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.
To calculate the derivative of an output layer of size N w.r.t an input of size M, we need a Jacobian matrix of size M x N. To calculate a complete gradient from loss to inputs using the chain rule, we would need a large number of such Jacobians stored in memory.
I assume that tensorflow does not calculate a complete Jacobian matrix for each step of the graph, but does something more efficient. How does it do it?
Thanks
TensorFlow uses Automatic Differentiation to compute gradients efficiently. Concretely, it defines a computation graph in which nodes are operations and each directed edge represents the partial derivative of a child with respect to its parent. The total derivative of an operation f with respect to x is then given by the sum over all path values from x to f, where each path value is the product of the partial derivatives of the operations on the edges.
More specifically, TensorFlow uses reverse differentiation, which involves a forward pass to compute the value of each node in the computation graph, and a backward pass to compute the partial derivative of the function f that we are differentiating with respect to every node in the graph. We need to repeat the backward pass for each dimension of function f, so the computational complexity is O(dim(f))*O(f), where dim(f) is the output dimensionality of function f.
Although this approach is memory intensive (it requires storing the values of all the nodes before running the backward pass), it is very efficient for machine learning, where we typically have a scalar function f (i.e. dim(f)=1).
You might find this resource useful.
I have a TensorFlow tensor t with shape (d,d), a square matrix. I define the trace tensor tr = tf.trace(t). Now tr is evaluated, using session.run(tr): Is TensorFlow smart enough to only evaluate the diagonal elements of t, or are all elements of t evaluated first, and only then the trace is computed?
TensorFlow will compute the matrix first, then run the trace op to extract/sum the diagonal. Potentially this is something that XLA could optimize away if no other ops consume the full matrix (not sure if it does or not currently), but TensorFlow itself sees these ops as more or less black boxes.
If there are no consumers of the full matrix, maybe just do computations on a vector representing that diagonal? You could also use sparse tensors to avoid unnecessary computation while keeping track of indices.
I have a very large sparse 2D matrix (200k x 100k) (sparsity around 0.002), which I would need to multiply with itself (after transposition).
A = 200k x 100k
and I need to calculate:
A x A.transpose()
The problem is, that in Tensorflow it seems there is no sparse matrix to sparse matrix multiplication.
tf.sparse_tensor_dense_matmul() requires one side to be turned into a dense matrix, which would hit the memory limit.
tf.sparse_matmul() seems to actually require dense matrices.
There is no way to fit the whole matrix (A) in the GPU memory as a dense vector.
Is there a way to multiple sparse matrices with other sparse matrices in Tensorflow? I can't seem to find one.
We are trying to find a solution to implement a stochastic gradient descent (or a coordinate gradient descent) on a very very very large and very sparse least-squares problem . That is to say, for the standard least squares problem:
min ||y - Ax||2
The matrix A on the order of a billion rows and a billion columns, but it is 99% sparse . Similarly, the coefficient vector x is about 98% sparse, and the observation vector Y is also about 97% sparse.
However, although I know that tensorflow has a stochastic gradient descent function, it's unclear to me whether the gradient descent functions accept sparse matrices/vectors, and even if they do, it's unclear if these gradient descent libraries behind the scenes eventually converts the sparse matrices and vectors into dense representations which would then defeat the point of having inputted the data in sparse format to begin with. If tensorflow's gradient descent method converts everything to dense format, we would blow up the memory easily and also blow up the performance (since we would have ~ 100 * 100 more computations needed for all those zero values.)
If the SGD algorithms implicitly convert sparse matrixes and vectors to their dense counterparts, how hard would it be to change that logic to a sparse only logic? As in, could a reasonably knowledgeable python/C++ engineer (with some tensorflow knowledge) do it or is a major architectural rewrite?