Python : Vectorized comparison of two ordered numpy arrays of int of the same size

Question

I want to compare two ordered numpy arrays of int of the same size in Python and output the common elements which are the same value at the same position :

import numpy as np
a = np.asarray([20, 35, 226, 62, 129, 108, 156, 225, 115, 35, 162, 43, 9, 120, 181, 220])
b = np.asarray([1, 35, 69, 103, 137, 171, 205, 239, 18, 52, 86, 120, 154, 188, 222, 240])

The element-wise comparison would give : [35]

Can you help me ?

Answer 1

If you're using NumPy, than you can use a boolean mask:

import numpy as np 
a = np.asarray([20, 35, 226, 62, 129, 108, 156, 225, 115, 35, 162, 43, 9, 120, 181, 220])
b = np.asarray([1, 35, 69, 103, 137, 171, 205, 239, 18, 52, 86, 120, 154, 188, 222, 240])
c = a[a == b]
print(c) # [35]

Answer 2

something like that will do the job, though using zip is more pythonic ...

for index, value in enumerate(a):
if value == b[index]:
    c.append(value)

Answer 3

Something like this can work:

l = []
for x,y in zip(a,b):
   if x == y: 
      l.append(x)

in terms of list comprehension in can be written like this:

l = [x for x,y in zip(a,b) if x == y]

Explanation

zip(a,b) will generate the following:

>>> zip(a,b)
[(20, 1), (35, 35), (226, 69), (62, 103), (129, 137), (108, 171), (156, 205), (225, 239), (115, 18), (35, 52), (162, 86), (43, 120), (9, 154), (120, 188), (181, 222), (220, 240)]
>>>

Then, you iterate through each element (x,y) of the results of zip(a,b) and compare x and y .

Reproducible example:

>>> a = [20, 35, 226, 62, 129, 108, 156, 225, 115, 35, 162, 43, 9, 120, 181, 220
>>> b = [1, 35, 69, 103, 137, 171, 205, 239, 18, 52, 86, 120, 154, 188, 222, 240
>>> zip(a,b)
[(20, 1), (35, 35), (226, 69), (62, 103), (129, 137), (108, 171), (156, 205), (2
>>> [x for x,y in zip(a,b) if x == y ]
[35]
>>>

Answer 4

You apparently don't need a set intersection. Zip the lists and compare items at the same index:

>>> [x for x, y in zip(a, b) if x==y]
[35]