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)

• 0x5453: 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.
• Wiktor Stribiżew: @0x5453 No, {{.*?}} is a valid pattern. No need to escape anything. It is not Android or C++ (MSVC).
• Wiktor Stribiżew: 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?

Answer 1

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)

Answer 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円', '\2円']), s)
# a and c and 0

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