remove text after final period in string

问题:

I have a grep puzzle that's eluding me: I'd like to remove the text following the final period in a collection of strings (i am using R, so perl syntax is available).

For example, say the string is ABCD.txt this grep would return ABCD, and if the text was abc.com.foo.bar, it would return abc.com.foo.

Any help greatfully appreciated (i don't think i can drink any more coffee!).

回答1:

Here are a few solutions:

sub("^(.*)[.].*", "\\1", "abc.com.foo.bar") # 1
## [1] "abc.com.foo"

library(tools)
file_path_sans_ext("abc.com.foo.bar") # 3
## [1] "abc.com.foo"

ADDED. Regarding your comment asking to remove leading periods, simplest is to just feed this into any of the above where x is the input string:

sub("^[.]*", "", x)

To do any of them in one line:

x <- c("abc.com.foo.bar", ".abc.com.foo.bar", ".vimrc")

sub("^[.]*(.*)[.]?.*$", "\\1", x) # 1a
## [1] "abc.com.foo.bar" "abc.com.foo.bar" "vimrc"          

file_path_sans_ext(sub("^[.]*", "", x))
## [1] "abc.com.foo" "abc.com.foo" "vimrc" 

回答2:

And a non-regex answer for no reason whatsoever:

test <- c("abc.com.foo.bar","ABCD.txt")
sapply(strsplit(test,"\\."), function(x) paste0(head(x,-1),collapse=".") )
#[1] "abc.com.foo" "ABCD"

回答3:

You can use sub for example like this:

sub('(.*)[.](.*)','\\1',c('abc.com.foo.bar','ABCD.txt'))
[1] "abc.com.foo" "ABCD"  

回答4:

I cannot help you with r and I almost forgot perl, but this works both in JS (proof) and PHP

/\.[A-Za-z]+$/     -->    replace this with empty string ""
  ^    ^    ^
  |    |    |
  |    |    end of line
  |    only chars (you can add 0-9 if numbers are also present)
  dot before last chars

the syntax of regex is rather common, so I'm sure you can adopt it (maybe just get rid of /)