I am analyzing medical data for a hospital study and if I am using a random forest with only one tree then the cross validation scores are quite bad (indicating overfitting) whereas if I am using a decision tree the score values are actually quiet good. Both classifier have the same depth parameter. So how can this behaviour be explained?
The construction procedure for decision trees usually includes pruning, which is a part that is done a posteriori in order to reduce the depth and avoid overfitting. Random Forest does not use this method, as it actually takes advantage of the high variance of the overfitted decision trees by averaging them.
Moreover, the decision tree would be built by training on the full dataset, while the "random forest" tree would be build on a bootstrap of the training dataset, which could likely translate into poorer performance since it will be biased towards records that have been included multiple times in the sampling. Again, Random Forest takes advantage of this by averaging over multiple trees, but here it's a disadvantage.
All and all, the difference in performance is not surprising.
Related
Does the accuracy necessarily increase when we increase the number of estimators in the Random Forest Classifier?
Decission trees tend to overfit on the training data (they have large variance on out of sample data). Random forests were designed to overcome this difficulty by creating an ensemble of independent decision trees, which all have large individual variances, but when the voting process takes place, they tend to perform better on unseen data.
It is impossible to generalize the relationship between accuracy and n_estimators given the info you gave us, but you can expect it to be positive and increasing at marginally decreasing rates. That is, the more trees the better, but at some point the effect of adding more trees will become negligible.
You can try using GridSearchCV with param_grid={'n_estimators':range(1, 101)} to see the effect of adding more estimators to a random forest.
Finally, you should probably ask this kind of question on this forum. Take a look at #desertnaut's comment on your post.
I am learning the Random Forest Regression Model. I know that it forms many Trees(models) and then we can predict our target variables by averaging the result of all Trees. I also have a descent understanding of Decision Tree Regression Algorithm. How can we form the best number of Trees?
For example i have a dataset where i am predicting person salary and i have only two input variables that are 'Years of Experience', 'Performance Score ' then how many random Trees can i form using such dataset? Are Random Forest Trees dependent upon the number of input variables? Any Good Example will highly be appreciated..
Thanks in Advance
A decision tree trains the model on the entire dataset and only one model is created. In random forest, multiple decision trees are created and each decision tree is trained on a subset of data by limiting the number of rows and the features. In your case, you only have two features so the model will create and train data on subset of data.
You can create any number of random trees for your data. Usually in random forest, more trees result in better performance but also more computation time. Experiment with your data and see the performance changes between different number of trees. If performance remains same, then use less trees to have faster computation. You can use grid search for this.
Also you can experiment with other ml models like linear regression, which migh† perform well in your case.
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 data，and 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.
This is a question about linear regression with ngrams, using Tf-IDF (term frequency - inverse document frequency). To do this, I am using numpy sparse matrices and sklearn for linear regression.
I have 53 cases and over 6000 features when using unigrams. The predictions are based on cross validation using LeaveOneOut.
When I create a tf-idf sparse matrix of only unigram scores, I get slightly better predictions than when I create a tf-idf sparse matrix of unigram+bigram scores. The more columns I add to the matrix (columns for trigram, quadgram, quintgrams, etc.), the less accurate the regression prediction.
Is this common? How is this possible? I would have thought that the more features, the better.
It's not common for bigrams to perform worse than unigrams, but there are situations where it may happen. In particular, adding extra features may lead to overfitting. Tf-idf is unlikely to alleviate this, as longer n-grams will be rarer, leading to higher idf values.
I'm not sure what kind of variable you're trying to predict, and I've never done regression on text, but here's some comparable results from literature to get you thinking:
In random text generation with small (but non-trivial) training sets, 7-grams tend to reconstruct the input text almost verbatim, i.e. cause complete overfit, while trigrams are more likely to generate "new" but still somewhat grammatical/recognizable text (see Jurafsky & Martin; can't remember which chapter and I don't have my copy handy).
In classification-style NLP tasks performed with kernel machines, quadratic kernels tend to fare better than cubic ones because the latter often overfit on the training set. Note that unigram+bigram features can be thought of as a subset of the quadratic kernel's feature space, and {1,2,3}-grams of that of the cubic kernel.
Exactly what is happening depends on your training set; it might simply be too small.
