In the evening the bakery updates its chalkboard. The shelf is a dict: name → how many are left. Write still_left(shelf) to return a new dict with only what's still there (more than 0), in the same order. An empty dict gives an empty dict.
still_left({"bagel": 12, "croissant": 0, "muffin": 3})→{'bagel': 12, 'muffin': 3}still_left({})→{}def still_left(shelf):
# your code here: a dict comprehension keeping only n > 0
return {}
Run tries the examples. When they look right, Submit.