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Hot answers tagged regularization

36

The function of dropout is to increase the robustness of the model and also to remove any simple dependencies between the neurons. Neurons are only removed for a single pass forward and backward through the network - meaning their weights are synthetically set to zero for that pass, and so their errors are as well, meaning that the weights are not updated. ...

22

I am unsure there will be a formal way to show which is best in which situations - simply trying out different combinations is likely best! It is worth noting that Dropout actually does a little bit more than just provide a form of regularisation, in that it is really adding robustness to the network, allowing it to try out many many different networks. ...

17

First, note that in logistic regression, using both an L1 and an L2 penalty is common enough to have its own name: ElasticNet. (Perhaps see https://stats.stackexchange.com/q/184029/232706 .) So using both isn't unprecedented. Second, XGBoost and LightGBM have quite a number of hyperparameters that overlap in their purpose. Tree complexity can be ...

12

The function you have described is a loss function. It is the function which we want to minimize in order to train our model. The loss function is called ordinary least squares. We can also see that the model that we are trying to fit is a linear function with model parameters $\beta$. This is an optimization function and regularizarion is often used in ...

8

Basically, we add a regularization term in order to prevent the coefficients to fit so perfectly to overfit. The difference between L1 and L2 is L1 is the sum of weights and L2 is just the sum of the square of weights. L1 cannot be used in gradient-based approaches since it is not-differentiable unlike L2 L1 helps perform feature selection in sparse ...

8

activity_regularizer are used to control the output of a neural network. They tend to make the output smaller. Suppose the loss function is give as : loss function = DataLoss + regularizationLoss Then for weight_regularizer, regularizationLoss = f(Weights in a network). But for activity_regularizer, regularizationLoss = f(Predicted outputs from a network). ...

8

That is generally not true, to be more accurate we can say that L1 promotes sparsity. if a weight is larger than 1 then L2 cares more about it than L1 while if a weight is less than 1 then L1 cares more about it than L2. For a quick example imagine two weights, $w_1 = 15$ and $w_2 = 0.02$, let's imagine that the model considers reducing both of those ...

7

There are not any strong, well-documented principles to help you decide between types of regularisation in neural networks. You can even combine regularisation techniques, you don't have to choose just one. A workable approach can be based on experience, and following literature and other people's results to see what gave good results in different problem ...

7

As your network is working without dropout, I think your problem is about how many epoches you run. In your code, it seems that only one epoch will be run. With dropout enabled, each neuron has 50% percent (for example) chance to be activated. Maybe there are some un-trained neurons in your network, which ruin your accuracy. I think it is worth trying more ...

7

These techniques are not mutually exclusive; combining dropout with weight decay has become pretty standard for deep learning. However, where weight decay applies a linear penalty, dropout can cause the penalty to grow exponentially. This property of dropout can lead to hypothetical failures as proposed and proven in section 4.2 of this paper. In general, ...

6

Wikipedia lists some well-known approaches to hyper-parameter searches. The brute-force scan/search, or a grid search across multiple parameters, is still a very common and workable approach. As is random search, just trying some variations of parameters automatically and picking the best result from cross validation. A guided search by intuitive feel of ...

6

Normally you use regularization. The exception is if you know the data generating process and can model it exactly. Then you merely estimate the model parameters. In general you will not know the process, so you will have to approximate with a flexible enough model. If the model is not flexible enough you will not need to regularize but you won't approximate ...

6

What can I understand/interpret from the training & test loss graph? This checking out the quality of the model. If the train and test set loss decreasing according to the number of epoch in the same way(i.e plots should overlap each other), that means that model is good. Otherwise, we have a problem. In your graph plots fastly separates(from x-value ...

6

First I did a visualization of what your code is doing (see code at the bottom) The model seems completely fine. The coefficient of the linear regression is close to 0, where it should be given how you have created the data. You are misunderstanding regularized_logistic_regression_model.scores_ {1: array([[0.47058824, 0.47058824, 0.47058824, 0....

6

This is a very general question, however, there are many different solutions as explained below. For your case, probably, item 2 is not the case because you can not gather a large number of data points. I would recommend using solutions 1, 3, 5, and 6 (I see you used this method but try to combine it with other solutions such as cross-validation, ...

