Converting an int to a byte of a certain size in python

Question

I'm trying to create an array of bits in python that are two binary numbers into a fixed width.

For eg.

[11,90] ----> 0000101101011010

What I've tried/ Problems I've encountered:

bin(11) => 0b1010 (which is not the fixed width I need)
String manipulation using format '{0:11b}'.format(8) (this works but I'd like to avoid converting to strings and back if I can. It seems like a lot of overhead to do essentially a shift and add)

bin returns a string. I wouldn't worry about the overhead unless you actually see it causing problems, it's unlikely to matter much. If you are worried, then you should build your own lists of True and False, which are as close as you are likely to get to a bit array. '{:0>8b}{:0>8b}'.format(x, y) should be sufficient — Patrick Haugh
– Patrick Haugh, Commented Dec 25, 2017 at 0:07
@Galen I see it now. 11 in binary is 1011, and 00001011 is the first half of the number, 90 is 1011010 and 01011010 is the second half. So it seems as if they want to convert numbers in a list to binary, add 0's to the front until each is 8 bits long, then append them together. — user3483203
– user3483203, Commented Dec 25, 2017 at 0:07
Try a = 11 << 8 | 90, then bin(a) produces 0b101101011010 — John Anderson
– John Anderson, Commented Dec 25, 2017 at 0:10

Galen · Accepted Answer · 2017-12-25 00:17:23Z

3

You can use struct:

bin(int.from_bytes(struct.pack('>bb', 11, 90), byteorder='big'))

If you want the leading 0s:

'{:016b}'.format(int.from_bytes(struct.pack('>bb', 11, 90), byteorder='big'))

answered Dec 25, 2017 at 0:17

Galen

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Daniel F. · Accepted Answer · 2017-12-25 00:18:15Z

2

Guessing what you want to do, since you're not very detailed on your question, you can use bin(x)[2:] removing the first two characters, then zero fill it with str.zfill(length) and then concatenating your two numbers, like so:

print(bin(11)[2:].zfill(8) + bin(90)[2:].zfill(8))

Will print: 0000101101011010

answered Dec 25, 2017 at 0:18

Daniel F.

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