I created training data of R&R from ground truth and noticed that each question of ground truth made 10 records of training data without depending on the number of candidate answers of ground truth.
Only the number of questions of ground truth affects the size of R&R training data? I would like to know it because there is size limitation of training data.
noticed that each question of ground truth made 10 records of training data without depending on the number of candidate answers of ground truth
If you are using the python train.py utility to prepare the training data for R&R, the number of candidate answers per question is controlled by the optional -r (--rows) argument which specifies the number of answer results that the query returns. The default value is 10, which is what you are seeing.
Similarly, if you are directly using the /fcselect API call to generate the training data, then you can similarly use the optional rows parameter to specify the number of candidate answers for which features are generated. Again, the default is 10.
If you can afford to do so, it is generally better to override this default and experiment with higher values as that provides the ranker with more room to learn and re-rank answers. The RnR web tooling uses a default of 30.
Only the number of questions of ground truth affects the size of R&R training data?
No, the size of the training data is proportional to all aspects: (1) the number of queries, (2) the number of candidate answers per query, and (3) the number of features (columns). The number of features is, in itself, proportional to the number of fields in the schema that are marked for feature generation (i.e. in the default schema, they are marked with type watson_text_en).
Related
I know the general rule that we should test a trained classifier only on the testing set.
But now comes the question: When I have an already trained and tested classifier ready, can I apply it to the same dataset that was the base of the training and testing set? Or do I have to apply it to a new predicting set that is different from the training+testing set?
And what if I predict a label column of a time series (edited later: I do not mean to create a classical time series analysis here, but just a broad selection of columns from a typical database, weekly, monthly or randomly stored data that I convert into separate feature columns, each for one week / month / year ...), do I have to shift all of the features (not just the past columns of the time series label column, but also all other normal features) of the training+testing set back to a point in time where the data has no "knowledge" interception with the predicting set?
I would then train and test the classifier on features shifted to the past by n months, scoring against a label column that is unshifted and most recent, and then predicting from most recent, unshifted features. Shifted and unshifted features have the same number of columns, I align shifted and unshifted features by assigning the column names of the shifted features to the unshifted features.
p.s.:
p.s.1: The general approach on https://en.wikipedia.org/wiki/Dependent_and_independent_variables
In data mining tools (for multivariate statistics and machine learning), the dependent variable is assigned a role as target variable (or in some tools as label attribute), while an independent variable may be assigned a role as regular variable.[8] Known values for the target variable are provided for the training data set and test data set, but should be predicted for other data.
p.s.2: In this basic tutorial we can see that the predicting set is made different: https://scikit-learn.org/stable/tutorial/basic/tutorial.html
We select the training set with the [:-1] Python syntax, which produces a new array that contains all > but the last item from digits.data: […] Now you can predict new values. In this case, you’ll predict using the last image from digits.data [-1:]. By predicting, you’ll determine the image from the training set that best matches the last image.
I think you are mixing up some concepts, so I will try to give a general explanation for Supervised Learning.
The training set is what your algorithm LEARNS on. You split it in X (features) and Y (target variable).
The test set is a set that you use to SCORE your model, and it must contain data that was not in the training set. This means that a test set also has X and Y (meaning that you know the value of the target). What happens is that you PREDICT f(Y) based on X, and compare it with the Y you have, and see how good your predictions are
A prediction set is simply new data! This means that usually you DO NOT have a target, since the whole point of supervised learning is predicting it. You will only have your X (features) and you will predict f(X) (your estimate of the target Y) and use it for whatever you need.
So, in the end a test set is simply a prediction set for which you have a target to compare your estimation to.
For time series, it is a bit more complicated, because often the features (X) are transformations on past data of the target variable (Y). For example, if you want to predict today's SP500 price, you might want to use the average of the last 30 days as a feature. This means that for every new day, you need to recompute this feature over the past days.
In general though, I would suggest starting with NON time series data if you're new to ML, as Time Series is much harder in terms of feature engineering and data management and it is easy to make mistakes.
The question above When I have an already trained and tested classifier ready, can I apply it to the same dataset that was the base of the training and testing set? has the simple answer: No.
The question above Do I have to shift all of the features has the simple answer: Yes.
In short, if I predict a month's class column: I have to shift all of the non-class columns also back in time in addition to the previous class months I converted to features, all data must have been known before the month in that the class is predicted.
This also means: the predicting set has to be different from the dataset that contains the testing set. If you included the testing set, the training set loses valuable up-to-date data of the latest month(s) available! The term of a final "predicting set" is meant to be the "most current input to be used without a testing set" to get the "most current results" for the prediction.
This is confirmed by the following overview offered by this user who seems to have made the image, using days instead of months here, but the idea is the same:
Source: Answer on "Cross Validated" - Splitting Time Series Data into Train/Test/Validation Sets; the whole Q/A is recommended (!).
See the last line of the image and the valuable comments of that answer on "Cross Validated" to understand this.
230106:
The image shows that the last step is a training on the whole dataset, this is the "predicting set" that is the newest and that does not have a testing set.
On that image, there is one "mistake" which shows that this seemingly easy question of taking former labels as features for upcoming labels seems to be hard to be understood. I myself did not see this and posted the image without this remark: The "T&V" is in the past of the "Test". And that would be a wrong validation for a model that shall predict the future, the V must be in the "future" test block (unless you have a dataset that is not dynamically changing over time, like in physics).
You would have to change it to a "walk-forward" model, with the validation set - if at all - split k-fold from the testing set, not from the training set. That would look like this:
See also:
Can / should I use past (e.g. monthly) label columns from a database as features in an ML prediction (no time-series!)? with the "walk-forward" main image,
Splitting Time Series Data into Train/Test/Validation Sets with more insight into this and the comment that brought up the model name "walk-forward".
