CODEWITHSUNDEEP
pythonregex
Deus_Lysenko
Deus_Lysenko

Reputation: 3

Python regular expression match number in string

I used regular expression in python2.7 to match the number in a string but I can't match a single number in my expression, here are my code

import re
import cv2

s = '858  1790 -156.25 2'
re_matchData = re.compile(r'\-?\d{1,10}\.?\d{1,10}')
data = re.findall(re_matchData, s)
print data

and then print:

['858', '1790', '-156.25']

but when I change expression from

re_matchData = re.compile(r'\-?\d{1,10}\.?\d{1,10}')

to 

re_matchData = re.compile(r'\-?\d{0,10}\.?\d{1,10}')

then print:

['858', '1790', '-156.25', '2']

is there any confuses between d{1, 10} and d{0,10} ? If I did wrong, how to correct it ? Thanks for checking my question !

Upvotes: 0

Views: 3802

Answers (3)

Waket Zheng
Waket Zheng

Reputation: 6371

I would rather do as follows:

import re
s = '858  1790 -156.25 2'
re_matchData = re.compile(r'\-?\d{1,10}\.?\d{0,10}')
data = re_matchData.findall(s)
print data

Output:

['858', '1790', '-156.25', '2']

Upvotes: 0

georgexsh
georgexsh

Reputation: 16634

try this:

r'\-?\d{1,10}(?:\.\d{1,10})?'

use (?:)? to make fractional part optional.

for r'\-?\d{0,10}\.?\d{1,10}', it is \.?\d{1,10} who matched 2.

Upvotes: 2

Barmar
Barmar

Reputation: 782407

The first \d{1,10} matches from 1 to 10 digits, and the second \d{1,10} also matches from 1 to 10 digits. In order for them both to match, you need at least 2 digits in your number, with an optional . between them.

You should make the entire fraction optional, not just the ..

r'\-?\d{1,10}(?:\.\d{1,10})?'

Upvotes: 1

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