在python中将bytearray转换为short int

Question

I have a long bytearray

barray=b'\x00\xfe\x4b\x00...

What would be the best way to convert it to a list of 2-byte integers?

Answer 1

You can use the struct package for that:

from struct import unpack

tuple_of_shorts = unpack('h'*(len(barray)//2),barray)

This will produce signed shorts. For unsigned ones, use 'H' instead:

tuple_of_shorts = unpack('H'*(len(barray)//2),barray)

This produces on a little-endian machine for your sample input:

>>> struct.unpack('h'*(len(barray)//2),barray)
(-512, 75)
>>> struct.unpack('H'*(len(barray)//2),barray)
(65024, 75)

In case you want to work with big endian , or little endian, you can put a > (big endian) or < (little endian) in the specifications. For instance:

# Big endian
tuple_of_shorts = unpack('>'+'H'*(len(barray)//2),barray)  # unsigned
tuple_of_shorts = unpack('>'+'h'*(len(barray)//2),barray)  # signed

# Little endian
tuple_of_shorts = unpack('<'+'H'*(len(barray)//2),barray)  # unsigned
tuple_of_shorts = unpack('<'+'h'*(len(barray)//2),barray)  # signed

Generating:

>>> unpack('>'+'H'*(len(barray)//2),barray)  # big endian, unsigned
(254, 19200)
>>> unpack('>'+'h'*(len(barray)//2),barray)  # big endian, signed
(254, 19200)
>>> unpack('<'+'H'*(len(barray)//2),barray)  # little endian, unsigned
(65024, 75)
>>> unpack('<'+'h'*(len(barray)//2),barray)  # little endian, signed
(-512, 75)

Answer 2

Using the struct module:

import struct

count = len(barray)/2
integers = struct.unpack('H'*count, barray)

Depending on the endianness you may want to prepend a < or > for the unpacking format. And depending on signed/unsigned, it's h , or H .

Answer 3

If memory efficiency is a concern, you may consider using an array.array :

>>> barr = b'\x00\xfe\x4b\x00'
>>> import array
>>> short_array = array.array('h', barr)
>>> short_array
array('h', [-512, 75])

This is like a space-efficient primitive array, with an OO-wrapper, so it supports sequence-type methods you would have on a list , like .append , .pop , and slicing!

>>> short_array[:1]
array('h', [-512])
>>> short_array[::-1]
array('h', [75, -512])

Also, recovering your bytes object becomes trivial:

>>> short_array
array('h', [-512, 75])
>>> short_array.tobytes()
b'\x00\xfeK\x00'

Note, if you want the opposite endianness from the native byte-order, use the in-place byteswap method:

>>> short_array.byteswap()
>>> short_array
array('h', [254, 19200])

Answer 4

Note, using the Python struct library to convert your array also allows you to specify a repeat count for each item in the format specifier. So 4H for example would be the same as using HHHH .

Using this approach avoids the need to create potentially massive format strings:

import struct

barray = b'\x00\xfe\x4b\x00\x4b\x00'
integers = struct.unpack('{}H'.format(len(barray)/2), barray)

print(integers)

Giving you:

(65024, 75, 75)