random.choices() in python

Question

I'd like to confirm that

a = [random.choices([0,1],weights=[0.2,0.8],k=1) for i in range(0,10)]

does probabilistically the same thing as

a = random.choices([0,1],weights=[0.2,0.8],k=10)

In particular, I expect both to make 10 independent draws from the set {0,1} with probability 0.2 on 0 and 0.8 on 1. Is this right?

Thanks!

Is there any particular reason you thought it might be otherwise? The documentation seems pretty clear. — user2357112
– user2357112, Commented Oct 17, 2019 at 1:32
If you have a specific reason you thought the behavior might be different, we can give a more useful answer addressing that reason, instead of just saying "yes" and expecting you and future readers to trust us. — user2357112
– user2357112, Commented Oct 17, 2019 at 1:33
Thanks all. I ask because I took ~360 draws using this method for initial production of an app and observed a very unlikely outcome: about 1/3000 under the hypothesis that the data is generated as intended. So I just wanted to make sure this method doesn't somehow mess with independence across draws. — strawberrycello
– strawberrycello, Commented Oct 17, 2019 at 10:53

kingkupps · Accepted Answer · 2019-10-17 01:41:01Z

3

The documentation seems to indicate the two are probabilistically the same and after running the following experiment:

from collections import defaultdict
import pprint
import random

results1 = defaultdict(int)
results2 = defaultdict(int)

for _ in range(10000):
    a = [random.choices([0,1],weights=[0.2,0.8],k=1) for i in range(0,10)]
    for sublist in a:
        for n in sublist:
            results1[n] += 1

for _ in range(10000):
    a = random.choices([0,1],weights=[0.2,0.8],k=10)
    for n in a:
        results2[n] += 1


print('first way 0s: {}'.format(results1[0]))
print('second way 0s: {}'.format(results2[0]))
print('first way 1s: {}'.format(results1[1]))
print('second way 1s: {}'.format(results2[1]))

I am seeing very similar results between the two methods.

answered Oct 17, 2019 at 1:41

kingkupps

3,5542 gold badges20 silver badges31 bronze badges

Sign up to request clarification or add additional context in comments.

Comments

Dani Mesejo · Accepted Answer · 2019-10-17 01:45:42Z

As others have mentioned, the documentation is clear in regard to this aspect, you can further verified by setting the seed before each call, for example:

import random

random.seed(42)
print([random.choices([0, 1], weights=[0.2, 0.8], k=1)[0] for i in range(0, 10)])

random.seed(42)
print(random.choices([0, 1], weights=[0.2, 0.8], k=10))

Output

[1, 0, 1, 1, 1, 1, 1, 0, 1, 0]
[1, 0, 1, 1, 1, 1, 1, 0, 1, 0]

Furthermore setting just once, does leads to different results, as one might expect:

random.seed(42)
print([random.choices([0, 1], weights=[0.2, 0.8], k=1)[0] for i in range(0, 10)])
print(random.choices([0, 1], weights=[0.2, 0.8], k=10))

Output

[1, 0, 1, 1, 1, 1, 1, 0, 1, 0]
[1, 1, 0, 0, 1, 1, 1, 1, 1, 0]

Collectives™ on Stack Overflow

random.choices() in python

2 Answers 2

Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related