convert each element in array to binary python

Question

I have a np array (100,100 dtype=unit8) and I would like to convert each element into binary represented by 8 bits.

I've looked over the two questions and answers below, but can't quite apply them to an n x m array:

Converting integer array to binary array in python

 a=clip_8bit #clip_8bit is a 100x100 unit8 array
 b=np.zeros_like(a)

for i in a:
    b[i]=bin(int(a[i]))[2:].zfill(8)

IndexError: index 109 is out of bounds for axis 0 with size 100

I tried vectorizing and flattening and received either type errors or size-1 array errors

vector = np.vectorize(np.int)
x=clip_8bit_int
x=vector(x)

y=np.zeros_like(x)
y=y.astype(int)
y=vector(y)

for i in y:
    y[i] = bin(i)[2:]  

TypeError: only integer scalar arrays can be converted to a scalar index

How do I convert an array of integers to binary?

This looks to be exactly what I would like to do, but with the issues I had applying the first answer, I'm not understanding exactly how the array is being represented and how the conversion is being applied - here I'm not sure how to extend this code to iterate over an array. I know the following is wrong, but I've been going in circles a bit.

int = clip_8bit  #clip_8bit is a 100x100 unit8 array

int maxLen = 0;
for (int i = 0; i < n; i++) {
    bin[i] = Integer.toBinaryString(arr[i]);
    maxLen = Math.max(maxLen, bin[i].length());
}
for (int i = 0; i < n; i++) {
    if (bin[i].length() != maxLen)
        bin[i] = String.format("%" + maxLen + "s", bin[i]).replace(' ', '0');
    System.out.println(bin[i]);
}

Thanks so much, and thanks for your patience with obviously novice questions!

**** edit ****

I did get this working, but the answer below is much better

flat = clip_8bit.flatten()
n = flat.shape[0]
a = flat
b = np.array(np.zeros((n)))

for i in range(n):
    b[i]=bin(int(a[i]))[2:].zfill(8)

c = np.reshape(b, (100, 100), order = 'C')

Mark · Accepted Answer · 2022-02-08 16:37:52Z

1

Numpy has a fast function np.binary_repr() that does what you want for an individual number. You can apply it to an array with np.vectorize():

import numpy as np

l = np.arange(12).reshape([3, 4])
# array([[ 0,  1,  2,  3],
#        [ 4,  5,  6,  7],
#        [ 8,  9, 10, 11]])

np.vectorize(np.binary_repr)(l, 8)
# array([['00000000', '00000001', '00000010', '00000011'],
#        ['00000100', '00000101', '00000110', '00000111'],
#        ['00001000', '00001001', '00001010', '00001011']], dtype='<U8')

answered Feb 8, 2022 at 16:37

Mark

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1 Comment

Haley Sapers Over a year ago

Brilliant - thank you! Just posting what I got to work, but your answer is obviously more elegant

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