使用向量中的元素的 R grep 正则表达式（FOLLOW UP）

跟进这个问题，我还有另一个例子，我无法使用已接受的答案。

再说一次，我想找到每一个确切的group中的元素lab向量...

labs <- c("Beijing -- T0 -- BC-89 + CN",
"Beijing -- T24 -- BC-89 + CN",
"Beijing -- T0 -- BC-89 + CN with 2% DD + 1.6% ZC",
"Beijing -- T24 -- BC-89 + CN with 2% DD + 1.6% ZC",
"Beijing -- T0 -- BC-89 with 2% Puricare + 5% Merquat + CN",
"Beijing -- T24 -- BC-89 with 2% Puricare + 5% Merquat + CN",
"Beijing -- T0 -- BC-89 + CN",
"Zhangjiakou -- T0 -- BC-89 + CN",
"Beijing -- T0 -- BC-89 + CN with 2% DD + 1.6% ZC",
"Zhangjiakou -- T0 -- BC-89 + CN with 2% DD + 1.6% ZC",
"Beijing -- T0 -- BC-89 with 2% Puricare + 5% Merquat + CN",
"Zhangjiakou -- T0 -- BC-89 with 2% Puricare + 5% Merquat + CN",
"Beijing -- T0 -- BC-89 + CN",
"Beijing -- T0 -- BC-89 + CN",
"Beijing -- T0 -- BC-89 + CN with 2% DD + 1.6% ZC",
"Beijing -- T24 -- BC-89 + CN",
"Beijing -- T24 -- BC-89 + CN",
"Beijing -- T24 -- BC-89 + CN with 2% DD + 1.6% ZC",
"Zhangjiakou -- T0 -- BC-89 + CN",
"Zhangjiakou -- T0 -- BC-89 + CN",
"Zhangjiakou -- T0 -- BC-89 + CN with 2% DD + 1.6% ZC",
"Zhangjiakou -- T24 -- BC-89 + CN",
"Zhangjiakou -- T24 -- BC-89 + CN",
"Zhangjiakou -- T24 -- BC-89 + CN with 2% DD + 1.6% ZC")
labs
groups <- c("BC-89 + CN", "BC-89 + CN with 2% DD + 1.6% ZC", "BC-89 with 2% Puricare + 5% Merquat + CN")
groups

我尝试以下但它不起作用......

grep(paste0(groups[1], "$"), labs, value=TRUE)
grep(paste0(groups[2], "$"), labs, value=TRUE)
grep(paste0(groups[3], "$"), labs, value=TRUE)

有什么帮助吗？

Try

lapply(groups, function(g)
  grep(gsub("\\+", "\\\\+", paste0(g, "$")), labs, value = TRUE))
# [[1]]
# [1] "Beijing -- T0 -- BC-89 + CN"     
# [2] "Beijing -- T24 -- BC-89 + CN"    
# [3] "Beijing -- T0 -- BC-89 + CN"     
# [4] "Zhangjiakou -- T0 -- BC-89 + CN" 
# [5] "Beijing -- T0 -- BC-89 + CN"     
# [6] "Beijing -- T0 -- BC-89 + CN"     
# [7] "Beijing -- T24 -- BC-89 + CN"    
# [8] "Beijing -- T24 -- BC-89 + CN"    
# [9] "Zhangjiakou -- T0 -- BC-89 + CN" 
# [10] "Zhangjiakou -- T0 -- BC-89 + CN" 
# [11] "Zhangjiakou -- T24 -- BC-89 + CN"
# [12] "Zhangjiakou -- T24 -- BC-89 + CN"
# 
# [[2]]
# [1] "Beijing -- T0 -- BC-89 + CN with 2% DD + 1.6% ZC"     
# [2] "Beijing -- T24 -- BC-89 + CN with 2% DD + 1.6% ZC"    
# [3] "Beijing -- T0 -- BC-89 + CN with 2% DD + 1.6% ZC"     
# [4] "Zhangjiakou -- T0 -- BC-89 + CN with 2% DD + 1.6% ZC" 
# [5] "Beijing -- T0 -- BC-89 + CN with 2% DD + 1.6% ZC"     
# [6] "Beijing -- T24 -- BC-89 + CN with 2% DD + 1.6% ZC"    
# [7] "Zhangjiakou -- T0 -- BC-89 + CN with 2% DD + 1.6% ZC" 
# [8] "Zhangjiakou -- T24 -- BC-89 + CN with 2% DD + 1.6% ZC"
# 
# [[3]]
# [1] "Beijing -- T0 -- BC-89 with 2% Puricare + 5% Merquat + CN"    
# [2] "Beijing -- T24 -- BC-89 with 2% Puricare + 5% Merquat + CN"   
# [3] "Beijing -- T0 -- BC-89 with 2% Puricare + 5% Merquat + CN"    
# [4] "Zhangjiakou -- T0 -- BC-89 with 2% Puricare + 5% Merquat + CN"

您的方法的问题在于，例如，groups[1] is "BC-89 + CN"， 其中包含+，在正则表达式中具有特殊含义。仅鉴于此，添加fixed = TRUE in grep会解决这个问题，但随后$就会失去作用。所以我所做的就是逃避+首先在组名称中。

或者，与您链接的答案相关，您可以这样做

lapply(groups, function(g)
  grep(paste0(g, "$"), paste0(labs, "$"), value = TRUE, fixed = TRUE))