Conditional Sum of 2D Numpy Arrays

Question

I have 4 (but in reality an arbitrary amount of) 2D arrays like so:

import numpy as np

c1 = np.ones((75, 100))
c2 = np.ones((75, 100))
c3 = np.ones((75, 100))
c4 = np.ones((75, 100))

c1[22:42, 5:35] = np.random.rand(20, 30) / 2
c2[25:45, 25:55] = np.random.rand(20, 30) / 2
c3[28:48, 45:75] = np.random.rand(20, 30) / 2
c4[31:51, 65:95] = np.random.rand(20, 30) / 2

What I'd like to do is sum the arrays everywhere except for where the arrays overlap. And where there is an overlap, the value ought to be the left array. My instinct is to use np.where, but I cannot think of a clever/neat way of doing so.

Hopefully the below image makes this clear

c_arrays = np.array([c1, c2, c3, c4])
result = c_arrays.sum(axis=0)

fig, ax = plt.subplots()
ax.imshow(result)

Edit: I came up with an awful, recursive solution that at least shows the result I'm looking for. My hope is that someone can offer a much cleaner approach, in particular that isn't recursive

c_arrays_1 = []
for ci, cj in zip(c_arrays, c_arrays[1:]):
    c = np.where(ci + cj < 1, ci, ci + cj - 1)
    c_arrays_1.append(c)

c_arrays_2 = []
for ci, cj in zip(c_arrays_1, c_arrays_1[1:]):
    c = np.where(ci + cj < 1, ci, ci + cj - 1)
    c_arrays_2.append(c)

c_arrays_3 = []
for ci, cj in zip(c_arrays_2, c_arrays_2[1:]):
    c = np.where(ci + cj < 1, ci, ci + cj - 1)
    c_arrays_3.append(c)

fig, ax = plt.subplots()
ax.imshow(c_arrays_3[0])

np.where(c1 != 1, c1, c1 + c2)? For example? Are you looking to generalise something like this? — cs95
– cs95, Commented Oct 31, 2017 at 11:32
So you array is initialized at 1, and some parts are random numbers, which could also evaluate to 1. How to distinguish the two? — Chiel
– Chiel, Commented Oct 31, 2017 at 11:32
@cᴏʟᴅsᴘᴇᴇᴅ I think that's the idea. @Divakar the overlaps are not defined explicitly, but could be found with something like np.where(result < 3, 1, 0). Better yet, result < n-1 where n is the number of arrays. — lanery
– lanery, Commented Oct 31, 2017 at 11:34

Daniel F · Accepted Answer · 2017-10-31 13:33:26Z

1

Here's a recursive method that I think fits your requirements:

def condsum(*arrs, r = 1):
    if len(arrs) == 1:
        return arrs[0]
    else:
        a = condsum(*arrs[1:], r = r)
        return np.where(a == r, arrs[0], a)

Then you'd just need to do

plt.imshow(condsum(c1, c2, c3, c4))

answered Oct 31, 2017 at 13:33

Daniel F

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1 Comment

lanery Over a year ago

This is great! Thank you for ignoring my request for a solution without recursion :) Clearly did not know what was best for me.

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