我在R中调整SVM并收到以下错误：
#Error in if (any(co)) { : missing value where TRUE/FALSE needed
我正在使用插入包
svmRTune <- train(x=dataTrain[,predModelContinuous],y=dataTrain[,outcome],method = "svmRadial", tuneLength = 14, trControl = trCtrl)
训练集结构
str(dataTrain)
'data.frame': 40001 obs. of 42 variables:
$ PolNum : num 2e+08 2e+08 2e+08 2e+08 2e+08 ...
$ sex : Factor w/ 2 levels "Male","Female": 1 1 1 2 1 2 1 1 1 2 ...
$ type : Factor w/ 6 levels "A","B","C","D",..: 3 1 1 2 2 4 3 3 3 2 ...
$ catgry : Ord.factor w/ 3 levels "Large"<"Medium"<..: 2 2 2 3 3 3 3 2 2 2 ...
$ occup : Factor w/ 5 levels "Employed","Housewife",..: 2 1 1 1 5 4 1 1 4 2 ...
$ age : num 48 23 23 39 24 39 28 43 45 38 ...
$ group : Factor w/ 20 levels "1","2","3","4",..: 15 16 12 16 14 8 16 9 12 8 ...
$ bonus : Ord.factor w/ 21 levels "-50"<"-40"<"-30"<..: 14 8 4 3 5 2 5 5 1 15 ...
$ poldur : num 7 1 1 14 2 4 11 2 8 5 ...
$ value : num 1120 21755 18430 11930 24850 ...
$ adind : Factor w/ 2 levels "No","Yes": 2 1 1 2 1 2 2 2 1 1 ...
$ Pcode : chr "SC22" "CT109" "MA1" "SA12" ...
$ Area : Factor w/ 10 levels "CT","JU","MA",..: 7 1 3 6 6 6 6 4 1 2 ...
$ Density : num 270.5 57.3 43.2 167.9 169.8 ...
$ Prem : num 1159 532 527 197 908 ...
$ Premad : num 53.1 413.7 410.7 61.6 824.6 ...
$ numclm : num 0 1 0 1 0 0 0 1 0 0 ...
$ Invite : num 1 1 1 1 1 1 1 1 1 1 ...
$ Renewaltp : num 1302 928 632 291 960 ...
$ Renewalad : num 58.4 599 440.4 71.3 682 ...
$ Markettp : num 1110 884 565 253 833 ...
$ Marketad : num 53.4 611.4 431.6 55.5 587 ...
$ Premtot : num 1212 532 527 259 908 ...
$ Renewaltot : num 1361 928 632 362 960 ...
$ Markettot : num 1163 884 565 309 833 ...
$ Renew : Ord.factor w/ 2 levels "No"<"Yes": 1 1 1 1 1 1 1 1 1 1 ...
$ Premchng : num 1.12 1.74 1.2 1.4 1.06 ...
$ Compmeas : num 1.17 1.05 1.12 1.17 1.15 ...
$ numclmRec : Ord.factor w/ 3 levels "None"<"One"<"Two or more": 1 2 1 2 1 1 1 2 1 1 ...
$ PremChngRec: Factor w/ 20 levels "[0.546,0.758)",..: 16 20 18 19 14 3 7 19 17 11 ...
$ ageRec : Factor w/ 20 levels "[19,22)","[22,25)",..: 14 2 2 9 2 9 4 11 12 9 ...
$ valueRec : Factor w/ 20 levels "[ 1005, 3290)",..: 1 15 13 9 17 5 12 12 19 1 ...
$ densityRec : Factor w/ 20 levels "[ 14.4, 25.0)",..: 19 6 5 15 15 13 15 1 5 11 ...
$ CompmeasRec: Factor w/ 20 levels "[0.716,0.869)",..: 12 6 10 13 12 18 11 16 18 14 ...
$ poldurRec : Ord.factor w/ 16 levels "1"<"2"<"3"<"4"<..: 7 1 1 14 2 4 11 2 8 5 ...
$ ageST : num 0.407 -1.34 -1.34 -0.222 -1.27 ...
$ numclmST : num -0.433 1.627 -0.433 1.627 -0.433 ...
$ PremchngST : num 0.591 3.709 0.98 1.985 0.265 ...
