使循环使代码变得简单

Question

i列出了所有可能性并将它们替换为另一个单词，还有另一种简单的方法来编写这些代码，例如loop？

abnormal$type1[grepl("\(2\)x3", abnormal$v1)] <- "T2"
abnormal$type1[grepl("\(4\)x3", abnormal$v1)] <- "T4"
abnormal$type1[grepl("\(7\)x3", abnormal$v1)] <- "T7"
abnormal$type1[grepl("\(8\)x3", abnormal$v1)] <- "T8"
abnormal$type1[grepl("\(9\)x3", abnormal$v1)] <- "T9"
abnormal$type1[grepl("\(10\)x3", abnormal$v1)] <- "T10"
abnormal$type1[grepl("\(12\)x3", abnormal$v1)] <- "T12"
abnormal$type1[grepl("\(13\)x3", abnormal$v1)] <- "T13"
abnormal$type1[grepl("\(14\)x3", abnormal$v1)] <- "T14"
abnormal$type1[grepl("\(15\)x3", abnormal$v1)] <- "T15"
abnormal$type1[grepl("\(16\)x3", abnormal$v1)] <- "T16"
abnormal$type1[grepl("\(17\)x3", abnormal$v1)] <- "T17"
abnormal$type1[grepl("\(18\)x3", abnormal$v1)] <- "T18"
abnormal$type1[grepl("\(20\)x3", abnormal$v1)] <- "T20"
abnormal$type1[grepl("\(21\)x3", abnormal$v1)] <- "T21"
abnormal$type1[grepl("\(22\)x3", abnormal$v1)] <- "T22"
abnormal$type1[grepl("\(X\)x3", abnormal$v1)] <- "TX"

Answer 1

一个基本r选项将是定义分配/置换两侧出现的数字向量，然后使用 sapply：

v <- as.character(c(2, 4, 7:10, 12:18, 20:22, 'X'))
sapply(v, function(x) {
    pattern <- paste0("\(", x, "\)x3")
    abnormal$type1[grepl(pattern, abnormal$v1)] <<- paste0("T", x)
})