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
Returnsselftensor 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
Answer from Shai on Stack OverflowI 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
Returnsselftensor 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.
I've been stuck on this error for a long time now, and have gone through multiple solutions on stack overflow etc., but the problem is I believe I do have the .detach().numpy() and I still keep getting the error, so I'm not really sure what's going on. Could someone please help me out?
class plot_diagram():
# Constructor
def __init__(self, X, Y, w, stop, go = False):
start = w.data
self.error = []
self.parameter = []
self.X = X.numpy()
self.Y = Y.numpy()
self.parameter_values = torch.arange(start, stop)
self.Loss_function = [criterion(forward(X), Y) for w.data in self.parameter_values]
w.data = start
# Executor
def __call__(self, Yhat, w, error, n):
self.error.append(error)
self.parameter.append(w.data)
plt.subplot(212)
plt.plot(self.X, Yhat.detach().numpy())
plt.plot(self.X, self.Y,'ro')
plt.xlabel("A")
plt.ylim(-20, 20)
plt.subplot(211)
plt.title("Data Space (top) Estimated Line (bottom) Iteration " + str(n))
s = [p.detach().numpy() for p in self.parameter_values]
plt.plot(s, self.Loss_function)
plt.plot(self.parameter, self.error, 'ro')
plt.xlabel("B")
plt.figure()
# Destructor
def __del__(self):
plt.close('all')
X = torch.arange(-3, 3, 0.1).view(-1, 1)
Y = f + 0.1 * torch.randn(X.size())
w = torch.tensor(-10.0, requires_grad = True)
gradient_plot = plot_diagram(X, Y, w, stop = 5)
def forward(x):
return w * xdef criterion(yhat, y):
return torch.mean((yhat - y) ** 2)lr = 0.1
LOSS = []def train_model(iter):
for epoch in range (iter):
# make the prediction as we learned in the last lab
Yhat = forward(X)
# calculate the iteration
loss = criterion(Yhat,Y)
# plot the diagram for us to have a better idea
gradient_plot(Yhat, w, loss.item(), epoch)
# store the loss into list
LOSS.append(loss.item())
# backward pass: compute gradient of the loss with respect to all the learnable parameters
loss.backward()
# updata parameters
w.data = w.data - lr * w.grad.data
# zero the gradients before running the backward pass
w.grad.data.zero_()
Is it possible to convert a torch tensor to a numpy array using the GPU for faster performance? Currently, I am using
input = input.cpu().detach().numpy()
to convert the tensor to a numpy array on the CPU. Is there a way to utilize the GPU to perform this conversion instead, potentially saving time?
Error reproduced
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))
which leads to the exact same error for the line tensor1 = tensor1.numpy():
tensor([1., 2.], requires_grad=True)
<class 'torch.Tensor'>
Traceback (most recent call last):
File "/home/badScript.py", line 8, in <module>
tensor1 = tensor1.numpy()
RuntimeError: Can't call numpy() on Variable that requires grad. Use var.detach().numpy() instead.
Process finished with exit code 1
Generic solution
this was suggested to you in your error message, just replace var with your variable name
import torch
tensor1 = torch.tensor([1.0,2.0],requires_grad=True)
print(tensor1)
print(type(tensor1))
tensor1 = tensor1.detach().numpy()
print(tensor1)
print(type(tensor1))
which returns as expected
tensor([1., 2.], requires_grad=True)
<class 'torch.Tensor'>
[1. 2.]
<class 'numpy.ndarray'>
Process finished with exit code 0
Some explanation
You need to convert your tensor to another tensor that isn't requiring a gradient in addition to its actual value definition. This other tensor can be converted to a numpy array. Cf. this discuss.pytorch post. (I think, more precisely, that one needs to do that in order to get the actual tensor out of its pytorch Variable wrapper, cf. this other discuss.pytorch post).
I had the same error message but it was for drawing a scatter plot on matplotlib.
There is 2 steps I could get out of this error message :
import the
fastai.basicslibrary with :from fastai.basics import *If you only use the
torchlibrary, remember to take off therequires_gradwith :with torch.no_grad(): (your code)