基于索引向量的张量约简 [英] Tensor reduction based off index vector

问题描述

作为一个例子，我有2个张量:A = [1;2;3;4;5;6;7]和B = [2;3;2].我的想法是我想基于B来减少A-这样B的值代表如何求和A的值-这样B = [2;3;2]意味着减少的A应该是前两个值，下一个3和最后一个2的总和: A' = [(1+2);(3+4+5);(6+7)].显然，B的总和应始终等于A的长度.我试图尽可能有效地做到这一点-最好是pytorch/python中包含的特定函数或矩阵运算.谢谢！

As an example, I have 2 tensors: A = [1;2;3;4;5;6;7] and B = [2;3;2]. The idea is that I want to reduce A based off B - such that B's values represent how to sum A's values- such that B = [2;3;2] means the reduced A shall be the sum of the first 2 values, next 3, and last 2: A' = [(1+2);(3+4+5);(6+7)]. It is apparent that the sum of B shall always be equal to the length of A. I'm trying to do this as efficiently as possible - preferably specific functions or matrix operations contained within pytorch/python. Thanks!

推荐答案

以下是解决方案.

• 首先，我们创建具有相同大小A的索引B_idx数组.
• 然后，使用index_add_根据索引B_idx累积(添加)A中的所有元素.

• First, we create an array of indices B_idx with the same size of A.
• Then, accumulate (add) all elements in A based on the indices B_idx using index_add_.

A = torch.arange(1, 8) 
B = torch.tensor([2, 3, 2])

B_idx = [idx.repeat(times) for idx, times in zip(torch.arange(len(B)), B)]
B_idx = torch.cat(B_idx) # tensor([0, 0, 1, 1, 1, 2, 2])

A_sum = torch.zeros_like(B)
A_sum.index_add_(dim=0, index=B_idx, source=A)
print(A_sum) # tensor([ 3, 12, 13])

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