Using numpy arithmetic based on position in numpy array

Question

I am not quite sure how to describe my question, but I will try.

I want to know if numpy has the functionality to do this:

Lets say I have a 2D array called grid:

grid = [ [0,0],
         [0,0] ]

I also have a second 2D array called aList:

aList = [ [1,2],
          [3,4] ]

I want to apply math to the first array based on the index of the first array.

So the math done at each iteration would look like this:

grid[i][j] = [(i - aList[k][0]) + (j - aList[k][1])]

Currently doing this in python with for loops is way to expensive so I need an alternative.

EDIT: more clarification, if I were not to use numpy I would write something like this:

for i in range(2):
    for j in range(2):
        num = 0 
        for k in range(2):
            num += (i-aList[k][0]) + (j-aList[k][1])
        grid[i][j] = num

This is however way to slow in python for the amount of data I have.

Could you please paste a small working copy of your script? It is hard to parse what is going on here. What is k? — wflynny
– wflynny, Commented Jul 2, 2013 at 19:17
If this is really what you are doing, and k is the iteration number, then notice that your expression can be simplified to [i + j - c] where c = aList[k][0] + aList[k][1]... — Floris
– Floris, Commented Jul 2, 2013 at 19:19
Sorry if I made this confusing, I knew I was going to be bad at explaining this. i, j, and k are all iterators. i and j are iterating over the entire 2D array grid. K is iterating over the array, aList at each spot in grid. — still learning
– still learning, Commented Jul 2, 2013 at 19:32
Can you please post your current code using for loops. This is very likely something numpy can help with. — Daniel
– Daniel, Commented Jul 2, 2013 at 19:36

Jaime · Accepted Answer · 2013-07-02 20:04:29Z

1

Your code can be reproduced and substantially sped up as follows:

i_s = np.arange(2)
j_s = np.arange(2)

fast_grid = (i_s + j_s[:, None])*len(aList) - aList.sum()

answered Jul 2, 2013 at 20:04

Jaime

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