I have a tensor:
t1 = torch.randn(564, 400)
I want to unroll it to a 1-d tensor that's 225600 long.
How can I do this?
Note the difference between view and reshape as suggested by Kris -
From reshape's docstring:
When possible, the returned tensor will be a view
of input. Otherwise, it will be a copy. Contiguous inputs and inputs with compatible strides can be reshaped without copying...
So in case your tensor is not contiguous calling reshape should handle what one would have had to handle had one used view instead; That is, call t1.contiguous().view(...) to handle non-contiguous tensors.
Also, one could use faltten: t1 = t1.flatten() as an equivalent of view(-1), which is more readable.
Pytorch is much like numpy so you can simply do,
t1 = t1.view(-1) or t1 = t1.reshape(-1)
Related
I have two 2-d tensors, which align via broadcasting, so if I add/subtract them, I incur a huge 3-d tensor. I don't really need that though, since I'll be performing a mean on one dimension. In this demo, I unsqueeze the tensors to show how they align, but they are 2-d otherwise.
x = torch.tensor(...) # (batch , 1, B)
y = torch.tensor(...) # (1, , A, B)
out = torch.cos(x - y).mean(dim=2) # (batch, B)
Possible Solutions:
An algebraic simplification, but for the life of me I haven't solved this yet.
Some PyTorch primitive that'll help? This is cosine similarity, but, a bit different than torch.cosine_similarity. I'm applying it to complex numbers' .angle()s.
Custom C/CPython code that loops efficiently.
Other?
It has been firmly established that my_tensor.detach().numpy() is the correct way to get a numpy array from a torch tensor.
I'm trying to get a better understanding of why.
In the accepted answer to the question just linked, Blupon states that:
You need to convert your tensor to another tensor that isn't requiring a gradient in addition to its actual value definition.
In the first discussion he links to, albanD states:
This is expected behavior because moving to numpy will break the graph and so no gradient will be computed.
If you don’t actually need gradients, then you can explicitly .detach() the Tensor that requires grad to get a tensor with the same content that does not require grad. This other Tensor can then be converted to a numpy array.
In the second discussion he links to, apaszke writes:
Variable's can’t be transformed to numpy, because they’re wrappers around tensors that save the operation history, and numpy doesn’t have such objects. You can retrieve a tensor held by the Variable, using the .data attribute. Then, this should work: var.data.numpy().
I have studied the internal workings of PyTorch's autodifferentiation library, and I'm still confused by these answers. Why does it break the graph to to move to numpy? Is it because any operations on the numpy array will not be tracked in the autodiff graph?
What is a Variable? How does it relate to a tensor?
I feel that a thorough high-quality Stack-Overflow answer that explains the reason for this to new users of PyTorch who don't yet understand autodifferentiation is called for here.
In particular, I think it would be helpful to illustrate the graph through a figure and show how the disconnection occurs in this example:
import torch
tensor1 = torch.tensor([1.0,2.0],requires_grad=True)
print(tensor1)
print(type(tensor1))
tensor1 = tensor1.numpy()
print(tensor1)
print(type(tensor1))
I think the most crucial point to understand here is the difference between a torch.tensor and np.ndarray:
While both objects are used to store n-dimensional matrices (aka "Tensors"), torch.tensors has an additional "layer" - which is storing the computational graph leading to the associated n-dimensional matrix.
So, if you are only interested in efficient and easy way to perform mathematical operations on matrices np.ndarray or torch.tensor can be used interchangeably.
However, torch.tensors are designed to be used in the context of gradient descent optimization, and therefore they hold not only a tensor with numeric values, but (and more importantly) the computational graph leading to these values. This computational graph is then used (using the chain rule of derivatives) to compute the derivative of the loss function w.r.t each of the independent variables used to compute the loss.
As mentioned before, np.ndarray object does not have this extra "computational graph" layer and therefore, when converting a torch.tensor to np.ndarray you must explicitly remove the computational graph of the tensor using the detach() command.
Computational Graph
From your comments it seems like this concept is a bit vague. I'll try and illustrate it with a simple example.