$ valueST : num -1.462 0.499 0.183 -0.434 0.793 ...
$ DensityST : num 1.918 -0.748 -0.924 0.636 0.659 ...
$ CompmeasST : num 0.224 -0.539 -0.098 0.248 0.113 ...
$ poldurST : num 0.097 -1.2 -1.2 1.61 -0.984 ...
和
sessionInfo()
R version 3.0.2 (2013-09-25)
Platform: x86_64-w64-mingw32/x64 (64-bit)
locale:
[1] LC_COLLATE=Italian_Italy.1252 LC_CTYPE=Italian_Italy.1252
[3] LC_MONETARY=Italian_Italy.1252 LC_NUMERIC=C
[5] LC_TIME=Italian_Italy.1252
attached base packages:
[1] parallel splines grid stats graphics grDevices utils
[8] datasets methods base
other attached packages:
[1] C50_0.1.0-16 kernlab_0.9-19 nnet_7.3-8 plyr_1.8.1
[5] gbm_2.1 randomForest_4.6-7 rpart_4.1-8 klaR_0.6-10
[9] MASS_7.3-31 doParallel_1.0.8 iterators_1.0.6 foreach_1.4.1
[13] pROC_1.7.1 mda_0.4-4 class_7.3-10 earth_3.2-7
[17] plotrix_3.5-5 plotmo_1.3-3 Formula_1.1-1 survival_2.37-7
[21] caret_6.0-24 ggplot2_0.9.3.1 lattice_0.20-29 rj_1.1.3-1
loaded via a namespace (and not attached):
[1] car_2.0-19 cluster_1.15.2 codetools_0.2-8
[4] colorspace_1.2-4 combinat_0.0-8 compiler_3.0.2
[7] dichromat_2.0-0 digest_0.6.4 gtable_0.1.2
[10] Hmisc_3.14-3 labeling_0.2 latticeExtra_0.6-26
[13] munsell_0.4.2 proto_0.3-10 RColorBrewer_1.0-5
[16] Rcpp_0.11.1 reshape2_1.2.2 rj.gd_1.1.3-1
[19] scales_0.2.3 stringr_0.6.2 tools_3.0.2
答案 0 :(得分：3)
只是发布以防其他人遇到此问题。 它似乎是由训练数据集中包含因子或字符变量引起的。
为什么svm不能采用因子变量，我不知道。我用手动编码的假人替换了我的因素，并且它工作得很好，但这种方法太不优雅了。
答案 1 :(得分：2)
我可以确认Dan Brown的答案，错误似乎是由数据中的因素引起的。我编写了以下代码将因子转换为虚拟变量。它不是特别漂亮，但它可以完成这项工作。
library("foreach")
# Helper function, use the other one
# takes a column name (pointing to a factor variable) and a dataset
# returns a dataframe containing a 1-in-K coding for this factor variable
col_to_dummy <- function(colname, data) {
# tmp is a dataframe of K columns, where K is the number of levels of the factor in colname
# it is a 1-in-K dummy variable coding
levelnames <- levels(data[[colname]])
dummy <- foreach(i=1:length(levelnames), .combine=cbind) %do% {
as.numeric(as.numeric(data[[colname]])==i)
}
dummy <- as.data.frame(dummy)
names(dummy) <- paste0(colname, ":", levelnames)
dummy
}
factor_to_dummy <- function(obsdata) {
# finding the columns containing a factor variable
col_factor <- unlist(lapply(FUN=is.factor, obsdata))
# if they are none, then nothing to do
if(!any(col_factor)) {
return(obsdata)
}
# otherwise
# for each of these, convert it to dummy variables using col_to_dummy
foreach(colname=names(which(col_factor)), .combine = cbind,
.init = obsdata[,-which(col_factor)]) %do% {
col_to_dummy(colname, obsdata)
}
# each resulting data.frame is c-bound with the dataset without factors
}
其中一些解决方案使用model.matrix，但意识到默认情况下，model.matrix使用参考级别（拦截），然后对所有因素使用1-of（K-1）编码方案。你需要修改对比度参数，以获得你想要的东西。
这段代码非常易于使用。一旦运行了函数定义，您就可以执行以下操作：
df_with_dummy_vars <- factor_to_dummy(original_df)
所有因子列都将转换为虚拟变量。