Wow...how far I have to go still.

I spent 15 minutes trying to understand how the part with '\n1'.join(... works. It was much easier than I thought.

the kata needs clarification but I think if it meant to return "bye" and not "e" then it should also return "hi" if "hi" is the only argument. otherwise it's inconsistent. what if the arguments were ("hello", [1, "world", [2, 3]])? Should it return [1, "world", [2, 3]], [2, 3], or 3 ?

Easiest, cleanest.

yes

and this will return 2 --> 2.0

Cool, couldn't figure out where to put if/else in join !!! Thanks

groupby() take iterable(sort of sequence) and return tuple of group key(str) and lazy iterator(itertols._grouper).
So because its lazy it will give you a value only as in needed. But you can get sum of iterators values.

for k,g in groupby(s): 
  print(type(k)) # str
  print(type(g)) # itertols._grouper
  x = sum(1 for _ in g) 
  print(f'This is a key {k} and these is a sum {x} of the each group'}

Then you need to create generator expression, which consist of

[[computing_sum_of_the_group=example above, key] for key, group in each_tuple]

Hope i did it good.

nice use of recursion!

A zero-liner, anyone?

Modified, thanks!

thanks

Now it works.
Thank you!

Have you tried again? I changed the tests.