假设我有一个这样的张量
w = [[0.1, 0.7, 0.7, 0.8, 0.3],
[0.3, 0.2, 0.9, 0.1, 0.5],
[0.1, 0.4, 0.8, 0.3, 0.4]]
现在,我想根据某些条件消除某些值(例如大于或小于0.5)
w = [[0.1, 0.3],
[0.3, 0.2, 0.1],
[0.1, 0.4, 0.3, 0.4]]
然后将其填充到相等的长度:
w = [[0.1, 0.3, 0, 0],
[0.3, 0.2, 0.1, 0],
[0.1, 0.4, 0.3, 0.4]]
这就是我在pytorch中实现它的方式:
w = torch.rand(3, 5)
condition = w <= 0.5
w = [w[i][condition[i]] for i in range(3)]
w = torch.nn.utils.rnn.pad_sequence(w)
但是显然,这将非常缓慢,主要是由于列表理解。有没有更好的方法呢?
这是使用boolean masking,tensor splitting,然后最终使用torch.nn.utils.rnn.pad_sequence(...)
填充分割张量的一种直接方法。
torch.nn.utils.rnn.pad_sequence(...)
关于效率的简短说明:使用# input tensor to work with
In [213]: w
Out[213]:
tensor([[0.1000, 0.7000, 0.7000, 0.8000, 0.3000],
[0.3000, 0.2000, 0.9000, 0.1000, 0.5000],
[0.1000, 0.4000, 0.8000, 0.3000, 0.4000]])
# values above this should be clipped from the input tensor
In [214]: clip_value = 0.5
# generate a boolean mask that satisfies the condition
In [215]: boolean_mask = (w <= clip_value)
# we need to sum the mask along axis 1 (needed for splitting)
In [216]: summed_mask = boolean_mask.sum(dim=1)
# a sequence of splitted tensors
In [217]: splitted_tensors = torch.split(w[boolean_mask], summed_mask.tolist())
# finally pad them along dimension 1 (or axis 1)
In [219]: torch.nn.utils.rnn.pad_sequence(splitted_tensors, 1)
Out[219]:
tensor([[0.1000, 0.3000, 0.0000, 0.0000],
[0.3000, 0.2000, 0.1000, 0.5000],
[0.1000, 0.4000, 0.3000, 0.4000]])
是超级有效的,因为它会将分割后的张量作为原始张量的view返回(即不进行任何复制)。