如何仅在小数分隔符之前插入千位分隔符

Question

我使用这个正则表达式将千位分隔符放在一个字符串中：

while matchstr(mystr, '\(\d\)\(\d\{3}\)\(\D\|\s\|$\)') != ''
    let mystr = substitute(mystr, '\(\d\)\(\d\{3}\)\(\D\|\s\|$\)', '\1.\2\3', 'g')
endwhile

对于

let mystr = '2000000'

上面的代码给出了

2.000.000

问题是当有小数分隔符时，它也会在小数分隔符后面的数字的小数部分放置千位分隔符（以下是逗号）。

例如，

let mystr = '2000000,2346'

导致

2.000.000,2.346

当我想要它的时候

2.000.000,2346

我尝试修改上面的代码，但没有找到令人满意的解决方案。 谁能帮帮我？

Answer 1

使用以下对substitute() 函数的调用而不是整个函数 问题中列出的循环。

substitute(s, '\(\d,\d*\)\@<!\d\ze\(\d\{3}\)\+\d\@!', '&.', 'g')

Answer 2

试试这个（仅适用于正整数）：

查找

(?<=[0-9])(?=(?:[0-9]{3})+(?![0-9]))

替换为

,

---

我还没有尝试过vim 是否可行。但该模式与 PCRE 兼容。

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说明

<!--
(?<=[0-9])(?=(?:[0-9]{3})+(?![0-9]))

Options: case insensitive; ^ and $ match at line breaks

Assert that the regex below can be matched, with the match ending at this position (positive lookbehind) «(?<=[0-9])»
   Match a single character in the range between “0” and “9” «[0-9]»
Assert that the regex below can be matched, starting at this position (positive lookahead) «(?=(?:[0-9]{3})+(?![0-9]))»
   Match the regular expression below «(?:[0-9]{3})+»
      Between one and unlimited times, as many times as possible, giving back as needed (greedy) «+»
      Match a single character in the range between “0” and “9” «[0-9]{3}»
         Exactly 3 times «{3}»
   Assert that it is impossible to match the regex below starting at this position (negative lookahead) «(?![0-9])»
      Match a single character in the range between “0” and “9” «[0-9]»
-->