通过欧几里得距离从 numpy 数组的平均值中选择最接近的值

Question

I'm sure there's a straightforward answer to this, but I'm very much a Python novice and trawling stackoverflow is getting me tantalisingly close but falling at the final hurdle, so apologies. I have an array of one dimensional arrays (in reality composed of >2000 arrays, each of ~800 values), but for representation sake:

group = [[0,1,3,4,5],[0,2,3,6,7],[0,4,3,2,5],...]

I'm trying to select the nearest n 1-d arrays to the mean (by Euclidian distance), but struggling to extract them from the original list. I can figure out the distances and sort them, but can't then extract them from the original group.

# Compute the mean
group_mean = group.mean(axis = 0)
     
distances = []
for x in group:
    # Compute Euclidian distance from the mean
    distances.append(np.linalg.norm(x - group_mean))
    # Sort distances
    distances.sort()

print(distances[0:5]) # Prints the five nearest distances

Any advice as to how to select out the five (or whatever) arrays from group corresponding to the nearest distances would be much appreciated.

Answer 1

you can put the array in with the dist array, and sort based on the distance to the mean:

import numpy as np 
group = np.array([[0,1,3,4,5],[0,2,3,6,7],[0,4,3,2,5]])
group_mean = group.mean(axis = 0)

distances = [[np.linalg.norm(x - group_mean),x] for x in group]
distances.sort(key=lambda a : a[0])

print(distances[0:5]) # Prints the five nearest distances

If your arrays get larger, it might be wise to only save the index instead of the whole array:

distances = [[np.linalg.norm(x - group_mean),i] for i,x in enumerate(group)]

If you don't want to save the distances themself, but just want to sort based on the distance, you can do this:

group = list(group)
group.sort(key=lambda group: np.linalg.norm(group - np.mean(group)))