蟒蛇; 结构; 24位整数的重复计数

Question

I want to unpack array of 2760 bytes, to new array of 920 24-bit integers. Unlike eg 16 bit integers, where would struct.unpack_from('920h',array,0) do the thing, you can't use 'repeat count' syntax with 24 bit integers:

struct.unpack_from('920<i', array,0 )

This gives the following error: 'bad count in struct format'.

So what's the syntax for 24 bit integers? I can't find anything in the documentation.

Answer 1

struct does not natively support 24-bit integers.

If these are unsigned integers, on trick you could use is to use the array.array() type to read the data as bytes, then process these per 3:

import array

b = array.array('B', yourdata)
result = [b[i] << 16 | b[i + 1] << 8 | b[i + 2] for i in xrange(0, len(b), 3)]

or you could use array.fromfile() to read the data from a file input as needed:

with open('somefilename', 'rb') as infh:
    b = array.array('B')
    b.fromfile(infh, 920)
    result = [b[i] << 16 | b[i + 1] << 8 | b[i + 2] for i in xrange(0, len(b), 3)]

Adjust as needed for byte order (swap the b[i] and b[i + 2] references).

If these are signed 24-bit integers, you may have to stick to struct , and pad the least significant side with a null-byte, then right-shift the result by 8. That way you don't have to worry about negative vs. positive numbers and how to pad either type on the most-significant side:

[struct.unpack('>i', yourdata[i:i+3] + '\x00')[0] >> 8
 for i in range(0, len(yourdata), 3)]

for big-endian, and

[struct.unpack('<i', '\x00' + yourdata[i:i+3])[0] >> 8
 for i in range(0, len(yourdata), 3)]

for little-endian signed 24-bit integers.

If you are using Python 3.2 or newer, you can read the data 3 bytes at a time and convert to integers with int.from_bytes() , which gives you more flexibility over endianess and if you are parsing signed or unsigned integers.