# Adding Binary Arrays With Different Size

## 13 November 2018 - 1 answer

I made binary arrays from number and its bit length per line, if 295289042101659 with 6 bits per line, number sizes 49 bit so in array, it would be 6 bits X 9 line, through the code and modified to 6-length zero filled lines:

``````def listify(a, bit = 5):
res = []
while a:
a, b = divmod(a,2**bit)
res.append(b)
return res[::-1]
``````

000001
000011
001001
000001
010110
011101
011110
010110
011011

Since it is binary array, I used binary addition code without carrying:

``````def binaryadd(one, other):
if one & other:
return False
return one | other
``````

If I get some array of 402(0b110010010) with size 3, then how could I add into the array at point(2,2) by up-to-down coordinate, or (3,6) from down-to-up, right-to-left coordinate? It should be seemed as:

000001
001111
001101
000101
010110
011101
011110
010110
011011

I've done it like this:

``````def array_add(one,another, point = (0,0)):
a = [a*2**point[0] for a in another[:]]
a+=[0]*point[1]
a = [0]*(len(one)-len(a))+a
res = [binaryadd(a,b) for a, b in  zip(one[::-1],a[::-1])][::-1]
if not all(res):
return False
return res
``````

Is the best way to do it, is to apply binary addition to every values of list, by modifying one list?

Or am I misunderstanding the basics of array?

Since you mention the numpy tag, you can use it to have high performance and readable code :

``````import numpy as np

def int_to_array(n,width):
v=np.zeros(64,np.uint8)
i,u,total=0,n,0
while(u):
if i%width == 0 : total += width
u,v[i],i = u//2,u%2,i+1
return v[:total][::-1].reshape(-1,width)

sx,sy = point
ex = sx+b.shape[0]
ey = sy+b.shape[1]
a[sx:ex,sy:ey] |= b

a=int_to_array(295289042101659,6)
b=int_to_array(402,3)
print(a)
print(b)
print(a)
``````

For :

``````[[0 0 0 0 0 1]
[0 0 0 0 1 1]
[0 0 1 0 0 1]
[0 0 0 0 0 1]
[0 1 0 1 1 0]
[0 1 1 1 0 1]
[0 1 1 1 1 0]
[0 1 0 1 1 0]
[0 1 1 0 1 1]]

[[1 1 0]
[0 1 0]
[0 1 0]]

[[0 0 0 0 0 1]
[0 0 0 0 1 1]
[0 0 1 1 0 1]
[0 0 0 1 0 1]
[0 1 0 1 1 0]
[0 1 1 1 0 1]
[0 1 1 1 1 0]
[0 1 0 1 1 0]
[0 1 1 0 1 1]]
``````