计算随机森林与休假一个ID交叉验证

我有一个数据帧df计算随机森林与休假一个ID交叉验证

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")

我想与休假一个ID交叉验证这个数据集运行随机森林。因此，我不希望交叉验证是一种随机的。对于每次运行，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。这意味着具有相同ID的数据将具有相同的交叉验证索引。例如，第一次运行将训练ID = 5,6,8,9的数据，并将在ID = 4的数据上进行测试，第二次运行将训练ID = 4,6的数据， 8,9，并将在ID = 5的数据上进行测试，依此类推。有人知道如何在R中实现它吗？以下是我尝试过的命令行，但不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,]) 
}

来源

2015-09-20 SimonB

看看：http://stats.stackexchange.com/questions/109340/leave-one-subject -out-cross-validation-in-caret –

虽然在我的情况下它似乎没有工作。我已经使用我一直在做的一些命令行编辑了我的问题，但是没有使用假一个ID进行交叉验证。也许这会有所帮助。 – SimonB

@SimonB也许这个帖子是相关的？ http://stats.stackexchange.com/questions/137000/correlated-cases-and-cross-validation – WhiteViking

我想你问的是：

我们需要有尽可能多的交叉验证运行在训练集为不同的ID。所以我们在矢量uniqueIDs中收集这些ID，然后将每个训练观察结果与矢量crossclass中的正确运行相关联。像这样：

uniqueIDs <- unique(train$ID)   
nruns <- length(uniqueIDs) # number of cross validation runs: one for each unique ID   
crossclass <- match(train$ID, uniqueIDs)

您写的主要交叉验证循环保持不变。（我只加了一些调试输出。）

nobs <- nrow(na.omit(train)) 
crossPredict <- rep(NA, nobs) 

for (i in 1:nruns) { 
    indtrain <- which(crossclass != i) 
    indvalidate <- setdiff(1:nobs, indtrain) 
    cat("Run", i, ": training only on observations with ID not", uniqueIDs[i], "\n") 
    cat("  IDs in training set:", train[indtrain,"ID"], "\n") 
    cat("  IDs in validation set:", train[indvalidate,"ID"], "\n") 

    rf_df_CV <- randomForest(Y ~ ., data = train[indtrain,], 
          ntree = 1000, importance = T, na.action = "na.omit") 

    crossPredict[indvalidate] <- predict(rf_df_CV, train[indvalidate,]) 
}

下面是一个例子输出：

Run 1 : training only on observations with ID not 4 
     IDs in training set: 5 8 6 6 5 5 8 6 
     IDs in validation set: 4 4 4 4 4 4 4 
Run 2 : training only on observations with ID not 5 
     IDs in training set: 4 4 8 4 6 6 4 4 4 4 8 6 
     IDs in validation set: 5 5 5 
Run 3 : training only on observations with ID not 8 
     IDs in training set: 4 4 5 4 6 6 5 5 4 4 4 4 6 
     IDs in validation set: 8 8 
Run 4 : training only on observations with ID not 6 
     IDs in training set: 4 4 5 8 4 5 5 4 4 4 4 8 
     IDs in validation set: 6 6 6

注意，在这个例子中的df随机分成训练和测试集，训练集发生因为没有任何ID = 9的观察。所以也没有这个ID的CV运行...

来源

2015-09-20 21:43:01 WhiteViking

@ WhiteViking.Thanks。那么如何定义火车物体？它也必须改编，对吗？ – SimonB

@ WhiteViking。我根据你的回答编辑了我的问题（参见最后的命令行段落），但不确定它是否在概念上正确。你怎么看？ – SimonB

@SimonB我把你的问题作为一个简单的小R任务，但如果这实际上是关于如何对关联数据进行简历的一个概念性问题，那么我可能会超越。我想我最好删除这个答案。我希望有其他人介入来阐明一些光明。 – WhiteViking

计算随机森林与休假一个ID交叉验证

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