如何使用正则表达式替换一定数量的空格？

Question

I'm using Calibre to convert a PDF to MOBI, but it has trouble interpreting space-indented code blocks. The blocks contain a lot of spaces, but in a lot of different amounts. Some lines are even indented by 31 spaces.

Calibre allows for 3 regexes to do search and replace in the book before it's converted.

This is what I've tried.

\n( *) ( *)([a-zA-Z{};\*\/\(\)&#0-9])

Replace with:

\n\1 \2\3

The problem, it only replaces one of the spaces. I want them all replaced with the same abount of   .

I've also tried lazy versions of the first group etc.

Is this one of the cases where regular expressions are insufficient? I think this regex engine is the python standard.

Answer 1

If this were Perl you could replace (\\G|\\n) with $1  , and if it were a regex engine that allowed limited-width lookbehinds (instead of fixed-width lookbehinds like Python's) you could replace (?<=\\n {0,30}) with   ; but as it is, the only way I can think of is to replace something like ((?<=\\n)|(?<=\\n )|(?<=\\n {2})|(?<=\\n {3})|(?<=\\n {4})|(?<=\\n {5})|...|(?<=\\n {30})) with   . . . and I suspect that at that point you'll reach a limit on how long Calibre allows the input regex to be. :-/

Another option is to take a completely different approach, and replace    (two spaces) with   (non-breaking-space + regular space), without bothering to restrict it to the beginning of a line. I'm guessing that that will satisfy your needs?

Answer 2

\\ s {31}恰好匹配31个空格，\\ s {14,31} 14到31

Answer 3

Any reason not to just replace ALL spaces by non-breaking spaces? ( r/ / / .)

It won't change the appearance of normal English text (except where the source had extraeneous double-spaces) and your code blocks will render correctly.

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For fun, my attempt in Python:

>>> eight_spaces = "        hello world!"
>>> re.sub(r"^(|(?: )*)\s",r"\1 ",eight_spaces)
'       hello world!'

The idea is to replace one space at a time. It doesn't work because the re engine doesn't go back to the start of the line after a match - it consumes the string working left to right.

Note the alternation of (?: )* with the empty string, (|(?: )*) , so that the capture group \\1 always captures something (even the empty string.)