如何在numpy Python中找到1D二进制数组的十进制值

Question

I have a boolean numpy array like this,

>>> np_arr
array([[1, 1, 1, 1, 0, 0, 1, 1],
       [1, 1, 1, 1, 0, 0, 1, 1],
       [1, 1, 1, 1, 0, 0, 1, 1],
       [1, 1, 1, 1, 0, 0, 1, 1],
       [1, 1, 1, 1, 0, 0, 1, 1],
       [1, 1, 1, 1, 0, 1, 0, 0],
       [1, 1, 1, 1, 0, 1, 0, 0],
       [1, 1, 1, 1, 0, 1, 0, 0]])

and another 1D array like this,

>>> another_arr
array([128,  64,  32,  16,   8,   4,   2,   1])

I want to somehow do some and or addition to get only values where that 1 is present something like,

>>> np_arr
array([[128,64,32,8, 0, 0, 2, 1],
       [128,64,32,8, 0, 0, 2, 1],
         ....................
       [128,64,32,8, 0,4, 0, 0],
        .....................)

So then I can then sum them to find the binary value of the each 1D array in the 2D array.. Or is some simple way to get decimal value numpy array as result?

Answer 1

This is one way. It works because numpy broadcasts implicitly.

import numpy as np

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

arr = np.array([128,  64,  32,  16,   8,   4,   2,   1])

arr2 = arr * mask

# array([[128,  64,  32,  16,   0,   0,   2,   1],
#        [128,  64,  32,  16,   0,   0,   2,   1],
#        [128,  64,  32,  16,   0,   0,   2,   1],
#        [128,  64,  32,  16,   0,   0,   2,   1],
#        [128,  64,  32,  16,   0,   0,   2,   1],
#        [128,  64,  32,  16,   0,   4,   0,   0],
#        [128,  64,  32,  16,   0,   4,   0,   0],
#        [128,  64,  32,  16,   0,   4,   0,   0]])

Answer 2

What you need is probably this:

import numpy as np


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

np.packbits(ar, axis=-1)

Result:

array([[243],
       [243],
       [243],
       [243],
       [243],
       [244],
       [244],
       [244]], dtype=uint8)

Answer 3

here a quick and dirty way.

import numpy as np

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


a = np.array([17, 16, 12, 41, 0, 0, 5, 12])
for _ in ar:
    m = np.multiply(_, a)
    print(m)

here i've printed the values but you can append them in an array or use as you like.