Python regex return binary value based on compiled pattern

Question

I have a regex to extract numbers from a given string

import re
compiled_pattern = re.compile(r'\d+')
sample = "Hello world 32"

print(compiled_pattern.findall(sample))

output:

['32']

But is it possible to return 1 if there is a number in the string and 0 otherwise? Essentially 1 if there is a match for the pattern in the string and 0 otherwise. So in this case, the op should be 1. Any suggestions would be helpful

@41686d6564 please post it as the answer and I can accept it. — data_person
– data_person, Commented Jul 8, 2021 at 3:12
You can go ahead and accept the existing answer if it helps you; that's fine :) — 41686d6564
– 41686d6564, Commented Jul 8, 2021 at 3:14

Hammurabi · Accepted Answer · 2021-07-08 03:08:57Z

1

Yes, you can test for whether regex found the pattern or not. For example:

def match_or_not(sample):
    compiled_pattern = re.compile(r'\d+')
    match = compiled_pattern.findall(sample)
    return 1 if match else 0

for sample in ["Hello world 32", "Hello again"]:
    print(match_or_not(sample))

answered Jul 8, 2021 at 3:08

Hammurabi

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