Python : Efficient Bytearray Incrementation

How to iterate all possible values of bytearray of length = n in Python ? in worst case n <= 40bytes For example, iterate for n = 4 : 00000000 00000000 00000000 00000000 0000000

Solution 1:

You can use itertools.product:

In [11]: from itertools import product

In [15]: for x in product('01',repeat=4): #for your n=4 change repeat to 32 
    print "".join(x)
   ....:     
0000
0001
0010
0011
0100
0101
0110
0111
1000
1001
1010
1011
1100
1101
1110
1111

Solution 2:

Inspired by https://stackoverflow.com/a/15538456/1219006

n = 2
[[[i>>k&1 for k in range(j, j-8, -1)] for j in range(8*n-1, 0, -8)] 
 for i in range(2**(8*n))]

You'll need to run this on Python 3 for large n cause xrange doesn't support big ints.

As a generator:

def byte_array(n):
    for i in range(2**(8*n)):
        yield [[i>>k&1 for k in range(j, j-8, -1)] for j in range(8*n-1, 0, -8)]

---

>>> i = byte_array(4)
>>> next(i)
[[0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0]]
>>> next(i)
[[0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 1]]

Or if you don't want them grouped it's simpler:

[[i>>j&1 for j in range(8*n-1, -1, -1)] for i in range(2**(8*n))]

Equivalent generator:

def byte_array(n):
    for i in range(2**(8*n)):
        yield [i>>j&1 for j in range(8*n-1, -1, -1)]