Issue
Is it possible to shuffle two 2D tensors in PyTorch by their rows, but maintain the same order for both? I know you can shuffle a 2D tensor by rows with the following code:
a=a[torch.randperm(a.size()[0])]
To elaborate: If I had 2 tensors
a = torch.tensor([[1, 1, 1, 1, 1],
[2, 2, 2, 2, 2],
[3, 3, 3, 3, 3]])
b = torch.tensor([[4, 4, 4, 4, 4],
[5, 5, 5, 5, 5],
[6, 6, 6, 6, 6]])
And ran them through some function/block of code to shuffle randomly but maintain correlation and produce something like the following
a = torch.tensor([[2, 2, 2, 2, 2],
[1, 1, 1, 1, 1],
[3, 3, 3, 3, 3]])
b = torch.tensor([[5, 5, 5, 5, 5],
[4, 4, 4, 4, 4],
[6, 6, 6, 6, 6]])
My current solution is converting to a list, using the random.shuffle() function like below.
a_list = a.tolist()
b_list = b.tolist()
temp_list = list(zip(a_list , b_list ))
random.shuffle(temp_list) # Shuffle
a_temp, b_temp = zip(*temp_list)
a_list, b_list = list(a_temp), list(b_temp)
# Convert back to tensors
a = torch.tensor(a_list)
b = torch.tensor(b_list)
This takes quite a while and was wondering if there is a better way.
Solution
You mean
indices = torch.randperm(a.size()[0])
a=a[indices]
b=b[indices]
?
Answered By - hkchengrex
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