How to get all combination of n binary value?

Question

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?

@eumiro, I think my question is also equivalent to this one, http://stackoverflow.com/questions/3252528/converting-a-number-to-binary-with-a-fixed-length , but that answer gives a string instead of a list. — LWZ, Feb 18 '13 at 08:52

Volatility · Accepted Answer · 2013-02-18T08:11:50.860

97

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.

edited Feb 18 '13 at 08:11

answered Feb 18 '13 at 08:05

Volatility

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3

+1 - `map(list, product([0, 1], repeat=3))` will return the same format in case the OP is interested. – Aesthete Feb 18 '13 at 08:12
This is great, tuple is ok too. – LWZ Feb 18 '13 at 08:14
@Volatility, just of curiosity, how do you know this? I had no clue how to find this function from the Python documentation. – LWZ Feb 18 '13 at 08:18
1

@LWZ It [comes with experience](http://stackoverflow.com/a/2576240/1907098). (see point 6) – Volatility Feb 18 '13 at 08:20
1

@Volatility, what a great post, thanks! – LWZ Feb 18 '13 at 08:23
Thank you! This was extremely helpful – armitus Mar 29 '19 at 19:25
@Volatility thank you for the solution, I am trying to the same thing with 26 variables but I think my Jupyter NB runs out of memory. Do you know how I can possibly tackle this issue. Thanks. – Aditya Lahiri Aug 09 '19 at 21:35
@aditya lahiri the array size will be 2^n x n. For 26 that's quite a long array. – Jed May 19 '21 at 18:29

score 17 · Answer 2 · edited Nov 18 '19 at 10:24

17

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]

edited Nov 18 '19 at 10:24

ZF007

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answered Feb 18 '13 at 08:44

Anil

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For me, this yields: – user1111929 Mar 11 '19 at 23:35
1

Have to call (list(map...)) to get actual lists out – WattsInABox Nov 24 '19 at 01:14
Great idea! here is a simple version with format function: ``` def per(n): for i in range(1<>> per(3) – Richard Li Dec 14 '20 at 16:06

score 4 · Answer 3 · answered Feb 18 '13 at 08:05

4

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 ]

answered Feb 18 '13 at 08:05

MoveFast

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He gives `n=3` as an example only, so he wants to have this parametrized. – eumiro Feb 18 '13 at 08:07
This works but n can't be large... – LWZ Feb 18 '13 at 08:07
2

This wont scale! – Freeze Francis Apr 09 '19 at 18:21

How to get all combination of n binary value?

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