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

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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().

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