Using slice/mask instead of a for-loop to find items in an array

Question

p= np.array([[ 1,  2,  3,  4,  5],
   [ 6,  7,  8,  9, 10],
   [11, 12, 13, 14, 15],
   [16, 17, 18, 19, 20]])

I have the above array and need to find the items that are divisible by 2 and 3 without converting it to a flat array using for-loops and then slicing/masking.

I was able to figure out the for-loops part and am running into some issues with the slicing/masking part.

# we know that 6, 12 and 18 are divisible by 2 and 3
# therefore we can use slicing to pull those numbers out of the array

print(p[1:2,0:1]) # slice array to return 6
print(p[2:3,1:2]) # slice array to return 12
print(p[3:4,2:3]) # slice array to return 18

m=np.ma.masked_where(((p[:, :]%2==0)&(p[:, :]%3==0)),p)
print(m)

mask=np.logical_and(p%2==0,p%3==0)
print(mask)

Is there a more efficient way of slicing the array to find 6, 12 and 18? Also, is there a way to make either of the two mask functions output just 6, 12, and 18? The first one shows the inverse of what I want while the other returns a Boolean answer.

Refer here

AnkurSaxena
– AnkurSaxena

2021-01-29 21:14:06 +00:00
Commented Jan 29, 2021 at 21:14 — AnkurSaxena
– AnkurSaxena, Commented Jan 29, 2021 at 21:14

pho · Accepted Answer · 2021-01-29 21:46:37Z

1

You nearly had it!

mask=np.logical_and(p%2==0,p%3==0)

gives you True where p % 2 == 0 and p % 3 == 0.

mask = array([[False, False, False, False, False],
       [ True, False, False, False, False],
       [False,  True, False, False, False],
       [False, False,  True, False, False]])

From this, you can get the values of p where mask is True by simply

p[mask]

Which gives the output:

array([ 6, 12, 18])

answered Jan 29, 2021 at 21:46

pho

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2 Comments

Tony Stark Over a year ago

Whoa thanks Pranav. I have tried alot of different things to do that but apparently not that one. I appreciate the help.

Mr. T Over a year ago

If this provided your preferred answer to your question, please consider accepting it.

Alain T. · Accepted Answer · 2021-01-29 22:27:59Z

1

Why not simply do p[p%6==0] ?

import numpy as np

p= np.array([[ 1,  2,  3,  4,  5],
   [ 6,  7,  8,  9, 10],
   [11, 12, 13, 14, 15],
   [16, 17, 18, 19, 20]])

p[p%6==0] # [ 6 12 18]

answered Jan 29, 2021 at 22:27

Alain T.

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2 Comments

Tony Stark Over a year ago

Good question but the assignment says 3 and 2. But I think this would be acceptable. :D

Alain T. Over a year ago

I hope it is. Leveraging basic math should be encouraged as it always gives better/faster algorithms.

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