How to sort dictionary by key in numerical order Python

Question

Here is the dictionary looks like:

{\'57481\': 50, \'57480\': 89, \'57483\': 110, \'57482\': 18, \'57485\': 82, \'57484\': 40}  

I would

Answer 1

In Python 3 sorted() has an optional parameter key. And in 3.6+ dict maintains insertion order.

key specifies a function of one argument that is used to extract a comparison key from each element in iterable (for example, key=str.lower). The default value is None (compare the elements directly).

Therefore, what OP wants can be accomplished this way.

>>> d = {'57481': 50, '57480': 89, '57483': 110, '57482': 18, '57485': 82, '57484': 40}
>>> for key, value in sorted(d.items(), key=lambda item: int(item[0])):
...     print(key, value)
57480 89
57481 50
57482 18
57483 110
57484 40
57485 82

Or if OP wants to create a new sorted dictionary.

>>> d = {'57481': 50, '57480': 89, '57483': 110, '57482': 18, '57485': 82, '57484': 40}
>>> d_sorted = {key:value for key, value in sorted(d.items(), key=lambda item: int(item[0]))}
>>> d_sorted
{'57480': 89, '57481': 50, '57482': 18, '57483': 110, '57484': 40, '57485': 82}

d.items() returns a list of tuples, e.g. ('57480': 89) and so on. The lambda functions takes this tuple and applies int function to the first value. And then the result is used for comparison.