正则表达式：如何捕获可能由空格分隔的 6-12 位数字序列而不捕获任何尾随空格

Question

I am attempting capture a sequence of 6-12 digits that may be separated by spaces like the ones below (The letter D at the end is just an example. It's possible that there is nothing at the end of the string, or there is some kind of punctuation or letter).

123 345 4567 89 D
123 345456789 D

My current attempts are as follows:

Attempt 1 : with the lazy quantifier *? :

"\b(?:\d *?){6,12}\b"

With this, it will successfully return all the digits in this string, 123 345456789 D , but fails to fully capture the digits in 123 345 4567 89 D (only the first two groups are captured) -- This I assume is because the first two groups of digits (ie, 123 345 ) fulfill the minimum requirement of 6 digits due to the lazy quantifier, so the regex stops once the minimum requirement is fulfilled.

Attempt 2 : without using the lazy quantifier (just using * ):

"\b(?:\d *){6,12}\b"

With this, all the groups of digits in the examples above are captured. However, this regex will also capture the trailing space that is right between the last digit and the letter D.

So I wonder if there is a way to capture all the digits without including the trailing space. I am doing this in Python, so one thought was to use the second regex but strip away any trailing space after a match is returned, but it seems really inelegant.

Answer 1

This will do it: ((?:\d\s*){5,11}\d?)

See: https://regex101.com/r/qcRbip/1

Answer 2

The quantifier in (?:\d *) is greedy, and will match a space if it is there, also matching it at the end.

In this part (?:\d *?) the quantifier for matching the space is non greedy so after the minimum requirement of 6 times there is a match.

\b\d(?: *\d){5,11}\b

• \b A word boundary
• Match the first digit
• (?: *\d){5,11} Repeat 5 - 11 times and optional spaces and a digit
• \b A word boundary

Regex demo

Answer 3

I cannot reproduce your problem. Your attempt 2 works find for me. Here is my code:

s = "123 345 4567 89 D"
re.findall("(?:\d *?){6,12}", s)

['123 345', '4567 89']


d = "123 345456789 D"
re.findall("(?:\d *?){6,12}", d)

['123 345456789']

Answer 4

"\b\d(?: *\d){5,11}\b"