regular expression on python picking only one-digit number

Question

I'd like that only one-digit number to be picked.
On windows python, using Japanese unicode, I've:

s = "17 1 27歳女1"
re.findall(r'[1-7]\b', s)

I need to match the second 1 and last 1 in s - not 17 initial 1.
Desired output:

['1', '1'] 

Answer 1

This is the regex you're looking for:

(?<!\d)[1-7](?!\d)

Test:

import re
s="17 1 27歳女1"
re.findall(r'(?<!\d)[1-7](?!\d)', s)

Output:

['1', '1']

Answer 2

Try using a negative-lookbehind (?<!\d). This will ignore matches where a digit is preceded by another one, i.e.:

import re

s = "17 1 27歳女1"
x = re.findall(r"(?<!\d)[1-7]\b", s)
print(x)
# ['1', '1']

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Regex Demo
Python Demo

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Regex Explanation: