Create array with indexes of array based on value

Question

I have a 2D array like this:

map = [[0, 1, 1], [0, 2, 1], [0, 2, 2]]

I would like to get array of indexes based on values. For example map[0][0] == 0 so I would like the pair (0, 0) to be on index 0 in the result array. The result array should look like this:

result[0] = [(0, 0), (1, 0), (2, 0)]
result[1] = [(0, 1), (0, 2), (1, 2)]
result[2] = [(1, 1), (2, 1), (2, 2)]

I've struggled to get a result like this without writting a really bad code:

    val0 = []
    val1 = []
    val2 = []
    for i in range(3):
        for j in range(3):
            if(map[i][j] == 0) : val0.append((i, j))
            if(map[i][j] == 1) : val1.append((i, j))
            if(map[i][j] == 2) : val2.append((i, j))

    data = []
    data.append(val0)
    data.append(val1)
    data.append(val2)

Why don't you show us your code so we can get an idea of where you are? — FChm
– FChm, Commented Apr 1, 2019 at 10:05
I'd be you i'd do 2-dimensionnal for loops to take value from your map and append them to result where you want. But, as @FChm said, if you want us to help you, you'll need to show us some code. — Nenri
– Nenri, Commented Apr 1, 2019 at 10:07
After your edit : It seems you struggle with your actual "position" within the 2-dimensionnal loops. Check Droid's answer — Nenri
– Nenri, Commented Apr 1, 2019 at 10:09

rdas · Accepted Answer · 2019-04-01 10:08:18Z

4

Iterate over the matrix and save where you see each value in a dict. That's your result.

retval = defaultdict(list)
for i in range(len(map)):
   for j in range(len(map[i])):
      val = map[i][j]
      retval[val].append((i,j))
print retval

answered Apr 1, 2019 at 10:08

rdas

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nicola · Accepted Answer · 2019-04-01 10:18:05Z

1

You can use some numpy methods:

import numpy as np
map = [[0, 1, 1], [0, 2, 1], [0, 2, 2]]
npmap = np.array(map)
[list(zip(*np.where(npmap==x))) for x in range(np.max(npmap)+1)]
#[[(0, 0), (1, 0), (2, 0)], [(0, 1), (0, 2), (1, 2)], [(1, 1), (2, 1), (2, 2)]]

edited Apr 1, 2019 at 10:18

answered Apr 1, 2019 at 10:13

nicola

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