我想使用R插入符号(参见this example)执行离开主题交叉验证,但仅在训练中使用数据的子集来创建CV模型。仍然,左侧的CV分区应该作为一个整体使用,因为我需要测试一个遗漏主题的所有数据(无论数百万个样本由于计算限制而无法在训练中使用)。
我使用subset
和index
的{{1}}和caret::train
参数创建了一个最小的2类分类示例来实现此目的。根据我的观察,这应该可以解决问题,但实际上我很难确保评估仍以离开主题的方式进行。也许有这方面经验的人可以对此有所了解:
caret::trainControl
情节(myRoc,main ='all')
l_ply(唯一(m3 $ pred $ Resample),. fun = function(cls){ pred_sub< - m3 $ pred [m3 $ pred $ Resample == cls,] myRoc< - roc(predictor = pred_sub [,3],response = pred_sub $ obs) 情节(myRoc,main = cls) })
谢谢你的时间!
答案 0 :(得分:1)
同时在index
中同时使用indexOut
和caret::trainControl
参数似乎可以解决问题(感谢Max提示in this question)。这是更新的代码:
library(plyr)
library(caret)
library(pROC)
library(ggplot2)
str(diamonds)
# with diamonds we want to predict cut and look at results for different colors = subjects
d <- diamonds
d <- d[d$cut %in% c('Premium', 'Ideal'),] # make a 2 class problem
d$cut <- factor(d$cut)
indexes_data <- c(1,5,6,8:10)
indexes_labels <- 2
# population independent CV partitions for training and left out partitions for evaluation
indexes_populationIndependence_subjects <- 3
index <- llply(unique(d[,indexes_populationIndependence_subjects]), function(cls) c(which(d[,indexes_populationIndependence_subjects]!=cls)))
names(index) <- paste0('sub_', unique(d[,indexes_populationIndependence_subjects]))
indexOut <- llply(index, function(part) (1:nrow(d))[-part])
names(indexOut) <- paste0('sub_', unique(d[,indexes_populationIndependence_subjects]))
# subsample partitions for training
index <- llply(index, function(i) sample(i, 1000))
m3 <- train(x = d[,indexes_data],
y = d[,indexes_labels],
method = 'glm',
metric = 'ROC',
trControl = trainControl(returnResamp = 'final',
savePredictions = T,
classProbs = T,
summaryFunction = twoClassSummary,
index = index,
indexOut = indexOut))
m3$resample # seems OK
str(m3$pred) # seems OK
myRoc <- roc(predictor = m3$pred[,3], response = m3$pred$obs)
plot(myRoc, main = 'all')
# analyze results per subject
l_ply(unique(m3$pred$Resample), .fun = function(cls) {
pred_sub <- m3$pred[m3$pred$Resample==cls,]
myRoc <- roc(predictor = pred_sub[,3], response = pred_sub$obs)
plot(myRoc, main = cls)
} )
尽管如此,我并不完全确定这是否实际上是以人口独立的方式进行估算,所以如果有人知道有关详细信息,请分享您的想法!