如何以随机顺序循环遍历 Python 字典的项目？

Question

D = {(1,1):2, (2,3):6, (3,4):12, (0,1):0, (4,9):36}

for (i,j),val in D.items():
    print(i,j,"-->",val)

When I loop over the (key,value) pairs of a dictionary, is the order deterministic? How can I loop over them in a random order? Below works when the dictionary is small, but incurs memory error when there are thousands of pairs.

from itertools import permutations

P = list(permutations(D.items()))
for (i,j),val in sample(P,1)[0]:
    print(i,j,"-->",val)

Answer 1

It should be relatively simple to shuffle your entries before iterating over them.

from random import shuffle

dict_as_list_of_entries = [*D.items()]
shuffle(dict_as_list_of_entries) # randomizes the order

for (i, j), value in dict_as_list_of_entries:
    # do something

This makes sense for cpython3.6 (python3.7) and greater because dictionaries remember their insertion order. On lower versions the result is not deterministic (it's a bit more nuianced than "not deterministic", actually, but this will do).

Note that the output of shuffle can also be made deterministic by seeding the randomizer. "True randomness" is still only a concept - computers come close to achieving it though. Most randomizer routines are a tradeoff between randomness and performance.

Note that you're OOM'ing because you're generating every single permutation of your dictionary (when in reality you just want one, random permutation, not all of them).

Answer 2

When I loop over the (key,value) pairs of a dictionary, is the order deterministic?

This depends on what version of Python you're using.

• For Python < 3.6, the order is inconsistent, but not truly random. It will likely be different for different Python implementations, but you can't count on it being the same or not the same for two different people or two different runs.
• For Python 3.6 ( specifically CPython ), the iteration order happens to be the same as the order of insertion, but it's still not "officially" guaranteed
• For Python > 3.6, iteration order is explicitly guaranteed to be the order of insertion.

How can I loop over them in a random order? Try randomizing the order of the keys, then looping over those:

import random

shuffled_keys = random.sample(D.keys(), len(D.keys()))
for k in shuffled_keys:
    print(f'{k} --> {D[k]}')

Answer 3

You can make a shuffled list of the dictionary's keys, and then iterate over that. This means you have to make a copy of all the keys, but it avoids making a copy of the values. Depending on what is in your dictionary, this may save you some memory.

import random

d = {(1, 1): 2, (2, 3): 6, (3, 4): 12, (0, 1): 0, (4, 9): 36}
for key in random.sample(d.keys(), k=len(d)):
    value = d[key]
    i, j = key
    print(i, j, "-->", value)

(Disclaimer: I haven't tested whether it actually saves memory over cs95's solution, or other solutions. Intuitions about memory use and performance can often be wrong, so you should test how this code works on your data to see how it compares to other solutions.)

Answer 4

A dict is an unordered set of key-value pairs. When you iterate a dict, it is effectively random. But to explicitly randomize the sequence of key-value pairs, you need to work with a different object that is ordered, like a list. dict.items(), dict.keys(), and dict.values() each return lists, which can be shuffled. hope this help you in any way