I'm confused as to why there are double parantheses instead of just torch.randn(1,5).
Is torch.randn(1,5) the same thing as torch.randn((1,5))?
You should check the definition of this function here.
size (int...) – a sequence of integers defining the shape of the output tensor. Can be a variable number of arguments or a collection like a list or tuple.
>>> import torch
>>> a = torch.randn(1,5)
>>> b = torch.randn((1,5))
>>> a.shape == b.shape
True
Therefore, you can use either a or b since they have the same shape.
You can use both variants: (1, 2) and 1, 2. Because of python asterisk magics:
torch.randn(*size, *, out=None, dtype=None, layout=torch.strided, device=None, requires_grad=False) → Tensor
the first *size captures all positional arguments passed to the function, when passing 1, 2 function will pack it to (1, 2).
the second * turns any parameters that follow it to be keyword-only parameters, to avoid situations like this: randn(1, 2 None, torch.strided, "cuda", True), forcing you to randn(1, 2, out=None, dtype=None, layout=torch.strided, device="cuda", requires_grad=True)
Related
In PyTorch, given a tensor of size=[3], how to expand it by several dimensions to the size=[3,2,5,5] such that the added dimensions have the corresponding values from the original tensor. For example, making size=[3] vector=[1,2,3] such that the first tensor of size [2,5,5] has values 1, the second one has all values 2, and the third one all values 3.
In addition, how to expand the vector of size [3,2] to [3,2,5,5]?
One way to do it I can think is by means of creating a vector of the same size with ones-Like and then einsum but I think there should be an easier way.
You can first unsqueeze the appropriate number of singleton dimensions, then expand to a view at the target shape with torch.Tensor.expand:
>>> x = torch.rand(3)
>>> target = [3,2,5,5]
>>> x[:, None, None, None].expand(target)
A nice workaround is to use torch.Tensor.reshape or torch.Tensor.view to do perform multiple unsqueezing:
>>> x.view(-1, 1, 1, 1).expand(target)
This allows for a more general approach to handle any arbitrary target shape:
>>> x.view(len(x), *(1,)*(len(target)-1)).expand(target)
For an even more general implementation, where x can be multi-dimensional:
>>> x = torch.rand(3, 2)
# just to make sure the target shape is valid w.r.t to x
>>> assert list(x.shape) == list(target[:x.ndim])
>>> x.view(*x.shape, *(1,)*(len(target)-x.ndim)).expand(target)
I have a tensor t of dim n x 3. When I apply torch.linalg.norm it returns one single value. What I need is a batch-wise norm function which will return a tensor with n norms, one for each vector in t.
Thanks for your help.
It seems the most relevant documentation place is:
https://pytorch.org/docs/stable/generated/torch.linalg.norm.html
In the terminal you could try: python3 and then the following python commands:
>>> from torch import linalg as LA
>>> c = torch.tensor([[1., 2., 3.],
... [-1, 1, 4]])
>>> LA.norm(c, dim=0)
tensor([1.4142, 2.2361, 5.0000])
>>> LA.norm(c, dim=1)
tensor([3.7417, 4.2426])
Conclusion:
In your specific case you will need to do:
torch.linalg.norm(t,dim=1)
I'm trying to slice a PyTorch tensor my_tensor of dimensions s x b x c so that the slicing along the first dimension varies according to a tensor indices of length b, to the effect of:
my_tensor[0:indices, torch.arange(0, b, dtype=torch.long), :] = something
The code above doesn't work and receives the error TypeError: tuple indices must be integers or slices, not tuple.
What I'm aiming for is, for example, if indices = torch.tensor([3, 5, 4]) then:
my_tensor[0:3, 0, :] = something
my_tensor[0:5, 1, :] = something
my_tensor[0:4, 2, :] = something
I'm hoping for a tensorized way to do this so I don't have to resort to a for loop. Also, the method needs to be compatible with TorchScript. Thanks very much.
This question already has answers here:
How do I add an extra column to a NumPy array?
(17 answers)
Closed 5 years ago.
l have the following vector
video_132.shape
Out[64]: (64, 3)
that l would to add to it a new 3D vector of three values
video_146[1][146][45]
such that
video_146[1][146][45].shape
Out[68]: (3,)
and
video_146[1][146][45]
Out[69]: array([217, 207, 198], dtype=uint8)
when l do the following
np.append(video_132,video_146[1][146][45])
l'm supposed to get
video_132.shape
Out[64]: (65, 3) # originally (64,3)
However l get :
Out[67]: (195,) # 64*3+3=195
It seems that it flattens the vector
How can l do the append by preserving the 3D structure ?
