ʞɔıu ʞɔıu - 10 months ago 38
Python Question

convert a string of bytes into an int (python)

How can I convert a string of bytes into an int in python?

Say like this:


I came up with a clever/stupid way of doing it:

sum(ord(c) << (i * 8) for i, c in enumerate('y\xcc\xa6\xbb'[::-1]))

I know there has to be something builtin or in the standard library that does this more simply...

This is different from converting a string of hex digits for which you can use int(xxx, 16), but instead I want to convert a string of actual byte values.


I kind of like James' answer a little better because it doesn't require importing another module, but Greg's method is faster:

>>> from timeit import Timer
>>> Timer('struct.unpack("<L", "y\xcc\xa6\xbb")[0]', 'import struct').timeit()
>>> Timer("int('y\xcc\xa6\xbb'.encode('hex'), 16)").timeit()

My hacky method:

>>> Timer("sum(ord(c) << (i * 8) for i, c in enumerate('y\xcc\xa6\xbb'[::-1]))").timeit()


Someone asked in comments what's the problem with importing another module. Well, importing a module isn't necessarily cheap, take a look:

>>> Timer("""import struct\nstruct.unpack(">L", "y\xcc\xa6\xbb")[0]""").timeit()

Including the cost of importing the module negates almost all of the advantage that this method has. I believe that this will only include the expense of importing it once for the entire benchmark run; look what happens when I force it to reload every time:

>>> Timer("""reload(struct)\nstruct.unpack(">L", "y\xcc\xa6\xbb")[0]""", 'import struct').timeit()

Needless to say, if you're doing a lot of executions of this method per one import than this becomes proportionally less of an issue. It's also probably i/o cost rather than cpu so it may depend on the capacity and load characteristics of the particular machine.

Answer Source

You can also use the struct module to do this:

>>> struct.unpack("<L", "y\xcc\xa6\xbb")[0]