How to remove multiple spaces between numbers using a single re.sub

Question

I would like to remove spaces between numbers using a single re.sub. With the following commands:

import re

print(re.sub('([0-9,.]) ([0-9,.])','\\1\\2',str("11 222")))

print(re.sub('([0-9,.]) ([0-9,.])','\\1\\2',str("11 222 33")))

print(re.sub('([0-9,.]) ([0-9,.])','\\1\\2',str("11 222 33 4")))

print(re.sub('([0-9,.]) ([0-9,.])','\\1\\2',str("11 222 33 4 55")))

print(re.sub('([0-9,.]) ([0-9,.])','\\1\\2',str("11 222 33 4 55 6")))

print(re.sub('([0-9,.]) ([0-9,.])','\\1\\2',str("11 222 33 4 55 6 77")))

I can however remove only spaces if there are more that one successive numbers:

11222
1122233
11222334
11222334 55
11222334 556
11222334 556 77

But how to remove also spaces with only one number, so that the result of command like

print(re.sub('([0-9,.]) ([0-9,.])','\\1\\2',str("11 222 33 4 55 6 77")))

would be

1122233455677

?

Answer 1

Try using lookarounds to detect the numbers surrounding a space:

print(re.sub('(?<=\\d) (?=\\d)','',str("11 222 33 4 55 6 77")))

1122233455677

The idea here is for each space we look behind and assert that a digit is present, and also we look ahead and assert that a digit is present.

Note that this answer won't remove spaces which might appear on either end of the string, but then again those spaces are not between numbers.

Answer 2

There are several expressions to grasp the digits. In this case, I find the simplest code in ref . It could get rid of other characters.

print(re.sub(r'\s','',str("11 222 33 4 55")))