regex - 消除句点前的空格，除非后跟一个数字

Question

除非句点后跟一个数字，否则如何使用 R 的正则表达式来消除句点之前的空格？

这是我所拥有的和我尝试过的:

x <- c("I have .32 dollars AKA 32 cents . ", 
    "I have .32 dollars AKA 32 cents .  Hello World .")

gsub("(\\s+)(?=\\.+)", "", x, perl=TRUE)
gsub("(\\s+)(?=\\.+)(?<=[^\\d])", "", x, perl=TRUE)

这给出(.32 之前没有空格):

## [1] "I have.32 dollars AKA 32 cents. "             
## [2] "I have.32 dollars AKA 32 cents.  Hello World."

我想得到:

## [1] "I have .32 dollars AKA 32 cents. "             
## [2] "I have .32 dollars AKA 32 cents.  Hello World."

我在这里背负着 gsub，但欢迎其他解决方案使这个问题对 future 的搜索者更有用。

Answer 1

您不需要复杂的表达式，您可以在此处使用 Positive Lookahead 。

> gsub(' +(?=\\.(?:\\D|$))', '', x, perl=T)
## [1] "I have .32 dollars AKA 32 cents. "             
## [2] "I have .32 dollars AKA 32 cents.  Hello World."

说明 :

 +        # ' ' (1 or more times)
(?=       # look ahead to see if there is:
  \.      #   '.'
  (?:     #   group, but do not capture:
    \D    #      non-digits (all but 0-9)
   |      #     OR
    $     #      before an optional \n, and the end of the string
  )       #   end of grouping
)         # end of look-ahead

注意: 如果这些空格字符可以是任何类型的空格，只需将 ' '+ 替换为 \s+
如果您对使用 (*SKIP)(*F) 回溯动词感到满意，这里是正确的表示:

> gsub(' \\.\\d(*SKIP)(*F)| +(?=\\.)', '', x, perl=T)
## [1] "I have .32 dollars AKA 32 cents. "             
## [2] "I have .32 dollars AKA 32 cents.  Hello World."