Asked By – ʞɔıu
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")', 'import struct').timeit() 0.36242198944091797 >>> Timer("int('y\xcc\xa6\xbb'.encode('hex'), 16)").timeit() 1.1432669162750244
My hacky method:
>>> Timer("sum(ord(c) << (i * 8) for i, c in enumerate('y\xcc\xa6\xbb'[::-1]))").timeit() 2.8819329738616943
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")""").timeit() 0.98822188377380371
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")""", 'import struct').timeit() 68.474128007888794
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.
Now we will see solution for issue: How to convert a string of bytes into an int?
You can also use the struct module to do this:
>>> struct.unpack("<L", "y\xcc\xa6\xbb") 3148270713L
This question is answered By – Greg Hewgill
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