python how to compute a simple checksum as quickly as zlib.adler32

Question

I wish to compute a simple checksum : just adding the values of all bytes.

The quickest way I found is:

checksum = sum([ord(c) for c in buf])

But for 13 Mb data buf, it takes 4.4 s : too long (in C, it takes 0.5 s)

If I use :

checksum = zlib.adler32(buf) & 0xffffffff

it takes 0.8 s, but the result is not the one I want.

So my question is: is there any function, or lib or C to include in python 2.6, to compute a simple checksum ?

Thanks by advance, Eric.

Answer 1

You could use sum(bytearray(buf)):

In [1]: buf = b'a'*(13*(1<<20))

In [2]: %timeit sum(ord(c) for c in buf)
1 loops, best of 3: 1.25 s per loop

In [3]: %timeit sum(imap(ord, buf))
1 loops, best of 3: 564 ms per loop

In [4]: %timeit b=bytearray(buf); sum(b)
10 loops, best of 3: 101 ms per loop

Here's a C extension for Python written in Cython, sumbytes.pyx file:

from libc.limits cimport ULLONG_MAX, UCHAR_MAX

def sumbytes(bytes buf not None):
    cdef:
        unsigned long long total = 0
        unsigned char c
    if len(buf) > (ULLONG_MAX // <size_t>UCHAR_MAX):
        raise NotImplementedError #todo: implement for > 8 PiB available memory
    for c in buf:
        total += c
    return total

sumbytes is ~10 time faster than bytearray variant:

name                    time ratio
sumbytes_sumbytes    12 msec  1.00 
sumbytes_numpy     29.6 msec  2.48 
sumbytes_bytearray  122 msec 10.19 

To reproduce the time measurements, download reporttime.py and run:

#!/usr/bin/env python
# compile on-the-fly
import pyximport; pyximport.install() # pip install cython
import numpy as np 
from reporttime import get_functions_with_prefix, measure    
from sumbytes import sumbytes # from sumbytes.pyx

def sumbytes_sumbytes(input):
    return sumbytes(input)

def sumbytes_bytearray(input):
    return sum(bytearray(input))

def sumbytes_numpy(input):
    return np.frombuffer(input, 'uint8').sum() # @root's answer

def main():
    funcs = get_functions_with_prefix('sumbytes_')
    buf = ''.join(map(unichr, range(256))).encode('latin1') * (1 << 16)
    measure(funcs, args=[buf])

main()

Answer 2

Use numpy.frombuffer(buf, "uint8").sum(), it seems to be about 70 times faster than your example:

In [9]: import numpy as np

In [10]: buf = b'a'*(13*(1<<20))

In [11]: sum(bytearray(buf))
Out[11]: 1322254336

In [12]: %timeit sum(bytearray(buf))
1 loops, best of 3: 253 ms per loop

In [13]: np.frombuffer(buf, "uint8").sum()
Out[13]: 1322254336

In [14]: %timeit np.frombuffer(buf, "uint8").sum()
10 loops, best of 3: 36.7 ms per loop

In [15]: %timeit sum([ord(c) for c in buf])
1 loops, best of 3: 2.65 s per loop