Pythonic way to access arbitrary element from dictionary

Question

I have a dictionary, full of items. I want to peek at a single, arbitrary item:

print("Amongst our dictionary\'s items are such diverse elements as: %s&q         


        

Answer 1

I believe the question has been significantly answered but hopefully this comparison will shed some light on the clean code vs time trade off:

from timeit import timeit
from random import choice
A = {x:[y for y in range(100)] for x in range(1000)}
def test_pop():
    k, v= A.popitem()
    A[k] = v

def test_iter(): k = next(A.iterkeys())

def test_list(): k = choice(A.keys())

def test_insert(): A[0] = 0

if __name__ == '__main__':
    print('pop', timeit("test_pop()", setup="from __main__ import test_pop", number=10000))
    print('iter', timeit("test_iter()", setup="from __main__ import test_iter", number=10000))
    print('list', timeit("test_list()", setup="from __main__ import test_list", number=10000))
    print('insert', timeit("test_insert()", setup="from __main__ import test_insert", number=10000))

Here are the results:

('pop', 0.0021750926971435547)
('iter', 0.002003908157348633)
('list', 0.047267913818359375)
('insert', 0.0010859966278076172)

It seems that using iterkeys is only marginal faster then poping an item and re-inserting but 10x's faster then creating the list and choosing a random object from it.

Answer 2

Why not use random?

import random

def arb(dictionary):
    return random.choice(dictionary.values())

This makes it very clear that the result is meant to be purely arbitrary and not an implementation side-effect. Until performance becomes an actual issue, always go with clarity over speed.

It's a shame that dict_values don't support indexing, it'd be nice to be able to pass in the value view instead.

Update: since everyone is so obsessed with performance, the above function takes <120ms to return a random value from a dict of 1 million items. Relying on clear code is not the amazing performance hit it's being made out to be.

Answer 3

Avoiding the whole values/itervalues/viewvalues mess, this works equally well in Python2 or Python3

dictionary[next(iter(dictionary))]

alternatively if you prefer generator expressions

next(dictionary[x] for x in dictionary)

Answer 4

Similar to your second solution, but slightly more obvious, in my opinion:

return next(iter(dictionary.values()))

This works in python 2 as well as in python 3, but in python 2 it's more efficient to do it like this:

return next(dictionary.itervalues())