Find x, y coordinates in two np arrays containing X and Y-CodePudding

I can't seem to find an efficient way to do this. The question is this: If I do coords = np.where(A==0), where A is a numpy 2d array, I get a tuple of two np arrays, an array containing the x coordinates (of the points that satisfy the condition) and an array of y coordinates, e.g., (array([0, 0, 0, 0, 1, 1, 2, 2, 2, 2, 3, 3, 3]), array([0, 1, 2, 3, 0, 1, 0, 1, 2, 3, 0, 1, 3])). How do I know if a given point, say (0,0) exists in this tuple (without having to change the tuple into another structure)?

CodePudding user response：

Don't use np.where if you don't want the two arrays of indexes. A==0 will return a boolean array, which may be exactly what you want, if you need to check specific indexes to see if the check is true or not.

mask = A==0
if mask[x,y]: # where x and y are set somewhere above...
   # do stuff

Of course, if the stuff you're doing is numpy related, checking one index at a time is likely to be less efficient than some kind of broader check that lets you do a broadcast operation. For instance, you could change all the 0 values in the array to 1 with this code:

A[mask] = 1

Or this, which does the equivalent of abs (it would be more efficient to save the mask as a separate variable, but I'm not doing that in this example to show how it's not strictly necessary):

A[A < 0] = -A[A < 0]

CodePudding user response：

You can use argwhere instead of where even if it is not optimal (read the answer of @Blccknght)

>>> (np.argwhere(A == 0) == [0, 0]).all(1).any()
True