I am trying to use machine learning to predict a dataset. It is a regression problem with 180 input features and 1 continuously-valued output. I try to compare deep neural networks, random forest regression, and linear regression.
As I expect, 3-hidden-layer deep neural networks outperform other two approaches with a root mean square error (RMSE) of 0.1. However, I unexpected to see that random forest even performs worse than linear regression (RMSE 0.29 vs. 0.27). In my expectation, the random forest can discover more complex dependencies between features to decrease error. I have tried to tune the parameters of random forest (number of trees, maximum features, max_depth, etc.). I also tried different K-cross validation, but the performance is still less than linear regression.
I searched online, and one answer says linear regression may perform better if features have a smooth, nearly linear dependence on the covariates. I do not fully get the point because if that is the case, should not deep neural networks give much performance gain?
I am struggling to give an explanation. Under what situation, random forest is worse than linear regression, but deep neural networks can perform much better?
If your features explain linear relation to the target variable then a Linear Model usually performs well than a Random Forest Model. It totally depends on the linear relations between your features.
That said, Linear models are not superior or the Random Forest is any inferior one.
Try scaling and transforming the data using MinMaxScaler() from scikit-learn to see if the linear model improves further
Pro Tips
If linear model is working like a charm you need to ask your self Why? and How? And get into the basics of both the models to understand why it worked on your data. These questions will lead you to feature engineer better. And as a matter of fact, Kaggle Grand Masters do use Linear Models in stacking to get that top 1% score by capturing the linear relations in the dataset.
So at the end of the day, linear models could wonders too.
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I am trying to solve a regression problem by predicting a continuous value using machine learning. I have a dataset which composed of 6 float columns.
The data come from low price sensors, this explain that very likely we will have values that can be considered out of the ordinary. To fix the problem, and before predicting my continuous target, I will predict data anomalies, and use him as a data filter, but the data that I have is not labeled, that's mean I have unsupervised anomaly detection problem.
The algorithms used for this task are Local Outlier Factor, One Class SVM, Isolation Forest, Elliptic Envelope and DBSCAN.
After fitting those algorithms, it is necessary to evaluate them to choose the best one.
Can anyone have an idea how to evaluate an unsupervised algorithm for anomaly detection ?
The only way is to generate synthetic anomalies which mean to introduce outliers by yourself with the knowledge of how a typical outlier will look like.
According to my understanding, RF selects features randomly and hence is hard to overfit. But, in sklearn Gradient boosting also offers the option of max_features which can help to prevent overfitting. So, why would anyone use Random forest?
Can anyone explain when to use Gradient boosting vs Random forest based on the given data?
Any help is highly appreciated.
According to my personal experience, Random Forest could be a better choice when..
You train a model on small data set.
Your data set has few features to learn.
Your data set has low Y flag count or you try to predict a situation that has low chance to occur or rarely occurs.
In these situations, Gradient Boosting algorithms like XGBoost and Light GBM can overfit (though their parameters are tuned) while simple algorithms like Random Forest or even Logistic Regression may perform better. To illustrate, for XGboost and Ligh GBM, ROC AUC from test set may be higher in comparison with Random Forest but shows too high difference with ROC AUC from train set.
Despite the sharp prediction form Gradient Boosting algorithms, in some cases, Random Forest take advantage of model stability from begging methodology (selecting randomly) and outperform XGBoost and Light GBM. However, Gradient Boosting algorithms perform better in general situations.
Similar question asked on Quora:
https://www.quora.com/How-do-random-forests-and-boosted-decision-trees-compare
I agree with the author at the link that random forests are more robust -- they don't require much problem-specific tuning to get good results. Besides that, a couple other items based on my own experience:
Random forests can perform better on small data sets; gradient boosted trees are data hungry
Random forests are easier to explain and understand. This perhaps seems silly but can lead to better adoption of a model if needed to be used by less technical people
I think that's also true. I have also read on this page How Random Forest Works
There explains the advantages of random forest. like this :
For applications in classification problems, Random Forest algorithm
will avoid the overfitting problem
For both classification and
regression task, the same random forest algorithm can be used
The Random Forest algorithm can be used for identifying the most
important features from the training dataset, in other words,
feature engineering.
My colleague and I are trying to wrap our heads around the difference between logistic regression and an SVM. Clearly they are optimizing different objective functions. Is an SVM as simple as saying it's a discriminative classifier that simply optimizes the hinge loss? Or is it more complex than that? How do the support vectors come into play? What about the slack variables? Why can't you have deep SVM's the way you can't you have a deep neural network with sigmoid activation functions?
I will answer one thing at at time
Is an SVM as simple as saying it's a discriminative classifier that simply optimizes the hinge loss?
SVM is simply a linear classifier, optimizing hinge loss with L2 regularization.
Or is it more complex than that?
No, it is "just" that, however there are different ways of looking at this model leading to complex, interesting conclusions. In particular, this specific choice of loss function leads to extremely efficient kernelization, which is not true for log loss (logistic regression) nor mse (linear regression). Furthermore you can show very important theoretical properties, such as those related to Vapnik-Chervonenkis dimension reduction leading to smaller chance of overfitting.
