FAQ: Sets - Creating a Frozenset

This community-built FAQ covers the “Creating a Frozenset” exercise from the lesson “Sets”.

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Why when we print frozensets we see the frozenset constructor on the screen? For example,

top_genres = ['rap', 'rock', 'pop']

# Write your code below!
frozen_top_genres = frozenset(top_genres)

This outputs this:

frozenset({'rock', 'pop', 'rap'})

Why is there still frozenset on the screen?

Hello, @ superpythoncoder12

We see that output simply because it’s a quirk of python to disambiguate two very similar data structures that perform almost the same tasks but vary ever so subtly.

In this case set() vs frozenset() are both “set-type” data structures, with the difference being the latter cannot be modified.

The same behaviour can be observed between dict() and OrderedDict(), which are both “dictionary-type” data structures. The latter now resides within the collections module so you’d have to import it to use it, and has its own uses that differentiate from a “normal”, and which I cannot even begin to cover here. Nonetheless here’s what the output would look like between the two:

from collections import OrderedDict

In [23]: normal_d = dict(b=2, d=4, a=1, c=3)
In [24]: print(normal_d)
{'b': 2, 'd': 4, 'a': 1, 'c': 3}

In [25]: ordered_d = OrderedDict(b=2, d=4, a=1, c=3)
In [26]: print(ordered_d)
OrderedDict([('b', 2), ('d', 4), ('a', 1), ('c', 3)])

Hope this answers your question.
Take care

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