从字符串中提取数字，同时保留空格

Question

I have some string like this

'    12    2    89   29   11    92     92     10'

(all the numbers are positive integers so no - and no . ), and I want to extract all numbers from it, edit some of the numbers, and then put them all together with the same whitespaces. For example, if I change the number 11 to 22 , I want the final string as

'    12    2    89   29   22    92     92     10'

I did some search and most questions disregard the whitespaces and only care about the numbers. I tried

match = re.match((\s*(\d+)){8}, str)

but match.group(0) gives me the whole string,, match.group(1) gives me the first match \\ 12 (I added the \\ otherwise the website won't show the leading whitespaces), and match.group(2) gives me 12 . But it won't give me any numbers after that, any index higher than 2 gives me an error. I don't think my approach is the correct one, what is the right way to do this?

I just tried re.split('(\\d+)', str) and that seems to be what I need.

Answer 1

I'd recommend using a regular expression with non-capturing groups , to get a list of 'space' parts and 'number' parts:

In [15]: text = '    12    2    89   29   11    92     92     10'
In [16]: parts = re.findall('((?: +)|(?:[0-9]+))', text)
In [17]: parts
Out[17]: ['    ', '12', '    ', '2', '    ', '89', '   ', '29', '   ',
  '11', '    ', '92', '     ', '92', '     ', '10']

Then you can do:

for index, part in enumerate(parts):
    if part == '11':
        parts[index] = '22'
replaced = ''.join(parts)

(or whatever match and replacement you want to do).

Answer 2

Match all numbers with spaces, change desired number and join array.

import re

newNum = '125'
text = '    12    2    89   29   11    92     92     10'
                                              ^^
marray = re.findall(r'\s+\d+', text)
marray[6] = re.sub(r'\d+', newNum, marray[6])

print(marray) 

['    12', '    2', '    89', '   29', '   11', '    92', '     125', '     10']