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For simplicity, take the small example below. Lets say we have 2 sets of numpy arrays, values and distances. I'd like to find values that are above 1 and sort them by its corresponding distances. If there are values with similar distances, I'd like to have it sorted with the higher value first.

v = np.array([[1.0,2.0,0.0],[1.0,0.0,0.0],[0.0,0.0,0.0]])
d = np.array([[1.5,1.0,1.5],[1.0,0.0,1.0],[1.5,1.0,1.5]])

indexes = np.argwhere(v >= 1)

list = ( ((d[r,c],v[r,c],(r,c))) for r, c in indexes)

closest_highest = sorted(list,key=lambda t: (t[0],-t[1]))
print(closest_highest)

output:

[(1.0, 2.0, (0, 1)), (1.0, 1.0, (1, 0)), (1.5, 1.0, (0, 0))]

Each tuple contains the distance, value, and its coordinates from the two arrays.

Is there a faster way to do the above using just numpy/vectorized computations? If not, is there a faster/more efficient way to do the following? I dont really need it to return a tuple, just the index is enough. Even just the index of the lowest distance with the highest value is enough.

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  • If there are values with similar distances, I'd like to have it sorted with the higher value first - Which higher values? Apart from the distances, there are only row and column indices. Commented Jan 3, 2018 at 19:31
  • Value from the 'v' array. As the example, indexes (0,1) and (1,0) both have a distance of 1.0. But (0,1) is first since its value '2' is higher than value '1' from index (1,0). Hope that helps Commented Jan 3, 2018 at 19:33
  • Right, that makes sense. Thanks. Commented Jan 3, 2018 at 19:34

1 Answer 1

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Approach #1 : Here's one approach to get index of the lowest distance with the highest value -

# Get row, col indices for the condition
r,c = np.where(v >= 1)

# Extract corresponding values off d and v
di = d[r,c]

# Get indices (indexable into r,c) corresponding to lowest distance
ld_indx = np.flatnonzero(di == di.min())

# Get max index (based off v) out of the selected indices
max_v_idx = v[r[ld_indx], c[ld_indx]].argmax()

# Get the index (indexable into r,c) with the max one based off v
max_idx = ld_indx[max_v_idx]

# Index into r,c with it
lowest_index_out = (r[max_idx], c[max_idx])

Think of it as a two-step filtering process - Once based off min di values and then in the next one the argmax() out of the first-step filtered ones to select that one winner. ld_indx and max_v_idx being the two filtering steps. max_idx is the step that traces back and gets us the index that could be used to get the final indexing tuple off r,c.

Approach #2 : Using more of masking -

indexes = np.argwhere(v >= 1)

di = d[indexes[:,0],indexes[:,1]]
valid_mask = di == di.min()
indexes_mask = indexes[valid_mask]

maxv_indx = v[indexes_mask[:,0],indexes_mask[:,1]].argmax()
lowest_index_out = indexes[valid_mask][maxv_indx]
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8 Comments

Question for the first approach. I understand all except the last one. I know what its doing, but i cant seem to understand how it works. For v[r,c][lidx].argmax(). You are passing r,c on the row index of v? How does that work?
Actually I am still a bit confused on that last statement :P
@user1179317 See if the added comments help out.
Not sure why I am having trouble grasping this. But for me, 'r' and 'c' are 1 dimensional arrays. 'di' is a 1D array as well, containing the distances where v>=1. 'ld_indx' is also 1D array containing the indices of the smallest distances. 'max_idx' should just be a single number where the max value is, based on indices from ld_indx. Which I still understand. But r[ld_indx][max_idx] is a bit confusing since 'r' is a 1D array. I know its working, but I am a little confused how 'r' can have 2 indices, ld_indx and max_idx
Nevermind....r[ld_indx] returns multiple values..duh. Sorry. And thanks for the solution
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