Efficiently build an integer from bits of a boolean Numpy array

Question

I am looking for a more efficient way to do the equivalent of

myarray * (2**arange(len(myarray))

Essentially I am after something like numpy.packbits that packs the bits into a single integer for any reasonable sized myarray yielding an appropriate size integer. I can implement this using numpy.packbits but I was wandering there is already a builtin that does this.

What do you mean by "builtin"? What is wrong with using numpy.packbits? — ali_m
– ali_m, Commented Jan 31, 2016 at 15:15

B. M. · Accepted Answer · 2016-02-01 06:37:56Z

2

Three versions:

from numpy import *
from numba import jit

myarray=random.randint(0,2,64).astype(uint64) 

def convert1(arr) : return (arr*(2**arange(arr.size,dtype=uint64))).sum()

pow2=2**arange(64,dtype=uint64)
def convert2(arr) : return (arr*pow2[:arr.size]).sum()

@jit("uint64(uint64[:])")
def convert3(arr):
    m=1
    y=0
    for i in range(arr.size):
        y=y + pow2[i] * arr[i]
    return y

with times:

In [44]: %timeit convert1(myarray)
10000 loops, best of 3: 62.7 µs per loop

In [45]: %timeit convert2(myarray)
10000 loops, best of 3: 11.6 µs per loop

In [46]: %timeit convert3(myarray)
1000000 loops, best of 3: 1.55 µs per loop

Precomputing and Numba allow big improvements.

edited Feb 1, 2016 at 6:37

answered Jan 31, 2016 at 22:10

B. M.

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1 Comment

Daniel Mahler Over a year ago

The Numba solution is pretty much what ended up going with. I used bitwise operations in the loop instead of arithmetic. There is no significant speed difference between the two.

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