
data(GermanCredit)
# Check tuning parameter via `modelLookup` (matches up with the web book)
modelLookup('rpart')
# model parameter label forReg forClass probModel
#1 rpart cp Complexity Parameter TRUE TRUE TRUE
# Observe that the `cp` parameter is tuned
set.seed(1)
model_rpart <- train(Class ~., data=GermanCredit, method='rpart')
model_rpart
#CART
#1000 samples
# 61 predictor
# 2 classes: 'Bad', 'Good'
#No pre-processing
#Resampling: Bootstrapped (25 reps)
#Summary of sample sizes: 1000, 1000, 1000, 1000, 1000, 1000, ...
#Resampling results across tuning parameters:
# cp Accuracy Kappa
# 0.01555556 0.7091276 0.2398993
# 0.03000000 0.7025574 0.1950021
# 0.04444444 0.6991700 0.1316720
#Accuracy was used to select the optimal model using the largest value.
#The final value used for the model was cp = 0.01555556.
我们看到已调整cp参数.现在让我们尝试glm.
We see that the cp parameter was tuned. Now let's try glm.
# Check tuning parameter via `modelLookup` (shows a parameter called 'parameter')
modelLookup('glm')
# model parameter label forReg forClass probModel
#1 glm parameter parameter TRUE TRUE TRUE
# Try out the train function to see if 'parameter' gets tuned
set.seed(1)
model_glm <- train(Class ~., data=GermanCredit, method='glm')
model_glm
#Generalized Linear Model
#1000 samples
# 61 predictor
# 2 classes: 'Bad', 'Good'
#No pre-processing
#Resampling: Bootstrapped (25 reps)
#Summary of sample sizes: 1000, 1000, 1000, 1000, 1000, 1000, ...
#Resampling results:
# Accuracy Kappa
# 0.7386384 0.3478527
在这种情况下，上面的glm不会执行任何参数调整.根据我的经验，名为parameter的parameter似乎只是一个占位符，而不是真正的调整参数.如以下代码所示，即使我们试图迫使它调整parameter，它基本上也只执行一个值.
In this case with glm above there was no parameter tuning performed. From my experience, it appears the parameter named parameter is just a placeholder and not a real tuning parameter. As demonstrated in the code that follows, even if we try to force it to tune parameter it basically only does a single value.
set.seed(1)
model_glm2 <- train(Class ~., data=GermanCredit, method='glm',
tuneGrid=expand.grid(parameter=c(0.001, 0.01, 0.1, 1,10,100, 1000)))
model_glm2
#Generalized Linear Model
#1000 samples
# 61 predictor
# 2 classes: 'Bad', 'Good'
#No pre-processing
#Resampling: Bootstrapped (25 reps)
#Summary of sample sizes: 1000, 1000, 1000, 1000, 1000, 1000, ...
#Resampling results across tuning parameters:
# Accuracy Kappa parameter
# 0.7386384 0.3478527 0.001
# 0.7386384 0.3478527 0.001
# 0.7386384 0.3478527 0.001
# 0.7386384 0.3478527 0.001
# 0.7386384 0.3478527 0.001
# 0.7386384 0.3478527 0.001
# 0.7386384 0.3478527 0.001
#Accuracy was used to select the optimal model using the largest value.
#The final value used for the model was parameter = 0.001.
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