How do you define an array in python whose input is a binary counter?

Question

Wrote this script to create a 4x16 matrix for all possible combinations of 4 bits:

import numpy as np
    a = []
    for x in range(2):
        for y in range(2):
            for z in range(2):
                for w in range(2):
                 a.append([x,y,z,w])
    a = np.array(a)
    print(a)

Output

It works.. however it's 4 loops for 4 bits. Raising the number of bits means more loops, does anyone have another way of doing this?

Haven't tried much.. new to programming

Vijayachandran Mariappan · Accepted Answer · 2022-12-20 20:56:42Z

2

with just numpy,we can use np.unpackbits

bits = 4
np.unpackbits(np.arange(2**bits, dtype=np.uint8)[...,None], count=bits, axis=1, bitorder='little')

edited Dec 20, 2022 at 20:56

answered Dec 20, 2022 at 6:02

Vijayachandran Mariappan

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2 Comments

Dordini Balbozhia Over a year ago

Works for 4 bits since 4^2 = 2^4 in this case, but if I change the bit number to 7, I get a (7^2, 7) matrix instead of a (2^7, 7) matrix

Vijayachandran Mariappan Over a year ago

i edited, it should be 2**bits in arange

Kraigolas · Accepted Answer · 2022-12-20 05:45:26Z

2

You can just use itertools.product:

list(itertools.product(range(2), repeat=4))

Here range(2) provides the 0 and 1, and then repeat=4 says 4 bits. If you want 20 bits, use repeat=20.

This result actually gives you a list, but if you want to iterate through each option one at a time, just use itertools.product(range(2), repeat=4) on its own, as this gives you a generator. For larger numbers of bits, the number of combinations might not fit in memory. Using the generator version means you only ever have one combination in memory at a time.

answered Dec 20, 2022 at 5:45

Kraigolas

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Comments

JayPeerachai · Accepted Answer · 2022-12-20 06:03:24Z

1

You can use recursion:

def create_matrix(n, arr=[]):
    if n == 0:
        print(arr)
    else:
        for i in range(2):
            create_matrix(n-1, arr + [I])

output:

> create_matrix(4)
[0, 0, 0, 0]
[0, 0, 0, 1]
[0, 0, 1, 0]
[0, 0, 1, 1]
[0, 1, 0, 0]
[0, 1, 0, 1]
[0, 1, 1, 0]
[0, 1, 1, 1]
[1, 0, 0, 0]
[1, 0, 0, 1]
[1, 0, 1, 0]
[1, 0, 1, 1]
[1, 1, 0, 0]
[1, 1, 0, 1]
[1, 1, 1, 0]
[1, 1, 1, 1]

This can be used to generate a matrix for all possible combinations of any bits.

It appends a 0 or 1 to arr and passes it on to the next recursive call until n leads to 0. When n becomes 0, it prints out arr.

UPDATE

As @Pranav Hosangadi suggestions, I've modified the code to get rid of mutable default argument and to has the return statement.

def create_matrix(n):
    if n == 0:
        return [[]]
    else:
        return [[i] + item for i in range(2) for item in create_matrix(n-1)]

P.S. It is good to learn about Recursion.

edited Dec 20, 2022 at 6:03

answered Dec 20, 2022 at 5:46

JayPeerachai

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2 Comments

pho Over a year ago

Beware the mutable default argument (doesn't really matter in this case because you don't modify it in the function)

pho Over a year ago

Also a function that builds something should return it instead of just printing it.

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