I want to iterate over all distinct permutations of a vector. I have tried doing this by using vecextract() in combination with numtoperm() to create a vector of permutations, and vecsort(,,,8) to remove equivalent permutations.
Unfortunately, this doesn't scale well: the maximum size of a vector within my current stack size of 4GB is less than 12!, and my machine only has 16GB.
Is there a way to do this without running out of memory, maybe by generating the k-th distinct permutation directly?
There is nothing built into PARI for this. I would suggest reading How to generate all the permutations of a multiset?.
Use forperm.
forperm([1,1,2,3], v, print(v))
Produces
Vecsmall([1, 1, 2, 3])
Vecsmall([1, 1, 3, 2])
Vecsmall([1, 2, 1, 3])
Vecsmall([1, 2, 3, 1])
Vecsmall([1, 3, 1, 2])
Vecsmall([1, 3, 2, 1])
Vecsmall([2, 1, 1, 3])
Vecsmall([2, 1, 3, 1])
Vecsmall([2, 3, 1, 1])
Vecsmall([3, 1, 1, 2])
Vecsmall([3, 1, 2, 1])
Vecsmall([3, 2, 1, 1])
Note that the input vector to forperm should be sorted for correct results.
Related
I have a huge torch Tensor and I'm looking for an efficient approach to subtract the elements of every pair of that Tensor.
Of course I could use two nested for but it wouldn't be efficient.
For example giving
[1, 2, 3, 4]
The output I want is
[1-2, 1-3, 1-4, 2-3, 2-4, 3-4]
You can do this easily:
>>> x = torch.tensor([1, 2, 3, 4])
>>> x[:, None] - x[None, :]
tensor([[ 0, -1, -2, -3],
[ 1, 0, -1, -2],
[ 2, 1, 0, -1],
[ 3, 2, 1, 0]])
see more details here.
I have 2 tensors of shape [2, 1, 9] and [2, 1, 3]. I'd like to concatenate across the 3rd dimension alternatively (once every 4).
For example:
a = [[[1,2,3,4,5,6,7,8,9]],[[11,12,13,14,15,16,17,18,19]]]
b = [[[10, 20, 30]], [[1, 2, 3]]]
result = [[[1,2,3,10,4,5,6,20,7,8,9,30]],[[11,12,13,1,14,15,16,2,17,18,19,3]]]
How can I do this in pytorch?
This would do the trick:
torch.concat([a.reshape((2, 1, 3, 3)), b.reshape(2, 1, 3, 1)], axis=-1).reshape((2, 1, -1))
There's probably a smarter way to do this, but hey, it works.
Suppose I have the following tensor: y = torch.randint(0, 3, (10,)). How would you go about counting the 0's 1's and 2's in there?
The only way I can think of is by using collections.Counter(y) but was wondering if there was a more "pytorch" way of doing this. A use case for example would be when building the confusion matrix for predictions.
You can use torch.unique with the return_counts option:
>>> x = torch.randint(0, 3, (10,))
tensor([1, 1, 0, 2, 1, 0, 1, 1, 2, 1])
>>> x.unique(return_counts=True)
(tensor([0, 1, 2]), tensor([2, 6, 2]))
Could someone help me how to perform bitwise AND operations on two tensors in Pytorch 1.4?
Apparently I could only find NOT and XOR operations in official document
I don't see them in the docs, but it looks like &, |, __and__, __or__, __xor__, etc are bit-wise:
>>> torch.tensor([1, 2, 3, 4]).__xor__(torch.tensor([1, 1, 1, 1]))
tensor([0, 3, 2, 5])
>>> torch.tensor([1, 2, 3, 4]) | torch.tensor([1, 1, 1, 1])
tensor([1, 3, 3, 5])
>>> torch.tensor([1, 2, 3, 4]) & torch.tensor([1, 1, 1, 1])
tensor([1, 0, 1, 0])
>>> torch.tensor([1, 2, 3, 4]).__and__(torch.tensor([1, 1, 1, 1]))
tensor([1, 0, 1, 0])
See https://github.com/pytorch/pytorch/pull/1556
Check this. There is no bitwise and/or operation for tensors in Torch. There are element-wise operations implemented in Torch, but not bite-wise ones.
However, if you could convert each bit as a separate Tensor dimension, you can use element-wise operation.
For an example,
a = torch.Tensor{0,1,1,0}
b = torch.Tensor{0,1,0,1}
torch.cmul(a,b):eq(1)
0
1
0
0
[torch.ByteTensor of size 4]
torch.add(a,b):ge(1)
0
1
1
1
[torch.ByteTensor of size 4]
Hope this will help you.
why is the output same every time?
a = torch.tensor([0, 1, 2, 3, 4])
a[-2:] = torch.tensor([[[5, 6]]])
a
tensor([0, 1, 2, 5, 6])
a = torch.tensor([0, 1, 2, 3, 4])
a[-2:] = torch.tensor([[5, 6]])
a
tensor([0, 1, 2, 5, 6])
a = torch.tensor([0, 1, 2, 3, 4])
a[-2:] = torch.tensor([5, 6])
a
tensor([0, 1, 2, 5, 6])
Pytorch is following Numpy here which allows assignment to slices as long as the shapes are compatible meaning that the two sides have the same shape or the right hand side is broadcastable to the shape of the slice. Starting with trailing dimensions, two arrays are broadcastable if they only differ in dimensions where one of them is 1. So in this case
a = torch.tensor([0, 1, 2, 3, 4])
b = torch.tensor([[[5, 6]]])
print(a[-2:].shape, b.shape)
>> torch.Size([2]) torch.Size([1, 1, 2])
Pytorch will perform the following comparisons:
a[-2:].shape[-1] and b.shape[-1] are equal so the last dimension is compatible
a[-2:].shape[-2] does not exist, but b.shape[-2] is 1 so they are compatible
a[-2:].shape[-3] does not exist, but b.shape[-3] is 1 so they are compatible
All dimensions are compatible, so b can be broadcasted to a
Finally, Pytorch will convert b to tensor([5, 6]) before performing the assignment thus producing the result:
a[-2:] = b
print(a)
>> tensor([0, 1, 2, 5, 6])