Question

I just started learning some numpy because of High performance calculation of least squares difference from all possible combinations (n lists):

Right now I’m stucked with the calculations and could use some help.

I have a numpy array object that looks like this:

>>> items
array([[ 246, 1143, 1491, ..., 1167,  325, 1158],
       [ 246, 1143, 1491, ..., 1167,  519, 1158],
       [ 246, 1143, 1491, ..., 1167,  507, 1158],
       ..., 
       [1491, 1143,  246, ..., 1167,  325, 1158],
       [1491, 1143,  246, ..., 1167,  519, 1158],
       [1491, 1143,  246, ..., 1167,  507, 1158]])

I would like to get the number of the array with the least square difference among all his members, a numpythonic version of:

for num,item in enumerate(items): #Calculate for each list of items
      for n in range(len(item)):
        for i in range(n, len(item)):
          dist += (item[n]-item[i])**2 #Key formula
          if dist>min_dist: #This is a shortcut
              break
          else:
              continue
          break               
      if min_dist is None or dist < min_dist:
        min_dist = dist
        best = num #We get the number of the combination we want

I would appreciate any hints.

Answer 1

Initialize your NxM array:

>>> import numpy as np
>>> items = np.random.random_sample((10,3))

Calculate the sum of squares between all elements of each of N M-dimensional vectors and store the results in a list:

>>> sq = [(np.subtract.outer(item,item) ** 2).sum() for item in items]

Find the index of the vector with the smallest sum-of-squares between all its elements:

>>> best_index = np.argmin(sq)

Or, to avoid the intermediate list:

best = np.inf
best_index = None
for i,item in enumerate(items):
    ls = (np.subtract.outer(item,item) ** 2).sum()
    if ls < best:
        best = ls
        best_index = i