Replace with a custom function with Python re

Question

For a string

s = '{{a,b}} and {{c,d}} and {{0,2}}'

I'd like to replace every {{...}} pattern by randomly one of the items in the list inside, i.e.:

"a and d and 2"  
"b and d and 0"  
"b and c and 0"
...

I remember that there's a way in module re to not simply replace like re.sub, but have a custom replacement function, but I can't find this anymore in the doc (maybe I'm searching with wrong keywords...)

This doesn't give any output:

import re

r = re.match('{{.*?}}', '{{a,b}} and {{c,d}} and {{0,2}}')
for m in r.groups():
    print(m)

The re.sub docs have an example of using a function for repl. Also be aware that {} have special meaning within regex and will need to be escaped if you want a literal match. — 0x5453
– 0x5453, Commented Apr 7, 2020 at 14:46
@0x5453 No, {{.*?}} is a valid pattern. No need to escape anything. It is not Android or C++ (MSVC). — Wiktor Stribiżew
– Wiktor Stribiżew, Commented Apr 7, 2020 at 14:46
You may consider ` re.sub(r'{{(\w+),(\w+)}}', lambda x: x.group(random.choice([1,2])), s)`, but it will support just two values. Do you want to produce a single result, or all possible permutations? — Wiktor Stribiżew
– Wiktor Stribiżew, Commented Apr 7, 2020 at 14:51

Jan · Accepted Answer · 2020-04-07 14:47:59Z

4

You could use

import random, re

def replace(match):
    lst = match.group(1).split(",")
    return random.choice(lst)

s = '{{a,b}} and {{c,d}} and {{0,2}}'

s = re.sub(r"{{([^{}]+)}}", replace, s)
print(s)

Or - if you're into one-liners (not advisable though):

s = re.sub(
    r"{{([^{}]+)}}", 
    lambda x: random.choice(x.group(1).split(",")), 
    s)

answered Apr 7, 2020 at 14:47

Jan

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Comments

Austin · Accepted Answer · 2020-04-07 14:52:09Z

2

You can avoid splitting using appropriate regex to fetch patterns:

import re, random

s = '{{a,b}} and {{c,d}} and {{0,2}}'
s = re.sub(r'{{(.*?),(.*?)}}', random.choice(['\\1', '\\2']), s)

# a and c and 0

answered Apr 7, 2020 at 14:52

Austin

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2 Comments

Basj Over a year ago

Thank you for your answer. I'd like the custom replacement function to work for any length of the list (here only 2 items), do you have any idea?

Austin Over a year ago

@Basj, Splitting is the option for any length list. But for small lengths, you can see using the approach in my answer.

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