我正在尝试使用 rpart 在插入符号中最大限度地提高模型选择的灵敏度。为此，我尝试复制此处给出的方法(向下滚动到使用用户定义函数 FourStat 的示例)caret's github page
# create own function so we can use "sensitivity" as our metric to maximise:
Sensitivity.fc <- function (data, lev = levels(data$obs), model = NULL) {
out <- c(twoClassSummary(data, lev = levels(data$obs), model = NULL))
c(out, Sensitivity = out["Sens"])
}
rpart_caret_fit <- train(outcome~pred1+pred2+pred3+pred4,
na.action = na.pass,
method = "rpart",
control=rpart.control(maxdepth = 6),
tuneLength = 20,
# maximise sensitivity
metric = "Sensitivity",
maximize = TRUE,
trControl = trainControl(classProbs = TRUE,
summaryFunction = Sensitivity.fc))
但是当我得到摘要时
rpart_caret_fit
表示它仍然使用ROC准则来选择最终模型:
CART
678282 samples
4 predictor
2 classes: 'yes', 'no'
No pre-processing
Resampling: Bootstrapped (25 reps)
Summary of sample sizes: 678282, 678282, 678282, 678282, 678282, 678282, ...
Resampling results across tuning parameters:
cp ROC Sens Spec Sensitivity.Sens
0.000001909738 0.7259486 0.4123547 0.8227382 0.4123547
0.000002864607 0.7259486 0.4123547 0.8227382 0.4123547
0.000005729214 0.7259489 0.4123622 0.8227353 0.4123622
0.000006684083 0.7258036 0.4123614 0.8227379 0.4123614
0.000007638953 0.7258031 0.4123576 0.8227398 0.4123576
0.000009548691 0.7258028 0.4123539 0.8227416 0.4123539
0.000010694534 0.7257553 0.4123589 0.8227332 0.4123589
0.000015277905 0.7257313 0.4123614 0.8227290 0.4123614
0.000032465548 0.7253456 0.4112838 0.8234272 0.4112838
0.000038194763 0.7252966 0.4112912 0.8234196 0.4112912
0.000076389525 0.7248774 0.4102792 0.8240339 0.4102792
0.000164237480 0.7244847 0.4093688 0.8246372 0.4093688
0.000194793290 0.7241532 0.4086596 0.8250930 0.4086596
0.000310650737 0.7237546 0.4087379 0.8250393 0.4087379
0.001625187154 0.7233805 0.4006570 0.8295729 0.4006570
0.001726403276 0.7233225 0.3983850 0.8308874 0.3983850
0.002173282000 0.7230906 0.3915758 0.8348320 0.3915758
0.002237258227 0.7230906 0.3915758 0.8348320 0.3915758
0.006140444689 0.7173854 0.4897494 0.7695558 0.4897494
0.055330843035 0.5730987 0.2710906 0.8545549 0.2710906
ROC was used to select the optimal model using the largest value.
The final value used for the model was cp = 0.000005729214.
如何覆盖 ROC 选择方法？
最佳答案
你把事情搞得太复杂了。
两个类摘要已经包含灵敏度作为输出。列名“Sens”。指定就足够了:
metric = "Sens" 到 train 和
summaryFunction = twoClassSummary 到 trainControl
完整示例:
library(caret)
library(mlbench)
data(Sonar)
rpart_caret_fit <- train(Class~.,
data = Sonar,
method = "rpart",
tuneLength = 20,
metric = "Sens",
maximize = TRUE,
trControl = trainControl(classProbs = TRUE,
method = "cv",
number = 5,
summaryFunction = twoClassSummary))
rpart_caret_fit
CART
208 samples
60 predictor
2 classes: 'M', 'R'
No pre-processing
Resampling: Cross-Validated (5 fold)
Summary of sample sizes: 167, 166, 166, 166, 167
Resampling results across tuning parameters:
cp ROC Sens Spec
0.0000000 0.7088298 0.7023715 0.7210526
0.0255019 0.7075400 0.7292490 0.6684211
0.0510038 0.7105388 0.7758893 0.6405263
0.0765057 0.6904202 0.7841897 0.6294737
0.1020076 0.7104681 0.8114625 0.6094737
0.1275095 0.7104681 0.8114625 0.6094737
0.1530114 0.7104681 0.8114625 0.6094737
0.1785133 0.7104681 0.8114625 0.6094737
0.2040152 0.7104681 0.8114625 0.6094737
0.2295171 0.7104681 0.8114625 0.6094737
0.2550190 0.7104681 0.8114625 0.6094737
0.2805209 0.7104681 0.8114625 0.6094737
0.3060228 0.7104681 0.8114625 0.6094737
0.3315247 0.7104681 0.8114625 0.6094737
0.3570266 0.7104681 0.8114625 0.6094737
0.3825285 0.7104681 0.8114625 0.6094737
0.4080304 0.7104681 0.8114625 0.6094737
0.4335323 0.7104681 0.8114625 0.6094737
0.4590342 0.6500135 0.8205534 0.4794737
0.4845361 0.6500135 0.8205534 0.4794737
Sens was used to select the optimal model using the largest value.
The final value used for the model was cp = 0.4845361.
另外，我认为您不能指定 这是不正确的- 插入符号使用 control = rpart.control(maxdepth = 6) 插入 train。... 向前传递任何参数。所以你几乎可以传递任何参数。
如果您想编写自己的摘要函数，这里有一个关于“Sens”的示例:
Sensitivity.fc <- function (data, lev = NULL, model = NULL) { #every summary function takes these three arguments
obs <- data[, "obs"] #these are the real values - always in column name "obs" in data
cls <- levels(obs) #there are the levels - you can also pass this to lev argument
probs <- data[, cls[2]] #these are the probabilities for the 2nd class - useful only if prob = TRUE
class <- as.factor(ifelse(probs > 0.5, cls[2], cls[1])) #calculate the classes based on some probability treshold
Sensitivity <- caret::sensitivity(class, obs) #do the calculation - I was lazy so I used a built in function to do it for me
names(Sensitivity) <- "Sens" #the name of the output
Sensitivity
}
现在:
rpart_caret_fit <- train(Class~.,
data = Sonar,
method = "rpart",
tuneLength = 20,
metric = "Sens", #because of this line: names(Sensitivity) <- "Sens"
maximize = TRUE,
trControl = trainControl(classProbs = TRUE,
method = "cv",
number = 5,
summaryFunction = Sensitivity.fc))
让我们检查两者是否产生相同的结果:
set.seed(1)
fit_sens <- train(Class~.,
data = Sonar,
method = "rpart",
tuneLength = 20,
metric = "Sens",
maximize = TRUE,
trControl = trainControl(classProbs = TRUE,
method = "cv",
number = 5,
summaryFunction = Sensitivity.fc))
set.seed(1)
fit_sens2 <- train(Class~.,
data = Sonar,
method = "rpart",
tuneLength = 20,
metric = "Sens",
maximize = TRUE,
trControl = trainControl(classProbs = TRUE,
method = "cv",
number = 5,
summaryFunction = twoClassSummary))
all.equal(fit_sens$results[c("cp", "Sens")],
fit_sens2$results[c("cp", "Sens")])
TRUE
all.equal(fit_sens$bestTune,
fit_sens2$bestTune)
TRUE
关于r - 优化插入符号的灵敏度似乎仍然优化 ROC，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/49265400/