Python regex fractions

Question

I'm trying to use regex to replace instances of a fraction (specifically 1/2) with the decimal equivalent

string = "7 1/2"
re.sub(r'[1/2]', '.5', string)

Firstly, I think the above replaces all instances of 1, /, 2 with .5, whereas I'm trying to find and replace the entire term "1/2"

Secondly, how do you handle the leading space before the fraction itself?

"I think the above..." Have you tried or is this just speculation? — Arc676
– Arc676, Commented Dec 16, 2015 at 5:38
well when I try it on 7 1/2 I get 7 .5.5.5 so I believe its true in that case, just not sure if thats the generalized outcome — Simon
– Simon, Commented Dec 16, 2015 at 5:42

Avinash Raj · Accepted Answer · 2015-12-16 05:52:15Z

8

A general solution should be like below which should use a lambda function in the replacement part of re.sub.

>>> import re
>>> from __future__ import division
>>> s = "7 1/2"
>>> re.sub(r'(\d+)/(\d+)', lambda m: str(int(m.group(1))/int(m.group(2))), s)
'7 0.5'
>>> re.sub(r'(\d+)/(\d+)', lambda m: str(int(m.group(1))/int(m.group(2))), '7 9/2')
'7 4.5'

Update:

>>> re.sub(r'^(\d+)\s+(\d+)/(\d+)$', lambda m: str(float(int(m.group(1)) + int(m.group(2))/int(m.group(3)))), '7 9/2')
'11.5'

edited Dec 16, 2015 at 5:52

answered Dec 16, 2015 at 5:42

Avinash Raj

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1 Comment

Simon Over a year ago

Thanks, how would you add the converted fraction part back to the integer though? "7 1/2" should be "7.5" and "7 9/2" should be "11.5"

mhawke · Accepted Answer · 2015-12-16 06:01:57Z

2

As an alternative to regex, if you always have a single number per string, and the fractional part is separated from the whole number part by whitespace, you could use the fractions module to perform a conversion:

from fractions import Fraction

for s in '7 1/2', '1 33/66', '22 21/22', '123', '123.45':
    f = sum(Fraction(x) for x in s.split())
    print(float(f))

Output

7.5
1.5
22.9545454545
123.0
123.45

answered Dec 16, 2015 at 6:01

mhawke

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