Consider a simple function of two (vector) variables, x and w:
x = torch.rand(4, requires_grad=True)
w = torch.rand(4, requires_grad=True)
y = x # w # inner-product of x and w
z = y ** 2 # square the inner product
If we are only interested in the value of z, we need not worry about any graphs, we simply moving forward from the inputs, x and w, to compute y and then z.
However, what would happen if we do not care so much about the value of z, but rather want to ask the question "what is w that minimizes z for a given x"?
To answer that question, we need to compute the derivative of z w.r.t w.
How can we do that?
Using the chain rule we know that dz/dw = dz/dy * dy/dw. That is, to compute the gradient of z w.r.t w we need to move backward from z back to w computing the gradient of the operation at each step as we trace back our steps from z to w. This "path" we trace back is the computational graph of z and it tells us how to compute the derivative of z w.r.t the inputs leading to z:
z.backward() # ask pytorch to trace back the computation of z
We can now inspect the gradient of z w.r.t w:
w.grad # the resulting gradient of z w.r.t w
tensor([0.8010, 1.9746, 1.5904, 1.0408])
Note that this is exactly equals to
2*y*x
tensor([0.8010, 1.9746, 1.5904, 1.0408], grad_fn=<MulBackward0>)
since dz/dy = 2*y and dy/dw = x.
Each tensor along the path stores its "contribution" to the computation:
z
tensor(1.4061, grad_fn=<PowBackward0>)
And
y
tensor(1.1858, grad_fn=<DotBackward>)
As you can see, y and z stores not only the "forward" value of <x, w> or y**2 but also the computational graph -- the grad_fn that is needed to compute the derivatives (using the chain rule) when tracing back the gradients from z (output) to w (inputs).
These grad_fn are essential components to torch.tensors and without them one cannot compute derivatives of complicated functions. However, np.ndarrays do not have this capability at all and they do not have this information.
please see this answer for more information on tracing back the derivative using backwrd() function.
Since both np.ndarray and torch.tensor has a common "layer" storing an n-d array of numbers, pytorch uses the same storage to save memory:
numpy() → numpy.ndarray
Returns self tensor as a NumPy ndarray. This tensor and the returned ndarray share the same underlying storage. Changes to self tensor will be reflected in the ndarray and vice versa.
The other direction works in the same way as well:
torch.from_numpy(ndarray) → Tensor
Creates a Tensor from a numpy.ndarray.
The returned tensor and ndarray share the same memory. Modifications to the tensor will be reflected in the ndarray and vice versa.
Thus, when creating an np.array from torch.tensor or vice versa, both object reference the same underlying storage in memory. Since np.ndarray does not store/represent the computational graph associated with the array, this graph should be explicitly removed using detach() when sharing both numpy and torch wish to reference the same tensor.
Note, that if you wish, for some reason, to use pytorch only for mathematical operations without back-propagation, you can use with torch.no_grad() context manager, in which case computational graphs are not created and torch.tensors and np.ndarrays can be used interchangeably.
with torch.no_grad():
x_t = torch.rand(3,4)
y_np = np.ones((4, 2), dtype=np.float32)
x_t # torch.from_numpy(y_np) # dot product in torch
np.dot(x_t.numpy(), y_np) # the same dot product in numpy
I asked, Why does it break the graph to to move to numpy? Is it because any operations on the numpy array will not be tracked in the autodiff graph?
Yes, the new tensor will not be connected to the old tensor through a grad_fn, and so any operations on the new tensor will not carry gradients back to the old tensor.
Writing my_tensor.detach().numpy() is simply saying, "I'm going to do some non-tracked computations based on the value of this tensor in a numpy array."
The Dive into Deep Learning (d2l) textbook has a nice section describing the detach() method, although it doesn't talk about why a detach makes sense before converting to a numpy array.
Thanks to jodag for helping to answer this question. As he said, Variables are obsolete, so we can ignore that comment.
