Given two arrays, `a` and `b`, how to find efficiently all combinations of elements in `b` that have equal value in `a`? - numpy

Question

Given two arrays, a and b, how to find efficiently all combinations of elements in b that have equal value in a?
here is an example:
Given
a = [0, 0, 0, 1, 1, 2, 2, 2, 2]
b = [1, 2, 4, 5, 9, 3, 7, 22, 10]
how would you calculate
c = [[1, 2],
[1, 4],
[2, 4],
[5, 9],
[3, 7],
[3, 22],
[3, 10],
[7, 22],
[7, 10],
[22, 10]]
?
a can be assumed to be sorted.
I can do this with loops, a la:
import torch
a = torch.tensor([0, 0, 0, 1, 1, 2, 2, 2, 2])
b = torch.tensor([1, 2, 4, 5, 9, 3, 7, 22, 10])
jumps = torch.cat((torch.tensor([0]),
torch.where(a.diff() > 0)[0] + 1,
torch.tensor([len(a)])))
cs = []
for i in range(len(jumps) - 1):
cs.append(torch.combinations(b[jumps[i]:jumps[i + 1]]))
c = torch.cat(cs)
Is there any efficient way to avoid the loop? The solution should work for CPU and CUDA.
Also, the solution should have runtime O(m * m), where m is the largest number of equal elements in a and not O(n * n) where n is the length of of a.
I prefer solutions for pytorch, but I am curious for solution for numpy as well.

Answer 1

I think the overhead of using torch is only justified for bigger datasets, as there is basically no computational difficulty in the function, imho you can achieve same results with:
from collections import Counter
def find_combinations1(a, b):
count_a = Counter(a)
combinations = []
for x in set(b):
if count_a[x] == b.count(x):
combinations.append(x)
return combinations
or even a simpler:
def find_combinations2(a, b):
return list(set(a) & set(b))
With pytorch I assume the most simple approach is:
import torch
def find_combinations3(a, b):
a = torch.tensor(a)
b = torch.tensor(b)
eq = torch.eq(a, b.view(-1, 1))
indices = torch.nonzero(eq)
return indices[:, 1]
This option has of course a time complexity of O(n*m) where n is the size of a and m is the size of b, and O(n+m) is the memory for the tensors.