Fast Bitwise Sum in Python

Question

Is there an efficient way to calculate sum of bits in each column over array in Python?

Example (Python 3.7 and Numpy 1.20.1):

Create numpy array with values 0 or 1

import numpy as np

array = np.array(
    [
     [1, 0, 1],   
     [1, 1, 1], 
     [0, 0, 1],    
    ]
)

Compress size by np.packbits

pack_array = np.packbits(array, axis=1)

Expected result: sum of bits in each position (column) without np.unpackbits to get the same as array.sum(axis=0):

array([2, 1, 3])

I found just very slow solution:

dim = array.shape[1]
candidates = np.zeros((dim, dim)).astype(int)
np.fill_diagonal(candidates, 1)

pack_candidates = np.packbits(candidates, axis=1)

np.apply_along_axis(lambda c:np.sum((np.bitwise_and(pack_array, c) == c).all(axis=1)), 1, pack_candidates)

Asked By: Lukáš Sýkora

||

Source

Answer 1

It seems there is no better option in numpy than numpy.unpackbits.

To be more clear, let’s take another example:

array = np.array([[1, 0, 1, 0, 1, 1, 1, 0, 1], 
                  [1, 1, 1, 1, 1, 1, 1, 1, 1], 
                  [0, 0, 1, 0, 0, 0, 0, 0, 0]])
pack_array = np.packbits(array, axis=1)
dim = array.shape[1]

Now, pack_array is calculated in this way:

[[1,0,1,0,1,1,1,0], [1,0,0,0,0,0,0,0]] -> [174, 128]
[[1,1,1,1,1,1,1,1], [1,0,0,0,0,0,0,0]] -> [255, 128]
[[0,0,1,0,0,0,0,0], [0,0,0,0,0,0,0,0]] -> [32, 0]

I’ve tested various algorithms and unpacking bits seems to be the fastest:

def numpy_sumbits(pack_array, dim):
    out = np.unpackbits(pack_array, axis=1, count=dim)
    arr = np.sum(out, axis=0)
    return arr

def manual_sumbits(pack_array, dim):
    arr = pack_array.copy()
    out = np.empty((dim//8+1) * 8, dtype=int)
    for i in range(8):
        out[7 - i%8::8] = np.sum(arr % 2, axis=0)
        arr = arr // 2
    return out[:dim]

def numpy_sumshifts(pack_array, dim):
    res = (pack_array.reshape(pack_array.size, -1) >> np.arange(8)) % 2
    res = res.reshape(*pack_array.shape, 8)
    return np.sum(res, axis=0)[:,::-1].ravel()[:dim]

print(numpy_unpackbits(pack_array, dim))
print(manual_unpackbits(pack_array, dim))
print(numpy_sumshifts(pack_array, dim))
>>>
[2 1 3 1 2 2 2 1 2]
[2 1 3 1 2 2 2 1 2]
[2 1 3 1 2 2 2 1 2]

%%timeit
numpy_sumbits(pack_array, dim)
>>> 3.49 ms ± 57.3 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

%%timeit
manual_sumbits(pack_array, dim)
>>> 10 ms ± 22.4 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

%%timeit
numpy_sumshifts(pack_array, dim)
>>> 20.1 ms ± 97.9 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)

Answered By: mathfux

Answer 2

Using np.unpackbits can be problematic if the input array is big since the resulting array can be too big to fit in RAM, and even if it does fit in RAM, this would be far from being efficient since the huge array have to be written and read from the (slow) main memory. The same thing apply for CPU caches: smaller arrays can generally be computed faster. Moreover, np.unpackbits have a quite big overhead for small arrays.

AFAIK, this is not possible to do this operation very efficiently in Numpy while using a small amount of RAM (ie. using np.unpackbits, as pointed out by @mathfux). However, Numba can be used to speed up this computation, especially for small arrays. Here is the code:

@nb.njit('int32[::1](uint8[:,::1], int_)')
def bitSum(packed, m):
    n = packed.shape[0]
    assert packed.shape[1]*8-7 <= m <= packed.shape[1]*8
    res = np.zeros(m, dtype=np.int32)
    for i in range(n):
        for j in range(m):
            res[j] += bool(packed[i, j//8] & (128>>(j%8)))
    return res

If you want a faster implementation, you can optimize the code by working on fixed-size tiles. However, this makes the code also more complex. Here is the resulting code:

@nb.njit('int32[::1](uint8[:,::1], int_)')
def bitSumOpt(packed, m):
    n = packed.shape[0]
    assert packed.shape[1]*8-7 <= m <= packed.shape[1]*8
    res = np.zeros(m, dtype=np.int32)
    for i in range(0, n, 4):
        for j in range(0, m, 8):
            if i+3 < n and j+7 < m:
                # Highly-optimized 4x8 tile computation
                k = j//8
                b0, b1, b2, b3 = packed[i,k], packed[i+1,k], packed[i+2,k], packed[i+3,k]
                for j2 in range(8):
                    shift = 7 - j2
                    mask = 1 << shift
                    res[j+j2] += ((b0 & mask) + (b1 & mask) + (b2 & mask) + (b3 & mask)) >> shift
            else:
                # Slow fallback computation
                for i2 in range(i, min(i+4, n)):
                    for j2 in range(j, min(j+8, m)):
                        res[j2] += bool(packed[i2, j2//8] & (128>>(j2%8)))
    return res

Here are performance results on my machine:

On the example array:
Initial code:    62.90 us   (x1)
numpy_sumbits:    4.37 us   (x14)
bitSumOpt:        0.84 us   (x75)
bitSum:           0.77 us   (x82)

On a random 2000x2000 array:
Initial code:  1203.8  ms   (x1)
numpy_sumbits:    3.9  ms   (x308)
bitSum:           2.7  ms   (x446)
bitSumOpt:        1.5  ms   (x802)

The memory footprint of the Numba implementations is much better too (at least 8 times smaller).

Answered By: Jérôme Richard

Answer 3

I parse binary for a living, and by your description, this is not a binary problem.


def test():
    a,b,c=0,0,0

    a_ray=[
     [1, 0, 1],   
     [1, 1, 1], 
     [0, 0, 1],    
     ]
    
    for i in a_ray:
        a+=i[0]
        b+=i[1]
        c+=i[2]

10000 loops, average of 7: 0.171 +- 0.21 usec per loop (using standard deviation)

With a_ray having a length of 30,000

10000 loops, average of 7: 206 +- 1.27 usec per loop (using standard deviation)

As far as timeit, I get the following warning when I call timeit from pypy3.

"
WARNING: timeit is a very unreliable tool. use pyperf or something else for real measurements
"

Answered By: Leroy Scandal

Fast Bitwise Sum in Python

Question:

Answers: