[英]How to check whether tensor values in a different tensor pytorch?
I have 2 tensors of unequal size我有 2 个大小不等的张量
a = torch.tensor([[1,2], [2,3],[3,4]])
b = torch.tensor([[4,5],[2,3]])
I want a boolean array of whether each value exists in the other tensor without iterating.我想要一个 boolean 数组,其中每个值是否存在于另一个张量中而不进行迭代。 something like
就像是
a in b
and the result should be结果应该是
[False, True, False]
as only the value of a[1] is in b因为只有 a[1] 的值在 b
I think it's impossible without using at least some type of iteration.我认为不使用至少某种类型的迭代是不可能的。 The most succinct way I can manage is using list comprehension:
我能管理的最简洁的方法是使用列表理解:
[True if i in b else False for i in a]
Checks for elements in b that are in a and gives [False, True, False].检查 b 中 a 中的元素并给出 [False, True, False]。 Can also be reversed to get elements a in b [False, True].
也可以反过来得到 b [False, True] 中的元素 a。
this should work这应该工作
result = []
for i in a:
try: # to avoid error for the case of empty tensors
result.append(max(i.numpy()[1] == b.T.numpy()[1,i.numpy()[0] == b.T.numpy()[0,:]]))
except:
result.append(False)
result
If you need to compare all subtensors across the first dimension of a
, use in
:如果您需要比较 a 的第一个维度上
a
所有子张量,请使用in
:
>>> [i in b for i in a]
[False, True, False]
Neither of the solutions that use tensor in tensor
work in all cases for the OP.在张量中使用
tensor in tensor
的解决方案都不适用于 OP。 If the tensors contain elements/tuples that match in at least one dimension, the aforementioned operation will return True
for those elements, potentially leading to hours of debugging.如果张量包含至少在一个维度上匹配的元素/元组,则上述操作将为这些元素返回
True
,这可能会导致数小时的调试。 For example:例如:
torch.tensor([2,5]) in torch.tensor([2,10]) # returns True
torch.tensor([5,2]) in torch.tensor([5,10]) # returns True
A solution for the above could be forcing the check for equality in each dimension, and then applying a Tensor Boolean add.上述解决方案可能是强制检查每个维度的相等性,然后应用张量 Boolean 添加。 Note, the following 2 methods may not be very efficient because
Tensors
are rather slow for iterating and equality checking, so converting to numpy
may be needed for large data:请注意,以下 2 种方法可能效率不高,因为
Tensors
在迭代和相等检查方面相当慢,因此对于大数据可能需要转换为numpy
:
[all(torch.any(i == b, dim=0)) for i in a] # OR
[any((i[0] == b[:, 0]) & (i[1] == b[:, 1])) for i in a]
That being said, @yuri's solution also seems to work for these edge cases, but it still seems to fail occasionally, and it is rather unreadable.话虽如此,@yuri 的解决方案似乎也适用于这些边缘情况,但它似乎仍然偶尔会失败,而且相当难以理解。
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