Python print all binary numbers from 000 - 111

Question

I needed to produce all the binary numbers from 000-111 in string form and my first approach is shown below. However, it struck me that this must be a very inefficient way to do this. Are there better ways please, and how are they more efficient?

 # print all binary numbers from 000 - 111
    for i in range(2):
        for j in range(2):
            for k in range(2):
                value = f'{i}{j}{k}'
                print(value)

I don't see how this is a duplicate of the linked question. I'm asking for a way to produce multiple binary values in an efficient way, while the other question just addresses a single conversion. — Robin Andrews, Dec 29 '18 at 15:02

score 3 · Answer 1 · answered Dec 29 '18 at 13:45

3

You could do:

for i in range(8):
    s = bin(i)[2:]
    print(s.zfill(3))

Output

answered Dec 29 '18 at 13:45

Dani Mesejo

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OK, but is it more efficient "under the hood" and if so why please? – Robin Andrews Dec 29 '18 at 13:47

Vicrobot · Answer 2 · 2018-12-29T14:02:57.447

1

You can use this;

>>> ['{:03b}'.format(i) for i in range(8)]
['000', '001', '010', '011', '100', '101', '110', '111']

>>> ['{:03b}'.format(i) for i in range(2,8)]
['010', '011', '100', '101', '110', '111']

Just will have to convert the range to decimal.

edited Dec 29 '18 at 14:02

answered Dec 29 '18 at 13:55

Vicrobot

3,795
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You can change the format specifier to `'{:03b}'` to exactly match op's output. – Wombatz Dec 29 '18 at 14:00
Thanks. I was looking for that. – Vicrobot Dec 29 '18 at 14:01

score 0 · Answer 3 · answered Dec 29 '18 at 14:04

Here are a few different ways to do it, and their results. The performance difference is minor in the grand scheme of things, so you shouldn't worry too much.

%%timeit
for i in range(2):
    for j in range(2):
        for k in range(2):
            value = f'{i}{j}{k}'
            print(value)

Original Result: 823 µs ± 49.2 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

%%timeit
for i in range(8):
    print("{0:03b}".format(i))

Result: 784 µs ± 35.6 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

%%timeit  
for i in range(8):    
    print(bin(i)) #note that this does not produce the output exactly as you'd want it

Result: 807 µs ± 36.7 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

%%timeit 
for i in range(8): 
    print(bin(i)[2:].rjust(3, '0')) #this one really surprised me, i didn't expect it to be fastest.

Result: 773 µs ± 43.6 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

And just an itertools version if you didn't want to use binary representations. Note that i have excluded library import for the sake of comparison.

from itertools import product
bits = ['0', '1']
%%timeit
for i in product(bits, repeat = 3):
    print(''.join(i))

Result: 801 µs ± 65.5 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

Python print all binary numbers from 000 - 111

3 Answers3