Argmin with the Euclidean distance condition-CodePudding

In Python, I have a vector v of 300 elements and an array arr of 20k 300-dimensional vectors. How do I get quickly the indices of the k closest elements to v from the array arr?

CodePudding user response：

Since 300 is a very small number, sorting all elements and then just using the `k first is not an expensive operation (usually; it depends on how many thousand times per second you need to do this).

so, sorted() is your friend; use the key= keyword argument, sorted_vector = sorted(v ,key=…) to implement sorting by euclidean distance.

Then, use the classic array[:end] syntax to select the first k.

CodePudding user response：

You can do this task with numpy

import numpy as np
v = np.array([[1,1,1,1]])
arr = np.array([
  [1,1,1,1],
  [2,2,2,2],
  [3,3,3,3]
])
dist = np.linalg.norm(v - arr, axis=1) # Euclidean distance
min_distance_index = np.argmin(dist) # Find index of minimum distance
closest_vector = arr[min_distance_index] # Get vector having minimum distance
closest_vector
# array([1, 1, 1])