两个张量之间的 Pytorch 差异

Question

我有两个这样的张量：

1st tensor
[[0,0],[0,1],[0,2],[1,3],[1,4],[2,1],[2,4]]

2nd tensor
[[0,1],[0,2],[1,4],[2,4]]

我希望结果张量是这样的：

[[0,0],[1,3],[2,1]] # differences between 1st tensor and 2nd tensor

我尝试使用 set、list、torch.where 等，但找不到任何好方法来实现这一点。有没有办法在两个不同大小的张量之间获得不同的行？（需要高效）

Answer 1

您可以执行成对比较以查看第一个张量的哪些元素存在于第二个向量中。

a = torch.as_tensor([[0,0],[0,1],[0,2],[1,3],[1,4],[2,1],[2,4]])
b = torch.as_tensor([[0,1],[0,2],[1,4],[2,4]])

# Expand a to (7, 1, 2) to broadcast to all b
a_exp = a.unsqueeze(1)

# c: (7, 4, 2) 
c = a_exp == b
# Since we want to know that all components of the vector are equal, we reduce over the last fim
# c: (7, 4)
c = c.all(-1)
print(c)
# Out: Each row i compares the ith element of a against all elements in b
# Therefore, if all row is false means that the a element is not present in b
tensor([[False, False, False, False],
        [ True, False, False, False],
        [False,  True, False, False],
        [False, False, False, False],
        [False, False,  True, False],
        [False, False, False, False],
        [False, False, False,  True]])
non_repeat_mask = ~c.any(-1)

# Apply the mask to a
print(a[non_repeat_mask])
tensor([[0, 0],
        [1, 3],
        [2, 1]])

如果你觉得很酷，你可以做一个班轮:)

a[~a.unsqueeze(1).eq(b).all(-1).any(-1)]