排除某个字符串第二次出现后的所有内容 [英] Exclude everything after the second occurrence of a certain string
问题描述
我有以下字符串
string <- c('a - b - c - d',
'z - c - b',
'y',
'u - z')
我想对它进行子集化,以便在第二次出现"- "之后的所有内容都被丢弃.
I would like to subset it such that everything after the second occurrence of ' - ' is thrown away.
结果是这样的:
> string
[1] "a - b" "z - c" "y" "u - z"
我使用了 substr(x = string, 1, regexpr(string, pattern = '[^ - ]*$') - 4)
,但它排除了最后一次出现的 ' - ',这不是我想要的.
I used substr(x = string, 1, regexpr(string, pattern = '[^ - ]*$') - 4)
, but it excludes the last occurrence of ' - ', which is not what I want .
推荐答案
请注意,您不能使用 否定字符类 否定字符序列.[^ - ]*$
匹配除空格以外的任何 0+ 个字符(是的,它也匹配 -
,因为 -
创建了一个一个空格和一个空格之间的范围)后跟字符串标记的结尾($
).
Note that you cannot use a negated character class to negate a sequence of characters. [^ - ]*$
matches any 0+ chars other than a space (yes, it matches -
, too, because the -
created a range between a space and a space) followed by the end of the string marker ($
).
您可以使用带有以下正则表达式的 sub
函数:
You may use a sub
function with the following regex:
^(.*? - .*?) - .*
替换为\1
.请参阅正则表达式演示.
R 代码:
> string <- c('a - b - c - d', 'z - c - b', 'y', 'u - z')
> sub("^(.*? - .*?) - .*", "\\1", string)
[1] "a - b" "z - c" "y" "u - z"
详情:
^
- 字符串的开始(.*? - .*?)
- 第 1 组(在替换模式中使用\1
反向引用引用)捕获任何 0+ 个字符 懒惰直到第一个空格,连字符,空格,然后是任何 0+ 个字符,直到下一个最左边出现的空格,连字符,空格-
- 一个空格、连字符和一个空格.*
- 直到字符串末尾的任何零个或多个字符.
^
- start of a string(.*? - .*?)
- Group 1 (referred to with the\1
backreference in the replacement pattern) capturing any 0+ chars lazily up to the first space, hyphen, space and then again any 0+ chars up to the next leftmost occurrence of space, hyphen, space-
- a space, hyphen and a space.*
- any zero or more chars up to the end of the string.
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