Imagine we have a list of variable names like following:
想象一下,我们有一个变量名列表,如下所示:
ls<-c("apple.mean", "orange.mean", "orange.sd", "apple.pie.mean", "orange.juice.n", "orange.juice.p%")
How can we remove the last part (after ".") in each element so we can get:
我们如何删除每个元素中的最后一部分(在“。”之后),以便我们得到:
"apple" "orange" "orange" "apple.pie" "orange.juice" "orange.juice"
Note that there might be "." inside the names but I don't want those words to be split.
请注意,可能有“。”在名称内,但我不希望这些单词被拆分。
I was trying to use gsub("\\..*$", "",ls) but it omits everything after the 1st dot. I'm not sure why the $ sign is not working here. Any ideas?
我试图使用gsub(“\\ .. * $”,“”,ls)但它在第一个点之后省略了所有内容。我不确定为什么$符号在这里不起作用。有任何想法吗?
> gsub("\\..*$", "",ls)
[1] "apple" "orange" "orange" "apple" "orange" "orange"
2 个解决方案
#1
6
You can try
你可以试试
sub('[.][^.]+$', '', ls)
#[1] "apple" "orange" "orange" "apple.pie" "orange.juice"
#[6] "orange.juice"
#2
2
Given that this is the equivalent of removing a file extension you could use
鉴于这相当于删除您可以使用的文件扩展名
library(tools)
file_path_sans_ext(ls)
#1
6
You can try
你可以试试
sub('[.][^.]+$', '', ls)
#[1] "apple" "orange" "orange" "apple.pie" "orange.juice"
#[6] "orange.juice"
#2
2
Given that this is the equivalent of removing a file extension you could use
鉴于这相当于删除您可以使用的文件扩展名
library(tools)
file_path_sans_ext(ls)