How to convert a string of bytes into an int?


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

Say like this: 'yxccxa6xbb'

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

sum(ord(c) << (i * 8) for i, c in enumerate('yxccxa6xbb'[::-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", "yxccxa6xbb")[0]', 'import struct').timeit()
>>> Timer("int('yxccxa6xbb'.encode('hex'), 16)").timeit()

My hacky method:

>>> Timer("sum(ord(c) << (i * 8) for i, c in enumerate('yxccxa6xbb'[::-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 structnstruct.unpack(">L", "yxccxa6xbb")[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", "yxccxa6xbb")[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.

Asked By: ʞɔıu



You can also use the struct module to do this:

>>> struct.unpack("<L", "yxccxa6xbb")[0]
Answered By: Greg Hewgill

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 = 'yxccxa6xbb'
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.

Answered By: James Antill
import array
integerValue = array.array("I", 'yxccxa6xbb')[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.

Answered By: Rafał Dowgird

In Python 3.2 and later, use

>>> int.from_bytes(b'yxccxa6xbb', byteorder='big')


>>> int.from_bytes(b'yxccxa6xbb', byteorder='little')

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.

Answered By: Mechanical snail

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')


Answered By: Jrm

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


Let x = 'xffx10x11'

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


data_bytes = struct.pack('<' + 'B'*len(data_ints), *data_ints) # 'xffx10x11'

That * is required!

See for a list of the format specifiers.

Answered By: Tetralux

I was struggling to find a solution for arbitrary length byte sequences that would work under Python 2.x. Finally I wrote this one, it’s a bit hacky because it performs a string conversion, but it works.

Function for Python 2.x, arbitrary length

def signedbytes(data):
    """Convert a bytearray into an integer, considering the first bit as
    sign. The data must be big-endian."""
    negative = data[0] & 0x80 > 0

    if negative:
        inverted = bytearray(~d % 256 for d in data)
        return -signedbytes(inverted) - 1

    encoded = str(data).encode('hex')
    return int(encoded, 16)

This function has two requirements:

  • The input data needs to be a bytearray. You may call the function like this:

    s = 'yxccxa6xbb'
    n = signedbytes(s)
  • The data needs to be big-endian. In case you have a little-endian value, you should reverse it first:

    n = signedbytes(s[::-1])

Of course, this should be used only if arbitrary length is needed. Otherwise, stick with more standard ways (e.g. struct).

Answered By: Andrea Lazzarotto

int.from_bytes is the best solution if you are at version >=3.2.
The “struct.unpack” solution requires a string so it will not apply to arrays of bytes.
Here is another solution:

def bytes2int( tb, order='big'):
    if order == 'big': seq=[0,1,2,3]
    elif order == 'little': seq=[3,2,1,0]
    i = 0
    for j in seq: i = (i<<8)+tb[j]
    return i

hex( bytes2int( [0x87, 0x65, 0x43, 0x21])) returns ‘0x87654321’.

It handles big and little endianness and is easily modifiable for 8 bytes

Answered By: user3435121
>>> reduce(lambda s, x: s*256 + x, bytearray("yxccxa6xbb"))

Test 1: inverse:

>>> hex(2043455163)

Test 2: Number of bytes > 8:

>>> reduce(lambda s, x: s*256 + x, bytearray("AAAAAAAAAAAAAAA"))

Test 3: Increment by one:

>>> reduce(lambda s, x: s*256 + x, bytearray("AAAAAAAAAAAAAAB"))

Test 4: Append one byte, say ‘A’:

>>> reduce(lambda s, x: s*256 + x, bytearray("AAAAAAAAAAAAAABA"))

Test 5: Divide by 256:

>>> reduce(lambda s, x: s*256 + x, bytearray("AAAAAAAAAAAAAABA"))/256

Result equals the result of Test 4, as expected.

Answered By: user3076105

As mentioned above using unpack function of struct is a good way. If you want to implement your own function there is an another solution:

def bytes_to_int(bytes):
    result = 0
    for b in bytes:
        result = result * 256 + int(b)
return result
Answered By: abdullahselek

A decently speedy method utilizing array.array I’ve been using for some time:

predefined variables:

offset = 0
size = 4
big = True # endian
arr = array('B')
arr.fromstring("x00x00xffx00") # 5 bytes (encoding issues) [0, 0, 195, 191, 0]

to int: (read)

val = 0
for v in arr[offset:offset+size][::pow(-1,not big)]: val = (val<<8)|v

from int: (write)

val = 16384
arr[offset:offset+size] = 
    array('B',((val>>(i<<3))&255 for i in range(size)))[::pow(-1,not big)]

It’s possible these could be faster though.

For some numbers, here’s a performance test (Anaconda 2.3.0) showing stable averages on read in comparison to reduce():

========================= byte array to =========================
5000 iterations; threshold of min + 5000ns:
    val = 0 nfor v in arr: val = (val<<8)|v |     5373.848ns |   850009.965ns |     ~8649.64ns |  62.128%
                  val = reduce( shift, arr ) |     6489.921ns |  5094212.014ns |   ~12040.269ns |  53.902%

This is a raw performance test, so the endian pow-flip is left out.
The shift function shown applies the same shift-oring operation as the for loop, and arr is just array.array('B',[0,0,255,0]) as it has the fastest iterative performance next to dict.

I should probably also note efficiency is measured by accuracy to the average time.

Answered By: Tcll

In python 3 you can easily convert a byte string into a list of integers (0..255) by

>>> list(b'yxccxa6xbb')
[121, 204, 166, 187]
Answered By: fhgd

For newer versions of Python a simple way is:

int(b'hello world'.hex(), 16)
Answered By: lnogueir
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