gradients | CSS module for quickly setting gradients | Data Visualization library
kandi X-RAY | gradients Summary
kandi X-RAY | gradients Summary
CSS module for quickly setting gradients with single purpose classes.
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gradients Key Features
gradients Examples and Code Snippets
def _ConcatGradHelper(op, grad, start_value_index, end_value_index, dim_index):
"""Gradient for concat op.
Args:
op: An operation.
grad: `Tensor` or `IndexedSlices` representing the gradients with respect to
each output of the op.
def gradients(ys,
xs,
grad_ys=None,
name="gradients",
colocate_gradients_with_ops=False,
gate_gradients=False,
aggregation_method=None,
stop_gradients=N
def _AggregatedGrads(grads,
op,
gradient_uid,
loop_state,
aggregation_method=None):
"""Get the aggregated gradients for op.
Args:
grads: The map of memoized
Community Discussions
Trending Discussions on gradients
QUESTION
This is my code so far:
...ANSWER
Answered 2022-Apr-04 at 20:14You can put your arrows inside the left/right gradient div
s. That way they will show/hide same way as the gradients.
EDIT
I cleaned up the code a bit since the original answer was kinda messy. (or 'weird' as mstephen19 put it :)).
QUESTION
I'm learning about policy gradients and I'm having hard time understanding how does the gradient passes through a random operation. From here: It is not possible to directly backpropagate through random samples. However, there are two main methods for creating surrogate functions that can be backpropagated through
.
They have an example of the score function
:
ANSWER
Answered 2021-Nov-30 at 05:48It is indeed true that sampling is not a differentiable operation per se. However, there exist two (broad) ways to mitigate this - [1] The REINFORCE way and [2] The reparameterization way. Since your example is related to [1], I will stick my answer to REINFORCE.
What REINFORCE does is it entirely gets rid of sampling operation in the computation graph. However, the sampling operation remains outside the graph. So, your statement
.. how does the gradient passes through a random operation ..
isn't correct. It does not pass through any random operation. Let's see your example
QUESTION
Suppose I have my custom loss function and I want to fit the solution of some differential equation with help of my neural network. So in each forward pass, I am calculating the output of my neural net and then calculating the loss by taking the MSE with the expected equation to which I want to fit my perceptron.
Now my doubt is: should I use grad(loss)
or should I do loss.backward()
for backpropagation to calculate and update my gradients?
I understand that while using loss.backward() I have to wrap my tensors with Variable and have to set the requires_grad = True for the variables w.r.t which I want to take the gradient of my loss.
So my questions are :
- Does
grad(loss)
also requires any such explicit parameter to identify the variables for gradient computation? - How does it actually compute the gradients?
- Which approach is better?
- what is the main difference between the two in a practical scenario.
It would be better if you could explain the practical implications of both approaches because whenever I try to find it online I am just bombarded with a lot of stuff that isn't much relevant to my project.
...ANSWER
Answered 2021-Sep-12 at 12:57TLDR; Both are two different interfaces to perform gradient computation: torch.autograd.grad
is non-mutable while torch.autograd.backward
is.
The torch.autograd
module is the automatic differentiation package for PyTorch. As described in the documentation it only requires minimal change to code base in order to be used:
you only need to declare
Tensor
s for which gradients should be computed with therequires_grad=True
keyword.
The two main functions torch.autograd
provides for gradient computation are torch.autograd.backward
and torch.autograd.grad
:
torch.autograd.backward
(source)
torch.autograd.grad
(source)
Description
Computes the sum of gradients of given tensors with respect to graph leaves.
Computes and returns the sum of gradients of outputs with respect to the inputs.
Header
torch.autograd.backward(
tensors,
grad_tensors=None,
retain_graph=None,
create_graph=False,
grad_variables=None,
inputs=None)
torch.autograd.grad(
outputs,
inputs,
grad_outputs=None,
retain_graph=None,
create_graph=False,
only_inputs=True,
allow_unused=False)
Parameters
- tensors
– Tensors of which the derivative will be computed.-
grad_tensors
– The "vector" in the Jacobian-vector product, usually gradients w.r.t. each element of corresponding tensors.-
retain_graph
– If False
, the graph used to compute the grad will be freed. [...]-
inputs
– Inputs w.r.t. which the gradient be will be accumulated into .grad
. All other Tensors will be ignored. If not provided, the gradient is accumulated into all the leaf Tensors that were used [...].
- outputs
– outputs of the differentiated function.-
inputs
– Inputs w.r.t. which the gradient will be returned (and not accumulated into .grad
).-
grad_tensors
– The "vector" in the Jacobian-vector product, usually gradients w.r.t. each element of corresponding tensors.-
retain_graph
– If False
, the graph used to compute the grad will be freed. [...].
Usage examples
In terms of high-level usage, you can look at torch.autograd.grad
as a non-mutable function. As mentioned in the documentation table above, it will not accumulate the gradients on the grad
attribute but instead return the computed partial derivatives. In contrast torch.autograd.backward
will be able to mutate the tensors by updating the grad
attribute of leaf nodes, the function won't return any value. In other words, the latter is more suitable when computing gradients for a large number of parameters.
In the following, we will take two inputs (x1
and, x2
), calculate a tensor y
with them, and then compute the partial derivatives of the result w.r.t both inputs, i.e. dL/dx1
and dL/dx2
:
QUESTION
I've updated angular cli and created a new project, with routing and scss.
When I run npm install i see:
...ANSWER
Answered 2022-Jan-10 at 11:25I'm afraid you just have to put up with the vulnerabilities. Angular has a very strict set of dependencies, and in changing the versions of those dependencies you've broken your app.
