删除最后一个空格之前的所有内容 [英] Remove everything before the last space
问题描述
我有以下字符串.我试图删除最后一个空格之前的所有字符串，但似乎无法实现.
I have a following string. I tried to remove all the strings before the last space but it seems I can't achieve it.
我试图关注这篇文章
str <- c("Veni vidi vici")
gsub("\\s*","\\1",str)
"Venividivici"
在删除最后一个空格之前的所有内容后，我只想剩下"vici"个字符串.
What I want to have is only "vici" string left after removing everything before the last space.
推荐答案
您的gsub("\\s*","\\1",str)代码使用对捕获组#1值的引用(由于您没有指定了模式中的任何捕获组.
Your gsub("\\s*","\\1",str) code replaces each occurrence of 0 or more whitespaces with a reference to the capturing group #1 value (which is an empty string since you have not specified any capturing group in the pattern).
您要匹配最后一个空格:
You want to match up to the last whitespace:
sub(".*\\s", "", str)
如果您不想在字符串末尾有空格时得到空白结果，请先修剪字符串:
If you do not want to get a blank result in case your string has trailing whitespace, trim the string first:
sub(".*\\s", "", trimws(str))
或者，使用方便的 stringi >包并带有简单的\S+模式(匹配1个或多个非空白字符):
Or, use a handy stri_extract_last_regex from stringi package with a simple \S+ pattern (matching 1 or more non-whitespace chars):
library(stringi)
stri_extract_last_regex(str, "\\S+")
# => [1] "vici"
请注意，.*尽可能匹配任何0+个字符(因为*是贪婪的量词，而TRE模式中的.匹配任何包含换行符的char)，并首先捕获整个字符串.然后，由于正则表达式引擎需要将空白与\s匹配，因此开始回溯. regex引擎从字符串的末尾开始逐个字符地产生字符，它在最后一个空白处绊倒，并将其称为一天，返回随后被删除的匹配项.
Note that .* matches any 0+ chars as many as possible (since * is a greedy quantifier and . in a TRE pattern matches any char including line break chars), and grabs the whole string at first. Then, backtracking starts since the regex engine needs to match a whitespace with \s. Yielding character by character from the end of the string, the regex engine stumbles on the last whitespace and calls it a day returning the match that is removed afterwards.
请参见 R演示和
此外，您可能希望在 regex调试器中查看回溯的工作方式: Also, you may want to see how backtracking works in the regex debugger: 那些红色箭头显示了回溯步骤. Those red arrows show backtracking steps. 这篇关于删除最后一个空格之前的所有内容的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！