5

L2 and L1 regularization are controlled via the lambda (=reg_lambda) and alpha (=reg_alpha) parameter respectively. Higher values of alpha mean more L1 regularization. See the documentation here.

5

That's correct. Without regularization your model would fit to an irrelevant noise present in your dataset. It means that training set will fit better but the overall predictive power will decrease. This is a good article about overfitting and regularization. Also you can get some intuition by watching these images: Without regularization With ...

5

In this medium post, you can find a concise and very clear explanation regarding these parameters https://medium.com/@gabrieltseng/gradient-boosting-and-xgboost-c306c1bcfaf5 Gabriel Tseng, Author of the blogpost: "These two regularization terms have different effects on the weights; L2 regularization (controlled by the lambda term) encourages the weights to ...

5

Regularization does decrease the capacity of the model in some sense, but as you already guessed, different capacity reductions result in models of different quality and are not interchangeable. L1 can be interpreted as making the assumption that the influence of different factors (represented by neurons) on each other shouldn’t be assumed without ...

5

The way most people gain an initial understanding of label smoothing (and what most common explanations have to say on the subject) plays a great role in how one would approach this question. At first glance, label smoothing is exactly what the name suggests: we modify the labels or some portion of them in order to get a better, more general, more robust ...

5

The convergence time is sensitive to the data you have and a random seed. Specifically, the convergence time is linear in expectation in all three cases. SGDClassifier uses the stochastic gradient descent for optimization. Since L1 loss is only subdifferential, the L1 penalty causes the algorithm to converge noticeably slower. Comparing with or without the ...

4

Yes, that is correct. For example, think of a polynomial $a_n x^n + a_{n-1} x^{n-1} + \dots + a_2 x^2 + a_1 x^1 + a_0 x^0$ which should fit 100 data points $(x_, y_i)$ where all $y_i$ were generated by one polynomial with some noise. Of course, you could always perfectly fit the model to the data making the MSE error MSE = \sum_{(x_i, y_i)} (y_i - \text{...

4

The most common sparse regularizer is sum of absolute values (so-called Lasso regression). With carefully chosen penalty coefficient, it makes some of less useful parameters exactly zero. Cardinality penalty exactly imposes sparsity, but it cannot be combined with gradient descent, and usually requires combinatorial optimization. Simply put, it is slow to ...

4

The question-asker resorted to scikit-learn until now, but statsmodels has just come out with its implementation of Lasso regression. The docs here are pretty self-explanatory and concise.

4

Another way of looking at what dropout does is that it is like a slab-and-spike prior for the coefficient for a covariate (that is some complex interaction term of the original covariates with some complicated functional transformations) in a Bayesian model. This is the interpretation proposed by Yarin Gal in his thesis (see his list of publications). Here ...

4

When p > n, the LASSO model can only sustain up to n variables (this can be proven using linear algebra, the rank of the data matrix in particular), leaving at least p - n variables out (some that might be predictive, consider a model where you use LASSO and all your variables are predictive). This is not necessairly a bad thing though, depending on your use ...

4

Unlike lasso, ridge does not have zeroing coefficients as a goal, and you shouldn't expect applying ridge penalty to have this effect. So the answer to your title question is "no." However, in your question body, you ask whether it is possible for the ridge penalty to produce a zero coefficient that was nonzero in an unpenalized solution. The answer here ...

4

Assuming your x axis is nrounds(Or ntrees) parameter, XGB is an ensemble of many many trees built on top of one another. Your XAxis indicates how many trees have been used. Consider 2 points at x = 100 and x= 200, When you had 100 trees the train and test loss were close to .15 and 0.26, but on building 100 more trees on top of this train loss reduced to 0....

3

The underlying true performance is likely convex or at least likely only has one minimum, but you don't know the true underlying performance, you only get a stochastic sample with a tiny sample size because you apply cross validation. The performance is a random variable where you are interested in the expected value, but you only get k samples where k is ...

3

This is because you have very little amount of data. If not enough data is provided to CNNs, they are very likely to overfit. You can do the following things in order to overcome this problem: Data Augmentation : It is a technique to create new examples from the training examples by doing some preprocessing on them e.g. Rotation, Scaling, etc. Rigorous ...

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