Convert string to json where some values are wrapped in square brackets

What would be the most efficient way (python 3) to convert a string to dict, where some of the keys and values may or may not be quoted and some values may contain sub objects wrapped in square brackets ([]) instead of curly ones.

Also values may contain colon (:) in them

Example:

string = "[key:value, key2:[key2a:val2:a, key2b:[foo:"bar"]]]"

the results needs to be a valid dict like:

{"key":"value", "key2":{"key2a":"val2:a", "key2b":{"foo":"bar"}}}

Solution

You can use a recursive generator function:

import re
string = "[key:value, key2:[key2a:val2:a, key2b:[foo:'bar']]]"
d = [i if not i.startswith("'") else i[1:-1] for i in re.findall("[\[\]]|:|'.*?'|\w+|,", string)[1:-1]]
def to_dict(d):
   while (n:=next(d, None)) not in {None, ']'}:
      _ = next(d)
      if (v:=next(d)) == '[':
         v = dict(to_dict(d))
      else:
         c = [v]
         while (j:=next(d)) not in {',', ']'}:
            c.append(j)
         if j == ']':
            d = iter([*d, j])
         v = ''.join(c)
      yield (n, v)

print(dict(to_dict(iter(d))))

Output:

{'key': 'value', 'key2': {'key2a': 'val2:a', 'key2b': {'foo': 'bar'}}}

Edit: solution without assignment expressions (:= walrus operator):

def to_dict(d):
   n = next(d, None)
   while n not in {None, ']'}:
      _, v = next(d), next(d)
      if v == '[':
         v = dict(to_dict(d))
      else:
         c, j = [v], next(d)
         while j not in {',', ']'}:
            c.append(j)
            j = next(d)
         if j == ']':
            d = iter([*d, j])
         v = ''.join(c)
      yield (n, v)
      n = next(d, None)


print(dict(to_dict(iter(d))))