As larsmans said, adding more variables / features makes it easier for the model to overfit hence lose in test accuracy. In the master branch of scikit-learn there is now a min_df parameter to cut-off any feature with less than that number of occurrences. Hence min_df==2 to min_df==5 might help you get rid of spurious bi-grams.
Alternatively you can use L1 or L1 + L2 penalized linear regression (or classification) using either the following classes:
sklearn.linear_model.Lasso (regression)
sklearn.linear_model.ElasticNet (regression)
sklearn.linear_model.SGDRegressor (regression) with penalty == 'elastic_net' or 'l1'
sklearn.linear_model.SGDClassifier (classification) with penalty == 'elastic_net' or 'l1'
This will make it possible to ignore spurious features and lead to a sparse model with many zero weights for noisy features. Grid Searching the regularization parameters will be very important though.
You can also try univariate feature selection such as done the text classification example of scikit-learn (check the SelectKBest and chi2 utilities.
I should decide between SVM and neural networks for some image processing application. The classifier must be fast enough for near-real-time application and accuracy is important too. Since this is a medical application, it is important that the classifier has the low failure rate.
which one is better choice?
A couple of provisos:
performance of a ML classifier can refer to either (i) performance of the classifier itself; or (ii) performance of the predicate step: execution speed of the model-building algorithm. Particularly in this case, the answer is quite different depending on which of the two is intended in the OP, so i'll answer each separately.
second, by Neural Network, i'll assume you're referring to the most common implementation--i.e., a feed-forward, back-propagating single-hidden-layer perceptron.
Training Time (execution speed of the model builder)
For SVM compared to NN: SVMs are much slower. There is a straightforward reason for this: SVM training requires solving the associated Lagrangian dual (rather than primal) problem. This is a quadratic optimization problem in which the number of variables is very large--i.e., equal to the number of training instances (the 'length' of your data matrix).
In practice, two factors, if present in your scenario, could nullify this advantage:
NN training is trivial to parallelize (via map reduce); parallelizing SVM training is not trivial, but it's also not impossible--within the past eight or so years, several implementations have been published and proven to work (https://bibliographie.uni-tuebingen.de/xmlui/bitstream/handle/10900/49015/pdf/tech_21.pdf)
mult-class classification problem SVMs are two-class classifiers.They can be adapted for multi-class problems, but this is never straightforward because SVMs use direct decision functions. (An excellent source for modifying SVMs to multi-class problems is S. Abe, Support Vector Machines for Pattern Classification, Springer, 2005). This modification could wipe out any performance advantage SVMs have over NNs: So for instance, if your data has
more than two classes and you chose to configure the SVM using
successive classificstaion (aka one-against-many classification) in
which data is fed to a first SVM classifier which classifiers the
data point either class I or other; if the class is other then
the data point is fed to a second classifier which classifies it
class II or other, etc.
Prediction Performance (execution speed of the model)
Performance of an SVM is substantially higher compared to NN. For a three-layer (one hidden-layer) NN, prediction requires successive multiplication of an input vector by two 2D matrices (the weight matrices). For SVM, classification involves determining on which side of the decision boundary a given point lies, in other words a cosine product.
Prediction Accuracy
By "failure rate" i assume you mean error rate rather than failure of the classifier in production use. If the latter, then there is very little if any difference between SVM and NN--both models are generally numerically stable.
Comparing prediction accuracy of the two models, and assuming both are competently configured and trained, the SVM will outperform the NN.
The superior resolution of SVM versus NN is well documented in the scientific literature. It is true that such a comparison depends on the data, the configuration, and parameter choice of the two models. In fact, this comparison has been so widely studied--over perhaps all conceivable parameter space--and the results so consistent, that even the existence of a few exceptions (though i'm not aware of any) under impractical circumstances shouldn't interfere with the conclusion that SVMs outperform NNs.
Why does SVM outperform NN?
These two models are based on fundamentally different learing strategies.
In NN, network weights (the NN's fitting parameters, adjusted during training) are adjusted such that the sum-of-square error between the network output and the actual value (target) is minimized.
Training an SVM, by contrast, means an explicit determination of the decision boundaries directly from the training data. This is of course required as the predicate step to the optimization problem required to build an SVM model: minimizing the aggregate distance between the maximum-margin hyperplane and the support vectors.
In practice though it is harder to configure the algorithm to train an SVM. The reason is due to the large (compared to NN) number of parameters required for configuration:
choice of kernel
selection of kernel parameters
selection of the value of the margin parameter