为什么re.finditer似乎忽略了所提供的模式？

Question

Maybe I'm misunderstanding how re.finditer works, but it seems to be ignoring the pattern I give it and matching on all instances of the character. For example, this code:

word = "aaaa"
for match in re.finditer(r"(?<=a)a(?=a)", word):
    word = re.sub(match.group(), "b", word)

is giving me "bbbb" , when I would expect it to give "abba" , like this code does:

word = "aaaa"
word = re.sub(r"(?<=a)a(?=a)", "b", word)

Any idea why matching through re.finditer seems to be working differently from doing it directly with re.sub ?

EDIT: To clarify, what I'm actually trying to do is lookup what substitution to apply in a dictionary, but

word = re.sub(r"(?<=a)(a)(?=a)", d[r"\1"], word)

(where d is a dictionary) doesn't work because \\1 is not a key, so I was hoping to iterate through and apply substitutions as I find them. I also can't just do a separate line for each substitution in the dictionary because I have a lot of changes that are circular, so a > b, b > c, and c > a, if that makes sense. In other words, if I try to apply the changes linearly, I'm going to end up back where I started. Maybe I'm approaching this the wrong way?

Answer 1

In the first case, the value of match.group() is "a" at both iterations. word = re.sub(match.group(), "b", word) becomes word = re.sub("a", "b", word) , which naturally replaces each "a" with a "b".

In the second example, only the middle "a"'s are replaced.

Answer 2

I found the answer I was looking for here: Dictionary lookup with key as a matched group in python re.sub module .

I didn't realize you can pass in a function as the second parameter to sub , so essentially, all you have to do is something like this to incorporate dictionary lookup into the substitution:

word = re.sub(r"(?<=a).(?=a)", lambda x: d[x.group()], word)