How to randomly discard a number of elements in a set using Python?

Question

for example, I have this Set object:

a = {0,1,2,3,4,5}

how to randomly remove fixed number of elements from this set?

Possible duplicate of [What is the most pythonic way to pop a random element from a list?](https://stackoverflow.com/questions/10048069/what-is-the-most-pythonic-way-to-pop-a-random-element-from-a-list) — vallentin, Dec 27 '17 at 04:33
@Vallentin: The questions are quite different - just list vs set is a huge difference. — Amadan, Dec 27 '17 at 04:35
@StefanPochmann it does not need to be cryptographically random, in my use case. but you're right, it might be a legit concern for someone. — Sajuuk, Dec 27 '17 at 05:53
But still "random", I guess, not just "arbitrary"? Just asking because of `for _ in range(2): a.pop()`. — Stefan Pochmann, Dec 27 '17 at 05:57

Amadan · Accepted Answer · 2017-12-27T04:42:24.703

7

To remove 2 random elements, sample 2 random elements, then remove them:

a.difference(random.sample(a, 2))

or just sample two elements less than the size of the set:

set(random.sample(a, len(a) - 2))

If you want a destructive operation (so that a changes), you can use difference_update on it instead:

a.difference_update(random.sample(a, 2))

edited Dec 27 '17 at 04:42

answered Dec 27 '17 at 04:34

Amadan

I wish `set(random.sample(a, -2))` would do what I'd obviously like it to do :-) – Stefan Pochmann Dec 27 '17 at 04:42

score 0 · Answer 2 · answered Dec 27 '17 at 04:34

0

A possible approach is to choose the number of elements to remove, iterate over the later range, and choose a value to remove:

import random
a = {0,1,2,3,4,5}
n = 2
for i in range(n):
   val = random.choice(list(a))
   a.remove(val)

Or, even shorter, a comprehension:

vals = [random.choice(list(a)) for i in range(n)]
a = {i for i in a if i not in vals}

answered Dec 27 '17 at 04:34

Ajax1234

Both are very inefficient, and the second one is wrong. – Stefan Pochmann Dec 27 '17 at 04:48

2 Answers2