convert array of numbers to byte array

[adetheheat]

I have an array of numbers - some are 8 bits, some are 16 bit little endian. eg
1,120,340, 54238, 65132,
How do I convert them to one byte array ? I know the offsets in the array where the 8bit numbers and 16 bit numbers are.

[DeaD_EyE]

With the struct module.

import struct

uint16be = struct.Struct(">I")
int16be = struct.Struct(">h")
int16le = struct.Struct("<h")
int8 = struct.Struct("b")

my_numbers = 1, 120, 340, 54238, 65132

for number in my_numbers:
    value = uint16be.pack(number)
    print(value)

Output:
b'\x00\x00\x00\x01'
b'\x00\x00\x00x'
b'\x00\x00\x01T'
b'\x00\x00\xd3\xde'
b'\x00\x00\xfel'

Some of your numbers are too big for 16-bit signed integers.

[adetheheat]

that's not quite what I want. I want them as numbers not strings.
So for instance for this small array of uint8 and then a uint16 : 14,54274
I want the bytes obtained to be:
14,2,212

I could do it procedureraly but there's got to be a faster way.

[bowlofred]

The output of the struct isn't "strings", it's just a different packing.

Perhaps you just need divmod()

>>> divmod(54274,256)
(212, 2)

Close might be something like:

>>> l = [1,120,340, 54238, 65132]
>>> [x if x < 256 else divmod(x,256)[::-1] for x in l]
[1, 120, (84, 1), (222, 211), (108, 254)]

But you'd have to flatten the list later to remove the tuples.