Question

我利用这一规章,将数千个分离者置于地狱中:

while matchstr(mystr,  (d)(d{3})(D|s|$) ) !=   
    let mystr = substitute(mystr,  (d)(d{3})(D|s|$) ,  1.23 ,  g )
endwhile

页: 1

let mystr =  2000000

上述法典

2.000.000

问题在于有一位精子的分离者,它还把几千名分离者放在离奇马尔的分离者(下称 com)之后的一小部分。

页: 1 example,

let mystr =  2000000,2346

导致

2.000.000,2.346

而我却希望

2.000.000,2346

I tried to adapt the above code but didn t find a satisfiable solution. Can anyone help me?

Answer 1

Use the following call to the substitute() function instead of the whole loop listed in the question.

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

Answer 2

www.un.org/spanish/ecosoc Try this (work for positive integer only):

<>http://europa-eu>

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

<>Replace with

I ve not tried over vim whether it would work or not. But the pattern is PCRE compatible.

<>Explanation>

<!--
(?<=[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]»
-->

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