插入符号包中的train（）返回有关名称的错误&；gsub

我正在使用插入符号包预测

improvementNoticed

变量

library(caret)
head(trainData)

     improvementNoticed                            V1               V2
681                   0                    0.06451613       0.006060769
1484                  0                    0.77924586       0.331009145
1356                  0                    0.22222222       0.017538684
541                   0                    0.21505376       0.011102470
2214                  1                    0.59195217       0.064764408
1111                  0                    0.97979798       0.036445064
               V3                                          V4       V5
681   0.008182531                                  0.05263158        0
1484  0.316603794                                  0.88825188        0
1356  0.016182822                                  0.20000000        0
541   0.012665610                                  0.10000000        0
2214  0.051008693                                  0.55000000        0
1111  0.034643632                                  0.93333333        0

我跑

myControl = trainControl(method='cv',number=5,repeats=2,returnResamp='none')
model1 = train(improvementNoticed~., data=trainData, method = 'glm', trControl=myControl)

我得到以下错误：

Error in names(out) <- paste("Fold", gsub(" ", "0", format(seq(along = out))),  : 
  'names' attribute [1] must be the same length as the vector [0]

请注意，

improvementNoticed

是一个二进制变量

如果我将

列车数据[，1]

转换为

整数

，我会得到与数字相同的错误

最后两件事：

traceback()
5: createFolds(y, trControl$number, returnTrain = TRUE)
4: train.default(x, y, weights = w, ...)
3: train(x, y, weights = w, ...)
2: train.formula(improvementNoticed ~ ., data = trainData, method = "glm", 
       trControl = myControl)
1: train(improvementNoticed ~ ., data = trainData, method = "glm", 
       trControl = myControl)

以及

sessionInfo（）的结果

碰巧，这个错误是一个非常基本的错误

我正在对数据执行标准化（我不怀疑这会导致问题），但结果发现其中一个变量中只有0；因此，我得到了所有的NaN，这导致了模型失败。
假设您试图适应一个分类模型，您应该将
改进通知编码为一个因子，而不是一个二进制变量。如果是二进制的，
train
不知道具有二进制结果的分类与回归的某些数值结果之间的区别。
traceback()
5: createFolds(y, trControl$number, returnTrain = TRUE)
4: train.default(x, y, weights = w, ...)
3: train(x, y, weights = w, ...)
2: train.formula(improvementNoticed ~ ., data = trainData, method = "glm", 
       trControl = myControl)
1: train(improvementNoticed ~ ., data = trainData, method = "glm", 
       trControl = myControl)

R version 3.0.1 (2013-05-16)
Platform: x86_64-redhat-linux-gnu (64-bit)

locale:
 [1] LC_CTYPE=en_GB.UTF-8       LC_NUMERIC=C              
 [3] LC_TIME=en_GB.UTF-8        LC_COLLATE=en_GB.UTF-8    
 [5] LC_MONETARY=en_GB.UTF-8    LC_MESSAGES=en_GB.UTF-8   
 [7] LC_PAPER=C                 LC_NAME=C                 
 [9] LC_ADDRESS=C               LC_TELEPHONE=C            
[11] LC_MEASUREMENT=en_GB.UTF-8 LC_IDENTIFICATION=C       

attached base packages:
[1] parallel  stats     graphics  grDevices utils     datasets  methods  
[8] base     

other attached packages:
 [1] elasticnet_1.1     lars_1.2           klaR_0.6-9         MASS_7.3-26       
 [5] kernlab_0.9-18     nnet_7.3-6         randomForest_4.6-7 doMC_1.3.0        
 [9] iterators_1.0.6    caret_5.17-7       reshape2_1.2.2     plyr_1.8          
[13] lattice_0.20-15    foreach_1.4.1      cluster_1.14.4    

loaded via a namespace (and not attached):
[1] codetools_0.2-8 grid_3.0.1      stringr_0.6.2   tools_3.0.1