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I have some trouble transforming a string into an array. The separators between numbers are spaces, and the thousands also.

My string is this:

'25 127,4 17 588,6 16 264,3 324,4 8,7'

I want it to look like this:

['25 127,4', '17 588,6', '16 264,3', '324,4', '8,7']

or better like this:

['25127,4', '17588,6', '16264,3', '324,4', '8,7']

I was trying to use regex and findall to do so, but the problem is that it only captures 5-digit numbers.

My code is kind of like this:

a = '25 127,4 17 588,6 16 264,3 324,4 8,7'
print(re.findall(r'\d+\s\d+,\d{1}', a))

which gives me this output:

['25 127,4', '17 588,6', '16 264,3', '4 8,7']

How to solve this?

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3
  • 1
    Try (?! )[\d ]+,\d Commented Jul 7, 2021 at 3:06
  • 1
    Maybe re.findall(r'[0-9\s]+,\d+', s) given a string s? Commented Jul 7, 2021 at 3:07
  • Thanks a lot @HaoWu and Mark. It worked!! Commented Jul 7, 2021 at 3:44

2 Answers 2

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1

The regex recommendation in the comment seems to work for your example, but having spaces represent delimiters and thousand separators in your data will cause a lot of problems. If possible you should see if a different sort of delimiter could be used!

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1 Comment

Thank you for the recommendations. Sure will change the delimeter
1

You can optionally repeat matching a space and matching 1+ digits. (Note that \s can also match a newline)

Then remove the spaces from the matches.

\b\d+(?: \d+)*,\d\b

Regex demo

import re

pattern = r"\b\d+(?: \d+)*,\d\b"
s = "25 127,4 17 588,6 16 264,3 324,4 8,7"
result = [m.replace(" ", "") for m in re.findall(pattern, s)]
print(result)

Output

['25127,4', '17588,6', '16264,3', '324,4', '8,7']
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