在 R 中使用留一 ID 交叉验证计算随机森林

Question

我有一个数据框df

dput(df)
    structure(list(ID = c(4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 
6, 6, 6, 6, 8, 8, 8, 9, 9), Y = c(2268.14043972082, 2147.62290922552, 
2269.1387550775, 2247.31983098201, 1903.39138268307, 2174.78291538358, 
2359.51909126411, 2488.39004804939, 212.851575751527, 461.398994384333, 
567.150629704352, 781.775113821961, 918.303706148872, 1107.37695799186, 
1160.80594193377, 1412.61328924168, 1689.48879626486, 685.154353165934, 
574.088067465695, 650.30821636616, 494.185166497016, 436.312162090908
), P = c(1750.51986303926, 1614.11541634798, 951.847023338079, 
1119.3682884872, 1112.38984390156, 1270.65773075982, 1234.72262170166, 
1338.46096616983, 1198.95775346458, 1136.69287367165, 1265.46480803983, 
1364.70149818063, 1112.37006707489, 1346.49240261316, 1740.56677791104, 
1410.99217295647, 1693.18871380948, 275.447173420805, 396.449789014179, 
251.609239829704, 215.432550271042, 55.5336257666349), A = c(49, 
50, 51, 52, 53, 54, 55, 56, 1, 2, 3, 4, 5, 14, 15, 16, 17, 163, 
164, 165, 153, 154), TA = c(9.10006221322572, 7.65505467142961, 
8.21480062559674, 8.09251754304318, 8.466220758789, 8.48094407814006, 
8.77304120569444, 8.31727518543397, 8.14410265791868, 8.80921738865237, 
9.04091478341757, 9.66233618146246, 8.77015716015164, 9.46037931956657, 
9.59702379240667, 10.1739258740118, 9.39524442215692, -0.00568604734662462, 
-2.12940164413048, -0.428603434930109, 1.52337963973006, -1.04714984064565
), TS = c(9.6499861763085, 7.00622420539595, 7.73511170298675, 
7.68006974050443, 8.07442411510912, 8.27687965909096, 8.76025039592727, 
8.3345638889156, 9.23658956753677, 8.98160722605782, 8.98234210211611, 
9.57066566368204, 8.74444401914267, 8.98719629775988, 9.18169205278566, 
9.98225438314085, 9.56196773059615, 5.47788158053928, 2.58106090926808, 
3.22420704848299, 1.36953555753786, 0.241334267522977), R = c(11.6679680423377, 
11.0166459173372, 11.1851268491296, 10.7404563561694, 12.1054055597684, 
10.9551321815546, 11.1975918244469, 10.7242192465965, 10.1661703705992, 
11.4840412725324, 11.1248456370953, 11.2529612597628, 10.7694642397996, 
12.3300887767583, 12.0478558531771, 12.3212362249214, 11.5650773932264, 
9.56070414783612, 9.61762902218185, 10.2076240621201, 11.8234628013552, 
10.9184029778985)), .Names = c("ID", "Y", "P", "A", "TA", "TS", 
"R"), na.action = structure(77:78, .Names = c("77", "78"), class = "omit"), row.names = c(NA, 
22L), class = "data.frame")

我想运行在此数据集上使用留一 ID 交叉验证的 RandomForest。因此，我不希望交叉验证是随机的。对于每个运行，我想省略具有相同 ID 值的数据，因为具有相同 ID 的数据不是独立的。例如，第一个运行将在 ID=5、6、8、9 的数据上进行训练，并将在 ID=4 的数据上进行测试，第二个运行将在ID=4,6,8,9 的数据，将在 ID=5 的数据上进行测试，依此类推。

我实现了下面的命令行，但我不太确定它在概念上是否正确。

# Create Training dataset
df<-na.omit(df)
tvec<-unique(df$ID)
nruns <- length(tvec)
crossclass<-sample(nruns,length(tvec),TRUE)
nobs<-nrow(df)
crossPredict<-rep(NA,nobs)

#Run a RandomForest with leave one out ID CV
for (i in 1:nruns) {
  indtrain<-which(df$ID %in% tvec[!crossclass==i])
  indvalidate<-setdiff(1:nobs,indtrain)
  rf<-randomForest(formula = Y ~ P + TA + TS + R + A, data=df, subset=indtrain,ntree=10000)
  crossPredict[indvalidate]<-predict(rf,df[indvalidate,])
}

有人可以帮我解决这个问题吗？

Answer 1

library(randomForest)
newIris <- data.frame(iris, id=1+c(1:nrow(iris))%%3)
id <- unique(newIris$id)

loo <- NULL
for(i in id){
    rf <- randomForest(Species~., data=newIris[newIris$id!=i,])
    loo[[i]] <- predict(rf, newdata=newIris[newIris$id==i,])
}

print(loo)

只需制作一个 ID 向量，然后依次省略每个 ID。

Answer 2

这似乎运作良好。

library (caret)
library(randomForest)
    # Create training datastet
    subs <- unique(df$ID)
    train<- vector(mode = "list", length = length(subs))
    test<- vector(mode = "list", length = length(subs))

    # Run a RandomForest with leave one out ID CV
    for(i in seq_along(subs))
    train[[i]] <- which(df$ID != subs[i])
    names(train) <- paste0("ID", subs)
    rfFit <- train(Y~ P + TA + TS + R + A,
                   data =df,
                   method = "rf",
                   ntree = 100,
                   prox=TRUE, allowParallel=TRUE,
                   importance = TRUE,
                   trControl = trainControl(method = "cv", 
                                             index = train))

    # Create test dataset
    for(i in seq_along(subs))
    test[[i]] <- which(df$ID == subs[i])
    names(test)<-paste0("ID", subs)

在 R 中使用留一 ID 交叉验证计算随机森林

Compute Random Forest with a leave one ID out cross validation in R

r

random-forest

cross-validation