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.
Related
As discussed here, Tensorflow's gradients are not true Jacobians -- the "gradient" of Y against X is actually just the gradient of sum(Y) against X. Similarly, Pytorch does not show Jacobians at all when the output is not a scalar.
But surely the full Jacobian is needed for computation (in the application of the chain rule).
Where, internally, is this stored? i.e. is there some object hidden within GradientTape that I can access?
I have a Tensorflow 2.0 tf.keras.Sequential model. Now, my technical specification prescribes using the Levenberg-Marquardt optimizer to fit the model. Tensorflow 2.0 doesn't provide it as an optimizer out of the box, but it is available in the Tensorflow Graphics module.
tfg.math.optimizer.levenberg_marquardt.minimize function accepts residuals ( a residual is a Python callable returning a tensor) and variables (list of tensors corresponding to my model weights) as parameters.
What would be the best way to convert my model into residuals and variables?
If I understand correctly how the minimize function works, I have to provide two residuals. The first residual must call my model for every learning case and aggregate all the results into a tensor. The second residuals must return all labels as a single constant tensor. The problem is that tf.keras.Sequential.predict function returns a numpy array instead of tensor. I believe that if I convert it to a tensor, the minimizer won't be able to calculate jacobians with respect to variables.
The same problem is with variables. It doesn't seem like there's a way to extract all weights from a model into a list of tensors.
There's a major difference between tfg.math.optimizer.levenberg_marquardt.minimize and Keras optimizers from the implementation/API perspective.
Keras optimizers, such as tf.keras.optimizers.Adam consume gradients as input and updates tf.Variables.
In contrast, tfg.math.optimizer.levenberg_marquardt.minimize essentially unrolls the optimization loop in graph mode (using a tf.while_loop construct). It takes initial parameter values and produces updated parameter values, unlike Adam & co, which only apply one iteration and actually change the values of tf.Variables via assign_add.
Stepping back a bit to the theoretical big picture, Levenberg-Marquardt is not a general gradient descent-like solver for any nonlinear optimization problem (such as Adam is). It specifically addresses nonlinear least-squares optimization, so it's not a drop-in replacement for optimizers like Adam. In gradient descent, we compute the gradient of the loss with respect to the parameters. In Levenberg-Marquardt, we compute the Jacobian of the residuals with respect to the parameters. Concretely, it repeatedly solves the linearized problem Jacobian # delta_params = residuals for delta_params using tf.linalg.lstsq (which internally uses Cholesky decomposition on the Gram matrix computed from the Jacobian) and applies delta_params as the update.
Note that this lstsq operation has cubic complexity in the number of parameters, so in case of neural nets it can only be applied for fairly small ones.
Also note that Levenberg-Marquardt is usually applied as a batch algorithm, not a minibatch algorithm like SGD, though there's nothing stopping you from applying the LM iteration on different minibatches in each iteration.
I think you may only be able to get one iteration out of tfg's LM algorithm, through something like
from tensorflow_graphics.math.optimizer.levenberg_marquardt import minimize as lm_minimize
for input_batch, target_batch in dataset:
def residual_fn(trainable_params):
# do not use trainable params, it will still be at its initial value, since we only do one iteration of Levenberg Marquardt each time.
return model(input_batch) - target_batch
new_objective_value, new_params = lm_minimize(residual_fn, model.trainable_variables, max_iter=1)
for var, new_param in zip(model.trainable_variables, new_params):
var.assign(new_param)
In contrast, I believe the following naive method will not work where we assign model parameters before computing the residuals:
from tensorflow_graphics.math.optimizer.levenberg_marquardt import minimize as lm_minimize
dataset_iterator = ...
def residual_fn(params):
input_batch, target_batch = next(dataset_iterator)
for var, param in zip(model.trainable_variables, params):
var.assign(param)
return model(input_batch) - target_batch
final_objective, final_params = lm_minimize(residual_fn, model.trainable_variables, max_iter=10000)
for var, final_param in zip(model.trainable_variables, final_params):
var.assign(final_param)
The main conceptual problem is that residual_fn's output has no gradients wrt its input params, since this dependency goes through a tf.assign. But it might even fail before that due to using constructs that are disallowed in graph mode.
Overall I believe it's best to write your own LM optimizer that works on tf.Variables, since tfg.math.optimizer.levenberg_marquardt.minimize has a very different API that is not really suited for optimizing Keras model parameters since you can't directly compute model(input, parameters) - target_value without a tf.assign.
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 free varaible (tf.variable) x, and I wish to minimize an error term with respect to subset of the tensor x (for example minimizing the error only with respect to the first row of 2D tensor).
One way is to compute the gradients and change the gradient to zero for the irrelevant parts of the tensor and apply the gradients. Is their another way?
You can use mask and tf.stop_gradient to selectively make the variable non-trainable: tf.stop_gradient(mask*x). The value in matrix mask 1 should denote parts to apply gradient and 0 otherwise.
I'm doing a Matrix Factorization in TensorFlow, I want to use coo_matrix from Spicy.sparse cause it uses less memory and it makes it easy to put all my data into my matrix for training data.
Is it possible to use coo_matrix to initialize a variable in tensorflow?
Or do I have to create a session and feed the data I got into tensorflow using sess.run() with feed_dict.
I hope that you understand my question and my problem otherwise comment and i will try to fix it.
The closest thing TensorFlow has to scipy.sparse.coo_matrix is tf.SparseTensor, which is the sparse equivalent of tf.Tensor. It will probably be easiest to feed a coo_matrix into your program.
A tf.SparseTensor is a slight generalization of COO matrices, where the tensor is represented as three dense tf.Tensor objects:
indices: An N x D matrix of tf.int64 values in which each row represents the coordinates of a non-zero value. N is the number of non-zeroes, and D is the rank of the equivalent dense tensor (2 in the case of a matrix).
values: A length-N vector of values, where element i is the value of the element whose coordinates are given on row i of indices.
dense_shape: A length-D vector of tf.int64, representing the shape of the equivalent dense tensor.
For example, you could use the following code, which uses tf.sparse_placeholder() to define a tf.SparseTensor that you can feed, and a tf.SparseTensorValue that represents the actual value being fed :
sparse_input = tf.sparse_placeholder(dtype=tf.float32, shape=[100, 100])
# ...
train_op = ...
coo_matrix = scipy.sparse.coo_matrix(...)
# Wrap `coo_matrix` in the `tf.SparseTensorValue` form that TensorFlow expects.
# SciPy stores the row and column coordinates as separate vectors, so we must
# stack and transpose them to make an indices matrix of the appropriate shape.
tf_coo_matrix = tf.SparseTensorValue(
indices=np.array([coo_matrix.rows, coo_matrix.cols]).T,
values=coo_matrix.data,
dense_shape=coo_matrix.shape)
Once you have converted your coo_matrix to a tf.SparseTensorValue, you can feed sparse_input with the tf.SparseTensorValue directly:
sess.run(train_op, feed_dict={sparse_input: tf_coo_matrix})