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Get a random boolean in python?

Posted by: admin November 1, 2017 Leave a comment

Questions:

I am looking for the best way (fast and elegant) to get a random boolean in python (flip a coin).

For the moment I am using random.randint(0, 1) or random.getrandbits(1).

Are there better choices that I am not aware of?

Answers:

Adam’s answer is quite fast, but I found that random.getrandbits(1) to be quite a lot faster. If you really want a boolean instead of a long then

bool(random.getrandbits(1))

is still about twice as fast as random.choice([True, False])

If utmost speed isn’t to priority then random.choice definitely reads better

$ python -m timeit -s "import random" "random.choice([True, False])"
1000000 loops, best of 3: 0.904 usec per loop
$ python -m timeit -s "import random" "random.choice((True, False))" 
1000000 loops, best of 3: 0.846 usec per loop
$ python -m timeit -s "import random" "random.getrandbits(1)"
1000000 loops, best of 3: 0.286 usec per loop
$ python -m timeit -s "import random" "bool(random.getrandbits(1))"
1000000 loops, best of 3: 0.441 usec per loop
$ python -m timeit -s "import random" "not random.getrandbits(1)"
1000000 loops, best of 3: 0.308 usec per loop
$ python -m timeit -s "from random import getrandbits" "not getrandbits(1)"
1000000 loops, best of 3: 0.262 usec per loop  # not takes about 20us of this

Added this one after seeing @Pavel’s answer

$ python -m timeit -s "from random import random" "random() < 0.5"
10000000 loops, best of 3: 0.115 usec per loop

Questions:
Answers:
random.choice([True, False])

would also work.

Questions:
Answers:

Found a faster method:

$ python -m timeit -s "from random import getrandbits" "not getrandbits(1)"
10000000 loops, best of 3: 0.222 usec per loop
$ python -m timeit -s "from random import random" "True if random() > 0.5 else False"
10000000 loops, best of 3: 0.0786 usec per loop
$ python -m timeit -s "from random import random" "random() > 0.5"
10000000 loops, best of 3: 0.0579 usec per loop

Questions:
Answers:

If you want to generate a number of random booleans you could use numpy’s random module. From the documentation

np.random.randint(2, size=10)

will return 10 random uniform integers in the open interval [0,2). The size keyword specifies the number of values to generate.

Questions:
Answers:

I like

 np.random.rand() > .5

Questions:
Answers:

A new take on this question would involve the use of Faker which you can install easily with pip.

from faker import Factory

#----------------------------------------------------------------------
def create_values(fake):
    """"""
    print fake.boolean(chance_of_getting_true=50) # True
    print fake.random_int(min=0, max=1) # 1

if __name__ == "__main__":
    fake = Factory.create()
    create_values(fake)