[FIXED] Numpy argmin() to find the nearest tuple

Issue

I have an array of tuples, I need to find the tuple from that array that is the closes to a given tuple(element wise), that is by the absolute value difference between each element of these two tuples. The array of tuples is like:

array_of_tuples = [(0.0, 6.5, 1),
     (0.0, 6.5, 4.5),
     (0.0, 6.5, 8.0),
     (0.0, 13.5, 1),
     (0.0, 13.5, 4.5),
     (0.0, 13.5, 8.0),
     (0.0, 21.0, 1),
     (0.0, 21.0, 4.5),
     (0.0, 21.0, 8.0),
     (7.0, 6.5, 1),
     (7.0, 6.5, 4.5),
     (13.5, 21.0, 8.0)]
 
 

While the query tuple is (13.1, 20.3, 8.4)

However doing: np.argmin(np.abs(array_of_tuples - (13.1, 20.3, 8.4)))

gives 8, while if I print the result without np.argmin() I clearly see that it is the last element in the array that has the least difference with the given tuple.

Solution

You have to consider all coordinates and aggregate them into a single number (per row).

Compute the sum of squares and get the (arg)min of that:

np.argmin(((array_of_tuples - (13.1, 20.3, 8.4))**2).sum(axis=1))

Output: 11

Intermediate before argmin:

array([416.81, 377.26, 362.21, 272.61, 233.06, 218.01, 226.86, 187.31,
       172.26, 282.41, 242.86,   0.81])

As suggested by @MadPhysicist, you can also use numpy.linalg.norm to compute the norm per row:

np.argmin(np.linalg.norm(array_of_tuples-(13.1, 20.3, 8.4), axis=1))

Answered By - mozway