How to clone or copy a set in Python?

Evgeni Sergeev picture Evgeni Sergeev · Apr 21, 2014 · Viewed 40.8k times · Source

For copying a list: shallow_copy_of_list = old_list[:].

For copying a dict: shallow_copy_of_dict = dict(old_dict).

But for a set, I was worried that a similar thing wouldn't work, because saying new_set = set(old_set) would give a set of a set?

But it does work. So I'm posting the question and answer here for reference. In case anyone else has the same confusion.

Answer

Evgeni Sergeev picture Evgeni Sergeev · Apr 21, 2014

Both of these will give a duplicate of a set:

shallow_copy_of_set = set(old_set)

Or:

shallow_copy_of_set = old_set.copy() #Which is more readable.

The reason that the first way above doesn't give a set of a set, is that the proper syntax for that would be set([old_set]). Which wouldn't work, because sets can't be elements in other sets, because they are unhashable by virtue of being mutable. However, this isn't true for frozensets, so e.g. frozenset(frozenset(frozenset([1,2,3]))) == frozenset([1, 2, 3]).

So a rule of thumb for replicating any of instance of the basic data structures in Python (lists, dict, set, frozenset, string):

a2 = list(a)      #a is a list
b2 = set(b)       #b is a set
c2 = dict(c)      #c is a dict
d2 = frozenset(d) #d is a frozenset
e2 = str(e)       #e is a string
#All of the above give a (shallow) copy.

So, if x is either of those types, then

shallow_copy_of_x = type(x)(x) #Highly unreadable! But economical.

Note that only dict, set and frozenset have the built-in copy() method. It would probably be a good idea that lists and strings had a copy() method too, for uniformity and readability. But they don't, at least in Python 2.7.3 which I'm testing with.