Can't use '\1' backreference to capture-group in a function call in re.sub() repr expression

Question

I have a string S = '02143' and a list A = ['a','b','c','d','e'] . I want to replace all those digits in 'S' with their corresponding element in list A .

For example, replace 0 with A[0] , 2 with A[2] and so on. Final output should be S = 'acbed' .

I tried:

S = re.sub(r'([0-9])', A[int(r'\g<1>')], S)

However this gives an error ValueError: invalid literal for int() with base 10: '\\g<1>' . I guess it is considering backreference '\g<1>' as a string. How can I solve this especially using re.sub and capture-groups, else alternatively?

Answer 1

The reason the re.sub(r'([0-9])',A[int(r'\\g<1>')],S) does not work is that \\g<1> (which is an unambiguous representation of the first backreference otherwise written as \\1 ) backreference only works when used in the string replacement pattern . If you pass it to another method, it will "see" just \\g<1> literal string, since the re module won't have any chance of evaluating it at that time. re engine only evaluates it during a match, but the A[int(r'\\g<1>')] part is evaluated before the re engine attempts to find a match.

That is why it is made possible to use callback methods inside re.sub as the replacement argument: you may pass the matched group values to any external methods for advanced manipulation.

See the re documentation :

re.sub(pattern, repl, string, count=0, flags=0)

If repl is a function, it is called for every non-overlapping occurrence of pattern . The function takes a single match object argument, and returns the replacement string.

Use

import re
S = '02143' 
A = ['a','b','c','d','e']
print(re.sub(r'[0-9]',lambda x: A[int(x.group())],S))

See the Python demo

Note you do not need to capture the whole pattern with parentheses, you can access the whole match with x.group() .