python - Binary numpy array to list of integers?

Question

I have a binary array, and I would like to convert it into a list of integers, where each int is a row of the array.

For example:

from numpy import *
a = array([[1, 1, 0, 0], [0, 1, 0, 0], [0, 1, 1, 1], [1, 1, 1, 1]])

I would like to convert a to [12, 4, 7, 15].

Answer 1

I once asked a similar question here. Here was my answer, adapted for your question:

def bool2int(x):
    y = 0
    for i,j in enumerate(x):
        y += j<<i
    return y

In [20]: a
Out[20]: 
array([[1, 1, 0, 0],
       [0, 1, 0, 0],
       [0, 1, 1, 1],
       [1, 1, 1, 1]])

In [21]: [bool2int(x[::-1]) for x in a]
Out[21]: [12, 4, 7, 15]

Answer 2

@SteveTjoa's answer is fine, but for kicks, here's a numpy one-liner:

In [19]: a
Out[19]: 
array([[1, 1, 0, 0],
       [0, 1, 0, 0],
       [0, 1, 1, 1],
       [1, 1, 1, 1]])

In [20]: a.dot(1 << arange(a.shape[-1] - 1, -1, -1))
Out[20]: array([12,  4,  7, 15])

(arange is numpy.arange.)

If the bits are in the opposite order, change the order of the values produced by arange:

In [25]: a.dot(1 << arange(a.shape[-1]))
Out[25]: array([ 3,  2, 14, 15])

Answer 3

You could also do this within numpy directly:

from numpy import *
a = array([[1, 1, 0, 0], [0, 1, 0, 0], [0, 1, 1, 1], [1, 1, 1, 1]])

b2i = 2**arange(a.shape[0]-1, -1, -1)

result = (a*b2i).sum(axis=1)  #[12  4  7 15]

Answer 4

If you like working directly with bitwise math, this one should work pretty well.

def bits2int(a, axis=-1):
    return np.right_shift(np.packbits(a, axis=axis), 8 - a.shape[axis]).squeeze()

bits2int(a)
Out: array([12,  4,  7, 15], dtype=uint8)

Answer 5

Another one:

def row_bits2int(arr):
    n = arr.shape[1]  # number of columns

    # shift the bits of the first column to the left by n - 1
    a = arr[:, 0] << n - 1  

    for j in range(1, n):
        # "overlay" with the shifted bits of the next column
        a |= arr[:, j] << n - 1 - j  
    return a