R-regex:匹配不以模式开头的字符串 [英] R-regex: match strings not beginning with a pattern

问题描述

我想使用正则表达式来查看字符串是否以某种模式开头.虽然我可以使用:[^ 将某些字符列入黑名单，但我不知道如何将某个模式列入黑名单.

I'd like to use regex to see if a string does not begin with a certain pattern. While I can use: [^ to blacklist certain characters, I can't figure out how to blacklist a pattern.

> grepl("^[^abc].+$", "foo")
[1] TRUE
> grepl("^[^abc].+$", "afoo")
[1] FALSE

我想做一些类似 grepl("^[^(abc)].+$", "afoo") 的事情，然后得到 TRUE，即如果字符串不以 abc 序列开头，则匹配.

I'd like to do something like grepl("^[^(abc)].+$", "afoo") and get TRUE, i.e. to match if the string does not start with abc sequence.

请注意，我知道这篇文章，我也尝试使用 perl = TRUE，但没有成功:

Note that I'm aware of this post, and I also tried using perl = TRUE, but with no success:

> grepl("^((?!hede).)*$", "hede", perl = TRUE)
[1] FALSE
> grepl("^((?!hede).)*$", "foohede", perl = TRUE)
[1] FALSE

有什么想法吗?

推荐答案

是的.将零宽度前瞻/外部/其他括号.这应该给你这个:

Yeah. Put the zero width lookahead /outside/ the other parens. That should give you this:

> grepl("^(?!hede).*$", "hede", perl = TRUE)
[1] FALSE
> grepl("^(?!hede).*$", "foohede", perl = TRUE)
[1] TRUE

我认为这是你想要的.

或者，如果您想捕获整个字符串，^(?!hede)(.*)$ 和 ^((?!hede).*)$是等价的，也是可接受的.

Alternately if you want to capture the entire string, ^(?!hede)(.*)$ and ^((?!hede).*)$ are both equivalent and acceptable.

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