R列表字符删除

Question

lst1是一个列表：

lst1 <- list(c("all the apples", "apples in the tables", "fashion prosthetics"),  c("meteorological concepts", "effects of climate change", "environmental"))

我想保留列表结构并从所有单词中删除最后一个。所需答案如下：

> lst1
[[1]]
[1] "all the apple"      "apple in the table" "nature"            

[[2]]
[1] "meteorological concept"   "effect of climate change"
[3] "environmental" 

我试过

gsub("\\'s|s$|s[[:space:]]{0}","",lst1)

但它没有保留列表结构。

怎么做？

Answer 1

您可以gsub与lapply一起使用循环列表元素

lapply(lst1, gsub, pattern= "\\'s|s$|s\\b", replacement='')
#[[1]]
#[1] "all the apple"       "apple  in the table" "fashion prosthetic" 

#[[2]]
#[1] "meteorological concept"    "effect  of climate change"
#[3] "environmental"    

Answer 2

相同的解决方案，不同的正则表达式，使用非捕获组按原样保留空格：

> lapply(lst1, gsub, pattern="s(?= |$)", replacement="", perl=TRUE)
[[1]]
[1] "all the apple"      "apple in the table" "fashion prosthetic"

[[2]]
[1] "meteorological concept"   "effect of climate change" "environmental"           

Answer 3

更简单的正则表达式：

lapply(lst1, function(x) gsub('s\\b', '', x))

结果：

[[1]]
[1] "all the apple"      "apple in the table" "fashion prosthetic"

[[2]]
[1] "meteorological concept"   "effect of climate change"
[3] "environmental"