Sort Dictionary by Reference? (Python 3.10)

June 26, 2023

I’m attempting to sort a dictionary in Python in ascending key order. I’m trying to do this inside of a function, with the dictionary being supplied as an argument.

I understand that there are mutable and immutable datatypes in Python, and I understand that a dictionary is supposed to be a mutable datatype. That is, I should be able to pass it by-reference when supplying it as a function argument.

Usually, I would use the dict(sorted(d.items())) method, however that does not appear to work when supplying a dictionary by reference.

Here’s an example:

def CreateDict(d):
    d[0] = "a"
    d[4] = "b"
    d[2] = "c"
    d[1] = "d"

def SortDict(d):
    d = dict(sorted(d.items()))

def main():
    d = {}

    print(d)
    CreateDict(d)
    print(d)
    SortDict(d)
    print(d)

if (__name__ == '__main__'):
    main()

Where I add some elements into the dictionary d in the CreateDict function, and I try to sort it in the SortDict function. However, the output I get is the following:

{}
{0: 'a', 4: 'b', 2: 'c', 1: 'd'}
{0: 'a', 4: 'b', 2: 'c', 1: 'd'}

With d remaining unsorted even after a call to SortDict.

What is the proper way of sorting a Python dictionary when supplied by-reference as a function argument?

Thanks for reading my post, any guidance is appreciated.

>Solution :

What is the proper way of sorting a Python dictionary when supplied by-reference as a function argument?

I’d say it’s to return the new dict, and have the caller use the return value.

But, if you insist on modifying the existing dict, here are a few ways:

def SortDict(d):
    items = sorted(d.items())
    d.clear()
    d |= items

def SortDict(d):
    for k in sorted(d):
        d[k] = d.pop(k)

def SortDict(d):
    k = sorted(d)
    d |= zip(k, map(d.pop, k))