Evaluate arrays and assign values in Python

Question

I have two multi-dimensional arrays and I would like to evaluate and assign values for those matching and non-matching values.

array_1 = array([[1,2,2], [1,2,3],[1,2,1]])
array_2 = array([[1,1,2], [2,2,3],[3,1,1]])

The idea is that for each member of array 1, if the value is the same, I want to assign 0 and it doesn't, I want to assign a different value. The logic is something like this.

(array_1 ==1 and array_2 ==1) = 0
(array_1 ==1 and array_2 ==2) = 10 
(array_1 ==1 and array_2 ==3) = 15
(array_1 ==2 and array_2 ==1) = 5
(array_1 ==2 and array_2 ==2) = 0
(array_1 ==2 and array_2 ==3) = 15
(array_1 ==3 and array_2 ==1) = 5
(array_1 ==3 and array_2 ==2) = 10
(array_1 ==3 and array_3 ==3) = 0

The over all goal is to detect a change. A brief description of this methodology is available on
https://www.e-education.psu.edu/geog883/node/496

John Ladasky · Accepted Answer · 2018-04-22 18:48:08Z

3

First: I am not familiar with the array format that you showed in your post. I have never seen an array instantiated in Python using just square brackets. That's not a function call.

Second: Your problem may not be fully specified. However: if you have shown us all the possible values that you can have in your input arrays, then only 1, 2, and 3 are possible. In that case, there are only 9 possible pairings of elements from array_1 and array_2, and the values you want as outputs can easily be stored in a 3 x 3 lookup table.

Finally: I use Numpy. You seem to want arrays, and Numpy is very common and handles arrays well.

import numpy as np
from itertools import product

arr1 = np.array([[1,2,2],[1,2,3],[1,2,1]], dtype=int)
arr2 = np.array([[1,1,2],[2,2,3],[3,1,1]], dtype=int)

lookup = np.array(((0,10,15),(5,0,15),(5,10,0)), dtype=int)
result = np.zeros_like(arr1)
for r, c in product(*[range(x) for x in arr1.shape]):
    a, b = arr1[r,c], arr2[r,c]
    result[r,c] = lookup[a-1,b-1]

print(arr1, "\n")
print(arr2, "\n")
print(result)

Here's the output:

[[1 2 2]
 [1 2 3]
 [1 2 1]] 

[[1 1 2]
 [2 2 3]
 [3 1 1]] 

[[ 0  5  0]
 [10  0  0]
 [15  5  0]]

answered Apr 22, 2018 at 18:48

John Ladasky

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2 Comments

Destaneon Over a year ago

Yep... you're right. I misplaced the open bracket. There are 9 possibilities three of which are 0 (no change) and the others are assigned based on the type of change. Thanks for your help!

John Ladasky Over a year ago

In hindsight, I offer a minor improvement. If you are willing to encode your conditions in arr1 and arr2 starting from zero rather than from one, then the lookup is simpler: just use result[r,c] = lookup[a,b]. I'm also wondering whether Numpy's fancy indexing can be brought to bear, eliminating both the need for zeros_like() and the for loop.

johnashu · Accepted Answer · 2018-04-22 22:24:33Z

0

You can do this in a list comprehension:

Use a conditional expression to produce your values.

You put the if, else statement first and then iterate through to replace the values.

I have created a list and pulled a number at random from that list.

import random

array_1 = [[1, 2, 2], [1, 2, 3], [1, 2, 1]]
array_2 = [[1, 1, 2], [2, 2, 3], [3, 1, 1]]

lst = [5, 10, 15, 20, 25, 30, 35, 40, 45, 50]

res = [[random.choice(lst) if i != j else 0 for i, j in zip(a, b)]
    for a, b in zip(array_1, array_2)]

for i in res:
    print(i)

Returns:

[0, 25, 0]
[50, 0, 0]
[35, 45, 0]

edited Apr 22, 2018 at 22:24

answered Apr 22, 2018 at 18:41

johnashu

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2 Comments

Destaneon Over a year ago

This would do the job but the idea was to print out one 3 x3 array.

johnashu Over a year ago

I did it both ways as i was not sure.although it was the same lol. I have update my answer for one 3x3 matrix..

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