How to get all combination of n binary value?

In Python, how can I get all combinations of n binary values 0 and 1?

For example, if n = 3, I want to have

[ [0,0,0], [0,0,1], [0,1,0], [0,1,1], ... [1,1,1] ]  #total 2^3 combinations

How can I do this?

Answers:

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Method 1

Use itertools.product

import itertools
lst = list(itertools.product([0, 1], repeat=3))

This will yield a list of tuples (see here)

You can easily change this to use a variable repeat:

n = 3
lst = list(itertools.product([0, 1], repeat=n))

If you need a list of lists, then you can use the map function (thanks @Aesthete).

lst = map(list, itertools.product([0, 1], repeat=n))

Or in Python 3:

lst = list(map(list, itertools.product([0, 1], repeat=n)))
# OR
lst = [list(i) for i in itertools.product([0, 1], repeat=n)]

Note that using map or a list comprehension means you don’t need to convert the product into a list, as it will iterate through the itertools.product object and produce a list.

Method 2

Without using any in-build functions or smart techniques we can get like this.

def per(n):
    for i in range(1<<n):
        s=bin(i)[2:]
        s='0'*(n-len(s))+s
        print (map(int,list(s)))
per(3)

output

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

Method 3

Following will give you all such combinations

bin = [0,1]
[ (x,y,z) for x in bin for y in bin for z in bin ]

All methods was sourced from stackoverflow.com or stackexchange.com, is licensed under cc by-sa 2.5, cc by-sa 3.0 and cc by-sa 4.0