How select random sampling from large dict whitout converting to list in Python

Question

I have a (potentially) huge dict in Python 3.10 and want to randomly sample a few values. Alas, random.sample(my_dict, k) says:

TypeError: Population must be a sequence.  For dicts or sets, use sorted(d).

and random.sample(my_dict.keys(), k) gives

DeprecationWarning: Sampling from a set deprecated
since Python 3.9 and will be removed in a subsequent version.

I don't want to pay the cost of converting my dictionary keys to a list, and I don't need them sorted.

There's an old question in a similar vein, but that's from before stuff got deprecated in Python and the person asking that question didn't mind converting to a list first.

I also tried running random.choice multiple times to simulate random.sample. But that's even worse: it just throws an exception when you use it on a dict. (Instead of giving you a reasonable error message.)

Answer 1

You need to use sample(list(dct)). (The example code select random 2 items from original dict.)

from random import sample
dct = {'a':1, 'b':2, 'c':3, 'd':4}

rnd_keys = sample(list(dct), 2)
# rnd_keys -> ['c', 'b']
rnd_dct = dict(sample(list(dct.items()), 2))
print(rnd_dct)

---

{'c': 3, 'b': 2}

Update without converting huge dict to list (this convert use O(n) space and question say, don't do this.). You can generate random number base len(dict) and use enumerate and only get k,v that idx match with random_idx and break from for-loop when reaching to zero base random_number that we want to select (this break helps you don't see all dict).

from random import sample
dct = {'a':1, 'b':2, 'c':3, 'd':4}
# idx -^0^----^1^----^2^----^3^---
number_rnd = 2
rnd_idx = set(sample(range(len(dct)), number_rnd))
print(rnd_idx)
# {0, 3}
res = {}
for idx, (k,v) in enumerate(dct.items()):
    if idx in rnd_idx:
        res[k] = v
        number_rnd -= 1
        if number_rnd == 0:
            break
print(res)
# {'b': 2, 'c': 3}

Third Approach By thanks Tomerikoo, We can use flip a coin idea, On each iterate over items() we cen generate a random 0 or 1 and if the random number is 1 save the item in the result dict. (Maybe we see all dict items but don't select all random numbers because, maybe we get many random 0.)

import random
dct = {'a':1, 'b':2, 'c':3, 'd':4}

number_rnd = 2
res = {}
for k,v in dct.items():
    rnd_ch = random.getrandbits(1)
    if rnd_ch:
        res[k] = v
        number_rnd -= 1
        if number_rnd == 0:
            break

print(res)

Answer 2

If you’re ok with getting the keys into a list, you can just pick the keys using a random set of integers off of the list. If you don’t want to store them into a list, you can generate your random integers based on the dict size, sort them, then iterate through the dict and sample as you get to indices matching your random integer picks.