I think the best answer I can find so far is in jodag's doc link:
To stop a tensor from tracking history, you can call .detach() to detach it from the computation history, and to prevent future computation from being tracked.
and in albanD's remarks that I quoted in the question:
If you don’t actually need gradients, then you can explicitly .detach() the Tensor that requires grad to get a tensor with the same content that does not require grad. This other Tensor can then be converted to a numpy array.
In other words, the detach method means "I don't want gradients," and it is impossible to track gradients through numpy operations (after all, that is what PyTorch tensors are for!)
This is a little showcase of a tensor -> numpy array connection:
import torch
tensor = torch.rand(2)
numpy_array = tensor.numpy()
print('Before edit:')
print(tensor)
print(numpy_array)
tensor[0] = 10
print()
print('After edit:')
print('Tensor:', tensor)
print('Numpy array:', numpy_array)
Output:
Before edit:
Tensor: tensor([0.1286, 0.4899])
Numpy array: [0.1285522 0.48987144]
After edit:
Tensor: tensor([10.0000, 0.4899])
Numpy array: [10. 0.48987144]
The value of the first element is shared by the tensor and the numpy array. Changing it to 10 in the tensor changed it in the numpy array as well.
I have a question how to efficiently apply a function which takes an m-dimensional slice of a n-dimensional array as an input.
For example, I have a n-dimensional array of shape (i,j,k,l). And on the dimensions (j,l), I want to apply the function, which gives me back a matrix of shape (j,l). The resulting numpy array should again have the shape (i,j,k,l).
For example I want to apply the following, normalisation function
def norm(arr2d):
return arr2d - np.mean(arr2d)
over the array
arrnd = np.arange(2*3*4*5).reshape(2,3,4,5) # Shape is (2,3,4,5)
on the slice (j,l).
The result I want to achieve I would get via a (slow?) Python list comprehension and moving axes.
result = np.asarray([ [ f(arrnd[:,j,:,l]) for l in range(5) ] for j in range(3)]) # Shape is (3,5,2,4)
result = np.moveaxis(np.moveaxis(result,2,0),2,3).shape # Shape is (2,3,4,5) again
Is there any better, more "numpyic" way to achieve this, without any involved loops?
I alreay looked at np.apply_along_axis() and np.apply_over_axes() but the former only works for 1-d functions, and the latter might only work, if my function is implemented as a ufunc.
The example I provided is just a toy example. The solution should work for any python function.
((If normalising a slice would be my specific problem, I could have circumenvented the python loop and moveaxis by using the ufunc's axes=(..).))
I have an array A of matrices (or a 3-dim tensor) and I want to do the following:
Denote each matrix with a number, so A is [1,2,3,4,...,], and let's say that we have a window of length 3, I want to pass as input to a TensorFlow graph the 4-dim array [[1,2,3],[2,3,4],[3,4,5],....]. What's the most efficient way of doing this? (It's a bit like a convolution with a constant kernel, but without summing over the resulting matrices).
At the moment this is what I'm doing:
input_NN = [data[t, t + window] for t in range(my_range)]
and then I pass it to a TF placeholder.
Shall I think of a better way of doing it in numpy and pass the result to a placeholder or is there a fast way of doing this in TensorFlow by passing A directly?
I have two (or sometimes more) matrixes, which I want to combine to a tensor. The matrixes e.g. have the shape (100, 400) and when they are combined, they should have the dimensions (2, 100, 400).
How do I do that? I tried it the same way I created matrixes from vectors, but that didn't work:
tensor = numpy.concatenate(list_of_matrixes, axis=0)
Probably you want
tensor = np.array(list_of_matrices)
np.array([...]) just loves to combine the inputs into a new array along a new axis. In fact it takes some effort to prevent that.:)
To use concatenate you need to add an axis to your arrays. axis=0 means 'join on the current 1st axis', so it would produce a (200,400) array.
np.concatentate([arr1[None,...], arr2[None,...], axis=0)
would do the the trick, or more generally
np.concatenate([arr[None,...] for arr in list_arr], axis=0)
If you look at the code for dstack, hstack, vstack you'll see that they do this sort of dimension adjustment before passing the task to concatenate.
The np.array solution is easy, but the concatenate solution is a good learning opportunity.