How to get an arbitrary element from a frozenset?

Question

I would like to get an element from a frozenset (without modifying it, of course, as frozensets are immutable). The best solution I have found so far is:

s = frozenset(['a'])
iter(s).next()

which returns, as expected:

'a'

In other words, is there any way of 'popping' an element from a frozenset without actually popping it?

Solution 1

(Summarizing the answers given in the comments)

Your method is as good as any, with the caveat that, from Python 2.6, you should be using next(iter(s)) rather than iter(s).next().

If you want a random element rather than an arbitrary one, use the following:

import random
random.sample(s, 1)[0]

Here are a couple of examples demonstrating the difference between those two:

>>> s = frozenset("kapow")
>>> [next(iter(s)) for _ in range(10)]
['a', 'a', 'a', 'a', 'a', 'a', 'a', 'a', 'a', 'a']

>>> import random
>>> [random.sample(s, 1)[0] for _ in range(10)]
['w', 'a', 'o', 'o', 'w', 'o', 'k', 'k', 'p', 'k']

Solution 2

If you know that there is but one element in the frozenset, you can use iterable unpacking:

s = frozenset(['a'])
x, = s

This is somewhat a special case of the original question, but it comes in handy some times.

If you have a lot of these to do it might be faster than next(iter..:

>>> timeit.timeit('a,b = foo', setup='foo = frozenset(range(2))', number=100000000)
5.054765939712524
>>> timeit.timeit('a = next(iter(foo))', setup='foo = frozenset(range(2))', number=100000000)
11.258678197860718

Solution 3

You could use with python 3:

>>> s = frozenset(['a', 'b', 'c', 'd'])
>>> x, *_ = s
>>> x
'a'
>>> _, x, *_ = s
>>> x
'b'
>>> *_, x, _ = s
>>> x
'c'
>>> *_, x = s
>>> x
'd'

Comments

[bbayles]

I think your method is as good as any. If you want a random element you might check out random.sample(fset, 1).

[ablondin]

I just want to get some arbitrary element from a frozenset. I shouldn't have used the word pop since the set remains unchanged. It is similar to peeking the first element of a stack without popping it.

[user2357112]

That's what I use (but with the next builtin instead of the method).

[Bakuriu]

Don't use the method .next(). There is a next() built-in function since at least python2.6 and using it means that your code will work also in python3 where the next method was renamed __next__.

[rynemccall]

Why should you call next(iter(s)) rather than iter(s).next()?

[Zero Piraeus]

@rynemccall Because in Python 3, iter.next() doesn't exist.

[A T]

ValueError: too many values to unpack on set with multiple elements.

[Dominic Kempf]

as said, this only works with exactly one element in the frozenset.

[wchargin]

The first of these is just a complicated way to write list(my_frozenset), which is quite wasteful for large sets. The second is just a complicated way to write next(iter(my_frozenset)), which is already the accepted answer.