Intuitively look at these three common losses:
hinge: max(0, 1-py)
log: y log p
mse: (p-y)^2
Only the first one has the property that once something is classified correctly - it has 0 penalty. All the remaining ones still penalize your linear model even if it classifies samples correctly. Why? Because they are more related to regression than classification they want a perfect prediction, not just correct.
How do the support vectors come into play?
Support vectors are simply samples placed near the decision boundary (losely speaking). For linear case it does not change much, but as most of the power of SVM lies in its kernelization - there SVs are extremely important. Once you introduce kernel, due to hinge loss, SVM solution can be obtained efficiently, and support vectors are the only samples remembered from the training set, thus building a non-linear decision boundary with the subset of the training data.
What about the slack variables?
This is just another definition of the hinge loss, more usefull when you want to kernelize the solution and show the convexivity.
Why can't you have deep SVM's the way you can't you have a deep neural network with sigmoid activation functions?
You can, however as SVM is not a probabilistic model, its training might be a bit tricky. Furthermore whole strength of SVM comes from efficiency and global solution, both would be lost once you create a deep network. However there are such models, in particular SVM (with squared hinge loss) is nowadays often choice for the topmost layer of deep networks - thus the whole optimization is actually a deep SVM. Adding more layers in between has nothing to do with SVM or other cost - they are defined completely by their activations, and you can for example use RBF activation function, simply it has been shown numerous times that it leads to weak models (to local features are detected).
To sum up:
there are deep SVMs, simply this is a typical deep neural network with SVM layer on top.
there is no such thing as putting SVM layer "in the middle", as the training criterion is actually only applied to the output of the network.
using of "typical" SVM kernels as activation functions is not popular in deep networks due to their locality (as opposed to very global relu or sigmoid)
Most classification algorithms are developed to improve the training speed. However, is there any classifier or algorithm focusing on the decision making speed(low computation complexity and simple realizable structure)? I can get enough training dataļ¼and endure the long training time.
There are many methods which classify fast, you could more or less sort models by classification speed in a following way (first ones - the fastest, last- slowest)
Decision Tree (especially with limited depth)
Linear models (linear regression, logistic regression, linear svm, lda, ...) and Naive Bayes
Non-linear models based on explicit data transformation (Nystroem kernel approximation, RVFL, RBFNN, EEM), Kernel methods (such as kernel SVM) and shallow neural networks
Random Forest and other committees
Big Neural Networks (ie. CNN)
KNN with arbitrary distance
Obviously this list is not exhaustive, it just shows some general ideas.
One way of obtaining such model is to build a complex, slow model, then use it as a black box label generator to train a simplier model (but on potentialy infinite training set) - thus getting a fast classifier at the cost of very expensive training. There are many works showing that one can do that for example by training a shallow neural network on outputs of deep nn.
In general classification speed should not be a problem. Some exceptions are algorithms which have a time complexity depending on the number of samples you have for training. One example is k-Nearest-Neighbors which has no training time, but for classification it needs to check all points (if implemented in a naive way). Other examples are all classifiers which work with kernels since they compute the kernel between the current sample and all training samples.
Many classifiers work with a scalar product of the features and a learned coefficient vector. These should be fast enough in almost all cases. Examples are: Logistic regression, linear SVM, perceptrons and many more. See #lejlot's answer for a nice list.
If these are still too slow you might try to reduce the dimension of your feature space first and then try again (this also speeds up training time).
Btw, this question might not be suited for StackOverflow as it is quite broad and recommendation instead of problem oriented. Maybe try https://stats.stackexchange.com/ next time.
I have a decision tree which is represented in the compressed form and which is at least 4 times faster than the actual tree in classifying an unseen instance.
I have been trying to figure out the correlation between the error rate and the number of features in both of these models. I watched some videos, and the creator of the video said that a simple model can be better than a complicated model. So I figured that the more features I had the greater the error rate would be. This did not prove to be true in my work, and when I had less features the error rate went up. I'm not sure if I'm doing this incorrectly, or if the guy in the video made a mistake. Can someone care to explain? I also am curious how features relate to Logistic Regression's error rate as well.
Naive Bayes and Logistic Regression are a "generative-discriminative pair," meaning they have the same model form (a linear classifier), but they estimate parameters in different ways.
For feature x and label y, naive Bayes estimates a joint probability p(x,y) = p(y)*p(x|y) from the training data (that is, builds a model that could "generate" the data), and uses Bayes Rule to predict p(y|x) for new test instances. On the other hand, logistic regression estimates p(y|x) directly from the training data by minimizing an error function (which is more "discrimative").
These differences have implications for error rate:
When there are very few training instances, logistic regression might "overfit," because there isn't enough data to estimate p(y|x) reliably. Naive Bayes might do better because it models the entire joint distribution.
When the feature set is large (and sparse, like word features in text classification) naive Bayes might "double count" features that are correlated with each other, because it assumes that each p(x|y) event is independent, when they are not. Logistic regression can do a better job by naturally "splitting the difference" among these correlated features.
If the features really are (mostly) conditionally independent, both models might actually improve with more and more features, provided there are enough data instances. The problem comes when the training set size is small relative to the number of features. Priors on naive Bayes feature parameters, or regularization methods (like L1/Lasso or L2/Ridge) on logistic regression can help in these cases.