[英]How can I remove leading zeros for two digits number (01, 02, etc.) from the middle of character string using R?
For the following string vector s
, I hope to remove leading zeros in each elements, which is reverse of the answer from this link :对于以下字符串向量
s
,我希望删除每个元素中的前导零,这与此链接的答案相反:
s <- c('week 01st', 'weeks 02nd', 'year2022week01st', 'week 4th')
The expected result will like:预期的结果如下:
s <- c('week 1st', 'weeks 2nd', 'year2022week1st', 'week 4th')
I test the following code, it's not working out since the regex syntax is not complete:我测试了以下代码,由于正则表达式语法不完整,因此无法正常工作:
s <- 'week 01st'
sub('^0+(?=[1-9])', '', s, perl=TRUE)
sub('^0+([1-9])', '\\1', s)
Out:出去:
[1] "week 01st"
How could I do that using R?我怎么能用 R 做到这一点?
Update: for the following code contributed by @dvantwisk, it works for year2022week01st
, but not suitable to other elements:更新:对于@dvantwisk 贡献的以下代码,它适用于
year2022week01st
,但不适用于其他元素:
s <- c('week 01st', 'weeks 02nd', 'year2022week01st', 'week 4th')
gsub('(year[0-9]{4,})(week)(0{0,})([1-9]{1})([0-9a-zA-Z]{1,})', '\\1\\2\\4\\5', s)
Out:出去:
[1] "week 01st" "weeks 02nd" "year2022week1st" "week 4th"
You might use:您可能会使用:
weeks?\h*\K0+(?=[1-9]\d*[a-zA-Z])
The pattern matches:模式匹配:
weeks?
Match week with optional s\h*\K
Match optional spaces and forget what is matched so far \h*\K
匹配可选空格并忘记到目前为止匹配的内容0+
Match 1+ times a zero 0+
匹配 1+ 次零(?=[1-9]\d*[a-zA-Z])
Positive lookahead, assert a char 1-9, optional digit and a char a-zA-Z to the right (?=[1-9]\d*[a-zA-Z])
正向前瞻,断言一个字符 1-9,可选数字和一个字符 a-zA-Z 到右边See a Regex demo and a R demo .请参阅Regex 演示和R 演示。
In the replacement use an empty string.在替换中使用空字符串。
For example例如
s <- c('week 01st', 'weeks 02nd', 'year2022week01st', 'week 4th')
gsub("weeks?\\h*\\K0+(?=[1-9]\\d*[a-zA-Z])", '', s, perl=T)
Output Output
[1] "week 1st" "weeks 2nd" "year2022week1st" "week 4th"
Or with 2 capture groups:或使用 2 个捕获组:
(weeks?\h*)0+([1-9]\d*[a-zA-Z])
Example:例子:
s <- c('week 01st', 'weeks 02nd', 'year2022week01st', 'week 4th')
gsub("(weeks?\\h*)0+([1-9]\\d*[a-zA-Z])", '\\1\\2', s,)
Output Output
[1] "week 01st" "weeks 02nd" "year2022week1st" "week 4th"
gsub('(week )(0{0,})([1-9]{1})([0-9a-zA-Z]{1,})', '\\1\\3\\4', week_string)
gsub()
takes three arguments as input: a pattern, a replacement, and a query character vector. gsub()
将三个 arguments 作为输入:一个模式、一个替换和一个查询字符向量。 Our strategy is to create a regular expression with four groups with ()
s.我们的策略是使用
()
创建一个包含四个组的正则表达式。
We fist match 'week '.我们拳头比赛'周'。
We then match zero or more zeros with the expression (0{0,})
.然后我们将零个或多个零与表达式
(0{0,})
匹配。 The first zero indicates the character we are trying to match and the expression {0,}
indicates we are trying to match zero (hence the 0) or more (hence the comma) times.第一个零表示我们尝试匹配的字符,表达式
{0,}
表示我们尝试匹配零次(因此是 0)或更多次(因此是逗号)。
Our third group is matching any number between 1 to 9 one time.我们的第三组匹配一次 1 到 9 之间的任何数字。
Out fourth group is to match any number between 0 to 9 or any letter 1 or more times第四组是匹配 0 到 9 之间的任何数字或任何字母 1 次或多次
Our replacement is '\\1\\3\\4'
.我们的替代品是
'\\1\\3\\4'
。 This indicates we only want to keep group one and three in our result.这表明我们只想在结果中保留第一组和第三组。 Thus the output is:
因此 output 是:
[1] "week 1st" "week 2nd" "week 3rd" "week 4th"
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