删除所有以R中特定字符串结尾的字符串的行-grepl

Question

我想删除所有以“ _bundle”结尾的行。我尝试了两种不同的方法，但没有任何效果

claimsVolumeSC <- basisPerClaim[!grepl( '$_bundle', basisPerClaim$subcoveragekey),]
levels(claimsVolumeSC$subcoveragekey)
claimsVolumeSC <- basisPerClaim[!grepl( '\\>_bundle', basisPerClaim$subcoveragekey),]
levels(claimsVolumeSC$subcoveragekey)

我如何实现我想要的？为什么到目前为止我尝试过的不起作用？

> claimsVolumeSC <- basisPerClaim[!grepl( '$_bundle', basisPerClaim$subcoveragekey),]
> levels(claimsVolumeSC$subcoveragekey)
[1] "DA_Chemo"         "Daily_cash"       "Funeral"          "IP_Accommodation" "IP_bundle"        "IP_Upgrade"       "OP_Dialysis"     
[8] "OP_Physio"       
> claimsVolumeSC <- basisPerClaim[!grepl( '\\>_bundle', basisPerClaim$subcoveragekey),]
> levels(claimsVolumeSC$subcoveragekey)
[1] "DA_Chemo"         "Daily_cash"       "Funeral"          "IP_Accommodation" "IP_bundle"        "IP_Upgrade"       "OP_Dialysis"     
[8] "OP_Physio"   

Answer 1

您可以将subset与grepl一起使用>

claimsVolumeSC <- subset(basisPerClaim, !grepl( '_bundle$', subcoveragekey))
如果您不想在数据中添加因子，请将其更改为字符。 

claimsVolumeSC$subcoveragekey <- as.character(claimsVolumeSC$subcoveragekey)

Answer 2

解决方案可能是基于正则表达式的解决方案：