Make sure you keep updating your Angular project as often as is feasible, as the Angular team regularly update Angular's dependencies to mitigate these issues.
QUESTION
I trained a model for sequence classification using transformers (BertForSequenceClassification) and I get the error:
Expected all tensors to be on the same device, but found at least two devices, cpu and cuda:0! (when checking argument for argument index in method wrapper__index_select)
I don't really get where is the problem, if it's on my model, on how I tokenize the data, or what.
Here is my code:
LOADING THE PRETRAINED MODEL
...ANSWER
Answered 2021-Nov-25 at 06:19You did not move your model to device
, only the data. You need to call model.to(device)
before using it with data located on device
.
QUESTION
I'm trying to create a repeatable background in CSS, using multiple gradients. However it does not really work as intended.
I got this JsFiddle to show my progress
The problem is that i don't get the diagonal lines to connect to long ones, as the vertical. How would i achieve that? The goal is to make it seamless.
Code:
...ANSWER
Answered 2021-Nov-08 at 10:51Maybe with repeating gradient:
QUESTION
I have a gradient exploding problem which I couldn't solve after trying for several days. I implemented a custom message passing graph neural network in TensorFlow which is used to predict a continuous value from graph data. Each graph is associated with one target value. Each node of a graph is represented by a node attribute vector, and the edges between nodes are represented by an edge attribute vector.
Within a message passing layer, node attributes are updated in a certain way (e.g., by aggregating other node/edge attributes), and these updated node attributes are returned.
Now, I managed to figure out where the gradient problem occurs in my code. I have the below snippet.
...ANSWER
Answered 2021-Oct-29 at 16:33Looks great, as you have already followed most of the solutions to resolve gradient exploding problem. Below is the list of all solutions you can try
Solutions to avoid Gradient Exploding problem
Appropriate Weight initialization: utilise appropriate weight Initialization based on the activation function used.
Initialization Activation Function He ReLU & variants LeCun SELU Glorot Softmax, Logistic, None, TanhRedesigning your Neural network: use fewer layers in neural network and/or use smaller batch size
Choosing Non Saturation activation function: choose the right activation function with reduced learning rates
- ReLU
- Leaky ReLU
- randomized leaky ReLU (RReLU)
- parametric leaky ReLU (PReLU)
- exponential linear unit (ELU)
Batch Normalisation: Ideally using batch normalisation before/after each layer, based on what works best for your dataset.
after each layer Paper reference
QUESTION
I am going through this tutorial on how to customize the training loop
The last example shows a GAN implemented with a custom training, where only __init__
, train_step
, and compile
methods are defined
ANSWER
Answered 2021-Sep-25 at 13:17These are different concepts and are used like this:
train_step
is called byfit
. Basically,fit
loops over the dataset and provide each batch totrain_step
(and then handles metrics, bookkeeping, etc., of course).call
is used when you, well, call the model. To be precise, writingmodel(inputs)
or in your caseself(inputs)
will use the function__call__
, but theModel
class has that function defined such that it will in turn usecall
.
Those are the technical aspects. Intuitively:
call
should define the forward-pass of your model. i.e. how is the input transformed to the output.train_step
defines the logic of a training step, usually with gradient descent. It will often make use ofcall
since the training step tends to include a forward pass of the model to compute gradients.
As for the GAN tutorial you linked, I would say that can actually be considered incomplete. It works without defining call
because the custom train_step
explicitly calls the generator/discriminator fields (as these are predefined models, they can be called as usual). If you tried to call the GAN model like gan(inputs)
, I would assume you get an error message (I did not test this). So you would always have to call gan.generator(inputs)
to generate, for example.
Finally (this part may be a bit confusing), note that you can subclass a Model
to define a custom training step, but then initialize it via the functional API (like model = Model(inputs, outputs)
), in which case you can make use of call
in the training step without ever defining it yourself because the functional API takes care of that.
QUESTION
I have a problem where I need to predict some integers from an image. The problem is that this includes some negative integers too. I have done some reasearch and came accross Poisson which does count regression, however this does not work due to me also needing to predict some negative integers too, resulting in Poisson output nan as its loss. I was thinking of using Lambda to round the output of my model however this resulted in this error:
...ANSWER
Answered 2021-Sep-17 at 08:59Add the smallest value (in this case is negative) so that everything is >= 0. Then use Poisson.
QUESTION
I am having trouble understanding the conceptual meaning of the grad_outputs
option in torch.autograd.grad
.
The documentation says:
grad_outputs
should be a sequence of length matching output containing the “vector” in Jacobian-vector product, usually the pre-computed gradients w.r.t. each of the outputs. If an output doesn’trequire_grad
, then the gradient can beNone
).
I find this description quite cryptic. What exactly do they mean by Jacobian-vector product? I know what the Jacobian is, but not sure about what product they mean here: element-wise, matrix product, something else? I can't tell from my example below.
And why is "vector" in quotes? Indeed, in the example below I get an error when grad_outputs
is a vector, but not when it is a matrix.
ANSWER
Answered 2021-Aug-23 at 23:08If we take your example we have function f
which takes as input x
shaped (n,)
and outputs y = f(x)
shaped (n, n)
. The input is described as column vector [x_i]_i for i ∈ [1, n]
, and f(x)
is defined as matrix [y_jk]_jk = [x_j*x_k]_jk for j, k ∈ [1, n]²
.
It is often useful to compute the gradient of the output with respect to the input (or sometimes w.r.t the parameters of f
, there are none here). In the more general case though, we are looking to compute dL/dx
and not just dy/dx
, where dL/dx
is the partial derivative of L
, computed from y
, w.r.t. x
.
The computation graph looks like:
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