For visual simplicity let's rename video_132 --> a, and video_146[1][146][45] --> b. The particular values aren't important so let's say
In [82]: a = np.zeros((64, 3))
In [83]: b = np.ones((3,))
Then we can append b to a using:
In [84]: np.concatenate([a, b[None, :]]).shape
Out[84]: (65, 3)
Since np.concatenate returns a new array, reassign its return value to a to "append" b to a:
a = np.concatenate([a, b[None, :]])
Code for append:
def append(arr, values, axis=None):
arr = asanyarray(arr)
if axis is None:
if arr.ndim != 1:
arr = arr.ravel()
values = ravel(values)
axis = arr.ndim-1
return concatenate((arr, values), axis=axis)
Note how arr is raveled if no axis is provided
In [57]: np.append(np.ones((2,3)),2)
Out[57]: array([1., 1., 1., 1., 1., 1., 2.])
append is really aimed as simple cases like adding a scalar to a 1d array:
In [58]: np.append(np.arange(3),6)
Out[58]: array([0, 1, 2, 6])
Otherwise the behavior is hard to predict.
concatenate is the base operation (builtin) and takes a list, not just two. So we can collect many arrays (or lists) in one list and do one concatenate at the end of a loop. And since it doesn't tweak the dimensions before hand, it forces us to do that ourselves.
So to add a shape (3,) to a (64,3) we have transform that (3,) into (1,3). append requires the same dimension adjustment as concatenate if we specify the axis.
In [68]: np.append(arr,b[None,:], axis=0).shape
Out[68]: (65, 3)
In [69]: np.concatenate([arr,b[None,:]], axis=0).shape
Out[69]: (65, 3)
I don't understand why do we need tensor.reshape() function in Theano. It is said in the documentation:
Returns a view of this tensor that has been reshaped as in
numpy.reshape.
As far as I understood, theano.tensor.var.TensorVariable is some entity that is used for computation graphs creation. And it is absolutely independent of shapes. For instance when you create your function you can pass there matrix 2x2 or matrix 100x200. As I thought reshape somehow restricts this variety. But it is not. Suppose the following example:
X = tensor.matrix('X')
X_resh = X.reshape((3, 3))
Y = X_resh ** 2
f = theano.function([X_resh], Y)
print(f(numpy.array([[1, 2], [3, 4]])))
As I understood, it should give an error since I passed matrix 2x2 not 3x3, but it computes element-wise squares perfectly.
So what is the shape of the theano tensor variable and where should we use it?
There is an error in the provided code though Theano fails to point this out.
Instead of
f = theano.function([X_resh], Y)
you should really use
f = theano.function([X], Y)
Using the original code you are actually providing the tensor after the reshape so the reshape command never gets executed. This can be seen by adding
theano.printing.debugprint(f)
which prints
Elemwise{sqr,no_inplace} [id A] '' 0
|<TensorType(float64, matrix)> [id B]
Note that there is no reshape operation in this compiled execution graph.
If one changes the code so that X is used as the input instead of X_resh then Theano throws an error including the message
ValueError: total size of new array must be unchanged Apply node that
caused the error: Reshape{2}(X, TensorConstant{(2L,) of 3})
This is expected because one cannot reshape a tensor with shape (2, 2) (i.e. 4 elements) into a tensor with shape (3, 3) (i.e. 9 elements).
To address the broader question, we can use symbolic expressions in the target shape and those expressions can be functions of the input tensor's symbolic shape. Here's some examples:
import numpy
import theano
import theano.tensor
X = theano.tensor.matrix('X')
X_vector = X.reshape((X.shape[0] * X.shape[1],))
X_row = X.reshape((1, X.shape[0] * X.shape[1]))
X_column = X.reshape((X.shape[0] * X.shape[1], 1))
X_3d = X.reshape((-1, X.shape[0], X.shape[1]))
f = theano.function([X], [X_vector, X_row, X_column, X_3d])
for output in f(numpy.array([[1, 2], [3, 4]])):
print output.shape, output