I am reading in a byte array/list from socket. I want Python to treat the first byte as an "unsigned 8 bit integer". How is it possible to get its integer value as an unsigned 8 bit integer?
3 Answers
Use the struct module.
import struct
value = struct.unpack('B', data[0:1])[0]
We have to specify a range of 1 (0:1), because Python 3 converts automatically to an integer otherwise.
Note that unpack always returns a tuple, even if you're only unpacking one item.
Also, have a look at this SO question.
3 Comments
bytes/bytearray is a sequence of integers. If you just access an element by its index you'll have an integer:
>>> b'abc'
b'abc'
>>> _[0]
97
By their very definition, bytes and bytearrays contain integers in the range(0, 256). So they're "unsigned 8-bit integers".
3 Comments
python-3.x I guessAnother very reasonable and simple option, if you just need the first byte’s integer value, would be something like the following:
value = ord(data[0])
If you want to unpack all of the elements of your received data at once (and they’re not just a homogeneous array), or if you are dealing with multibyte objects like 32-bit integers, then you’ll need to use something like the struct module.
numpy, some of their functions take adtypearg, which you can set tonumpy.uint8. Example:bytestream = numpy.fromfile(infile, dtype=numpy.uint8)