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How can I convert a string of bytes into an int in python? Say like this: 'y\xcc\xa6\xbb' 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. UPDATE: 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() 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 FURTHER UPDATE: 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() 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")[0]""", 'import struct').timeit() 68.474128007888794 Needless to say, if you're doing a lo

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