How to convert a string of bytes into an int in Py

2019-01-01 05:22发布

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 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.

10条回答
旧人旧事旧时光
2楼-- · 2019-01-01 05:40

As Greg said, you can use struct if you are dealing with binary values, but if you just have a "hex number" but in byte format you might want to just convert it like:

s = 'y\xcc\xa6\xbb'
num = int(s.encode('hex'), 16)

...this is the same as:

num = struct.unpack(">L", s)[0]

...except it'll work for any number of bytes.

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还给你的自由
3楼-- · 2019-01-01 05:42

You can also use the struct module to do this:

>>> struct.unpack("<L", "y\xcc\xa6\xbb")[0]
3148270713L
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旧人旧事旧时光
4楼-- · 2019-01-01 05:46

In Python 3.2 and later, use

>>> int.from_bytes(b'y\xcc\xa6\xbb', byteorder='big')
2043455163

or

>>> int.from_bytes(b'y\xcc\xa6\xbb', byteorder='little')
3148270713

according to the endianness of your byte-string.

This also works for bytestring-integers of arbitrary length, and for two's-complement signed integers by specifying signed=True. See the docs for from_bytes.

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其实,你不懂
5楼-- · 2019-01-01 05:49

I use the following function to convert data between int, hex and bytes.

def bytes2int(str):
 return int(str.encode('hex'), 16)

def bytes2hex(str):
 return '0x'+str.encode('hex')

def int2bytes(i):
 h = int2hex(i)
 return hex2bytes(h)

def int2hex(i):
 return hex(i)

def hex2int(h):
 if len(h) > 1 and h[0:2] == '0x':
  h = h[2:]

 if len(h) % 2:
  h = "0" + h

 return int(h, 16)

def hex2bytes(h):
 if len(h) > 1 and h[0:2] == '0x':
  h = h[2:]

 if len(h) % 2:
  h = "0" + h

 return h.decode('hex')

Source: http://opentechnotes.blogspot.com.au/2014/04/convert-values-to-from-integer-hex.html

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骚的不知所云
6楼-- · 2019-01-01 05:51
import array
integerValue = array.array("I", 'y\xcc\xa6\xbb')[0]

Warning: the above is strongly platform-specific. Both the "I" specifier and the endianness of the string->int conversion are dependent on your particular Python implementation. But if you want to convert many integers/strings at once, then the array module does it quickly.

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后来的你喜欢了谁
7楼-- · 2019-01-01 05:52

In Python 2.x, you could use the format specifiers <B for unsigned bytes, and <b for signed bytes with struct.unpack/struct.pack.

E.g:

Let x = '\xff\x10\x11'

data_ints = struct.unpack('<' + 'B'*len(x), x) # [255, 16, 17]

And:

data_bytes = struct.pack('<' + 'B'*len(data_ints), *data_ints) # '\xff\x10\x11'

That * is required!

See https://docs.python.org/2/library/struct.html#format-characters for a list of the format specifiers.

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