Python find and replace numbers based on pattern

Question

I've got a string and I'd like to find and replace some numbers in it. I.e., there are multiple repetitions of "v = 324\n" in there, with different values. Now I want to divide all those numbers by n (rounded to nearest integer) and save it as a new string.

At the moment I'm using parse package:

n = 10
s = "this is v = 2342\n and another v = 231\n and some stuff..."
for r in findall("v = {:d}\\n", s):
    print r

This gives me the list of Results, but I don't know how to make changes to the string. How can I do it?

Jason Sperske · Accepted Answer · 2012-11-20 01:03:46Z

3

You can pass a function to re.sub that takes your matched pattern samples = {:d}\\n (which I had to update) and computes it in some way. Here is a demo:

import re

def sampleRounder(match):
    return str(int(float(match.group(1)))) #base=10

s = "this is v = 2342.2\n and another v = 231.003\n and some stuff..."

print(re.sub("v = ([0-9]*\.[0-9]+|[0-9]+)\\n", sampleRounder, s))

edited Nov 20, 2012 at 1:03

answered Nov 20, 2012 at 0:54

Jason Sperske

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3 Comments

sashkello Over a year ago

Perfect! Thank you, great answer and easy to reuse :)

abarnert Over a year ago

This truncates instead of rounding to the nearest int, but as long as the OP can figure out how to to do that himself (hint: the round builtin does what it says, try help(round) for more), this should be all he needs.

Jason Sperske Over a year ago

I reread the OP's question and I can see that my sample code was way off of the request. I'll fix it when I get back to a desktop browser, but the idea still holds :)

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