删除字母之间的空格[A-Za-z]

How to remove whitespaces between letters NOT numbers

如何删除字母NOT数字之间的空格

For example:

Input

I ES P 010 000 000 000 000 000 001 001 000 000 IESP 000 000

Output

IESP 010 000 000 000 000 000 001 001 000 000 IESP 000 000

I tried something like this

我试过这样的事

gsub("(?<=\\b\\w)\\s(?=\\w\\b)", "", x,perl=T)

But wasn't able to arrive at the output I was hoping for

但是无法达到我希望的输出

2 个解决方案

#1

You need to use

你需要使用

Input <- "I ES P E ES P 010 000 000 000 000 000 001 001 000 000 IESP 000 000"
gsub("(?<=[A-Z])\\s+(?=[A-Z])", "", Input, perl=TRUE, ignore.case = TRUE)
## gsub("(*UCP)(?<=\\p{L})\\s+(?=\\p{L})", "", Input, perl=TRUE) ## for Unicode

See the R demo online and a regex demo.

在线查看R演示和正则表达式演示。

NOTE: The ignore.case = TRUE will make the pattern case insensitive, if it is not expected, remove this argument.

注意:ignore.case = TRUE将使模式不敏感,如果不期望,则删除此参数。

Details

(?<=[A-Z]) (or (?<=\p{L})) - a letter must appear immediately to the left of the current location (without adding it to the match)

(?<= [A-Z])(或(?<= \ p {L})) - 一个字母必须立即出现在当前位置的左侧(不将其添加到匹配项中)

\\s+ - 1 or more whitespaces

\\ s + - 一个或多个空格

(?=[A-Z]) (or (?=\\p{L})) - a letter must appear immediately to the right of the current location (without adding it to the match).

(?= [A-Z])(或(?= \\ p {L})) - 一个字母必须立即出现在当前位置的右侧(不将其添加到匹配中)。

#2

Use gsub to replace whitespace " " with nothing "" between letters then return replacement and letters.

使用gsub替换字母之间的空格“”,然后返回替换和字母。

Input <- "I ES P 010 000 000 000 000 000 001 001 000 000 IESP 000 000"
gsub("([A-Z]) ([A-Z])", "\\1\\2", Input)
[1] "IESP 010 000 000 000 000 000 001 001 000 000 IESP 000 000"

Edit after @Wiktor Stribiżew comment (replaced [A-z] to [a-zA-Z]):

在@WiktorStribiżew评论后编辑(将[A-z]替换为[a-zA-Z]):

For lower and upper case use [a-zA-Z]

对于小写和大写使用[a-zA-Z]

Input <- "I ES P 010 000 000 000 000 000 001 001 000 000 IESP 000 000 aaa ZZZ"
gsub("([a-zA-Z]) ([a-zA-Z])", "\\1\\2", Input)
[1] "IESP 010 000 000 000 000 000 001 001 000 000 IESP 000 000 aaaZZZ"

#1