Convert a 64 bit integer into 8 separate 1 byte integers in python

Question

In python, I have been given a 64 bit integer. This Integer was created by taking several different 8 bit integers and mashing them together into one giant 64 bit integer.

Answer 1

bn = "0010001111111011001000000101100010101010000101101011111000000000"

print([int(bn[i:i+8], 2) for i in range(0,len(bn), 8)])
[35, 251, 32, 88, 170, 22, 190, 0]

If you are using the binary representation of n then the output would be different:

n = 2592701575664680373
bn = bin(n)

print([int(bn[i:i+8], 2) for i in range(0,len(bn), 8)])
[35, 251, 32, 88, 170, 22, 189, 181]

Some timings:

In [16]: %%timeit                                                
numbers = list((n >> i) & 0xFF for i in range(0,64,8))
list(reversed(numbers))
   ....: 
100000 loops, best of 3: 2.97 µs per loop

In [17]: timeit [(n >> (i * 8)) & 0xFF for i in range(7, -1, -1)]
1000000 loops, best of 3: 1.73 µs per loop

In [18]: %%timeit                                                
bn = bin(n)
[int(bn[i:i+8], 2) for i in range(0,len(bn), 8)]
   ....: 
100000 loops, best of 3: 3.96 µs per loop

You can also just divmod:

out = []
for _ in range(8):
    n, i = divmod(n, 256)
    out.append(i) 
out = out[::-1]

Which is almost as efficient:

In [31]: %%timeit
   ....: n = 2592701575664680411
   ....: out = []
   ....: for _ in range(8):
   ....:     n, i = divmod(n, 1 << 8)
   ....:     out.append(i)
   ....: out[::-1]
   ....: 
100000 loops, best of 3: 2.35 µs per loop

There is very little advantage in bit shifting with python, I would be more inclined to use whatever you and others find more readable.