I used the caret package to train a random forest, including repeated cross-validation. I’d like to know whether the OOB, as in the original RF by Breiman, is used or whether this is replaced by the cross-validation. If it is replaced, do I have the same advantages as described in Breiman 2001, like increased accuracy by reducing the correlation between input data? As OOB is drawn with replacement and CV is drawn without replacement, are both procedures comparable? What is the OOB estimate of error rate (based on CV)?
How are the trees grown? Is CART used?
As this is my first thread, please let me know if you need more details. Many thanks in advance.
There are a lot of basic questions here and you would be better served by reading a book on machine learning or predictive modeling. Thats probably why you haven't gotten much of a response.
For caret you should also consult the package website where some of these questions are answered.
Here are some notes:
CV and OOB estimation for RF are somewhat different. This post might help explain how. For this application, the OOB rate from random forest is computed while the model is being build whereas CV uses holdout samples that are predicted after the random forest model is computed.
The original random forest model (used here) uses unpruned CART trees. Again, this is in many text books and papers.
Max
I recently got a little confused with this too, but reading chapter 4 in Applied Predictive Modeling by Max Kuhn helped me to understand the difference.
If you use randomForest in R, you grow a number of decision trees by sampling N cases with replacement (N is the number of cases in the training set). You then sample m variables at each node where m is less than the number of predictors. Each tree is then grown fully and terminal nodes are assigned to a class based on the mode of cases in that node. New cases are classified by sending them down all the trees and then taking a vote; the majority vote wins.
The key points to note here are:
how the trees are grown - sampling WITH replacement (a bootstrap). This means that some cases will be represented many times in your bootstrap sample and others may not be represented at all. The bootstrap sample will be the same size as your training dataset.
The cases that are not selected for building trees are referred to as the OOB samples- an OOB error estimate is calculated by classifying the cases that aren't selected when building a tree. About 63% of the data points in the bootstrap sample are represented at least once.
If you use caret in R, you will normally use caret::train(....) and specify the method as "rf" and trControl="repeatedcv". You can change trControl to "oob" if you want out of the bag. The way this works is as follows (I'm going to use a simple example of a 10 fold cv repeated 5 times): the training dataset is split into 10 folds of roughly equal size, a number of trees will be built using only 9 samples - so omitting the 1st fold (which is held out). The held out sample is predicted by running the cases through the trees and used to estimate performance measures. The first subset is returned to the training set and the procedure repeats with the 2nd subset held out, and so on. The process is repeated 10 times. This whole procedure can be repeated multiple times (in my example, I do this 5 times); for each of the 5 runs, the training dataset with be split into 10 slightly different folds. It should be noted that 50 different held out samples are used to calculate model efficacy.
The key points to note are:
this involves sampling WITHOUT replacement - you split the training data and build a model on 9 samples and predict the held out sample (the remaining 1 sample of the 10) and repeat this process as above
the model is built using a dataset that is smaller than the training dataset; this is different to the bootstrap method discussed above
You are using 2 different resampling techniques which will yield different results therefore they are not comparable. The k fold repeated cv tends to have low bias (for k large); where k is 2 or 3, bias is high and comparable to the bootstrap method. K fold cv tends to have high variance though...
Is it possible to apply RandomForests to very small datasets?
I have a dataset with many variables but only 25 observation each. Random forests produce reasonable results with low OOB errors (10-25%).
Is there any rule of thumb regarding the minimum number of observations to use?
In fact one of the response variable is unbalanced, and if I'm going to subsample it I will end up with an even smaller number of observations.
Thanks in advance
Absolutely RF can be used on these type of datasets (i.e. p>n). In fact they use RF in fields like genomics where the number of fields >= 20000 and there are only a very small number of rows - say 10-12. The entire problem is figuring out which of the 20k variables would make up a parsimonious marker (i.e. feature selection is the entire problem).
I don't have any ROTs about minimum size other than if your model doesn't work well on a held back sample (or Hold-One-Back cross validation might work well in your case) well then you should try something else.
Hope this helps
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Is there a rule-of-thumb for how to best divide data into training and validation sets? Is an even 50/50 split advisable? Or are there clear advantages of having more training data relative to validation data (or vice versa)? Or is this choice pretty much application dependent?
I have been mostly using an 80% / 20% of training and validation data, respectively, but I chose this division without any principled reason. Can someone who is more experienced in machine learning advise me?
There are two competing concerns: with less training data, your parameter estimates have greater variance. With less testing data, your performance statistic will have greater variance. Broadly speaking you should be concerned with dividing data such that neither variance is too high, which is more to do with the absolute number of instances in each category rather than the percentage.
If you have a total of 100 instances, you're probably stuck with cross validation as no single split is going to give you satisfactory variance in your estimates. If you have 100,000 instances, it doesn't really matter whether you choose an 80:20 split or a 90:10 split (indeed you may choose to use less training data if your method is particularly computationally intensive).
Assuming you have enough data to do proper held-out test data (rather than cross-validation), the following is an instructive way to get a handle on variances:
Split your data into training and testing (80/20 is indeed a good starting point)
Split the training data into training and validation (again, 80/20 is a fair split).
Subsample random selections of your training data, train the classifier with this, and record the performance on the validation set
Try a series of runs with different amounts of training data: randomly sample 20% of it, say, 10 times and observe performance on the validation data, then do the same with 40%, 60%, 80%. You should see both greater performance with more data, but also lower variance across the different random samples
To get a handle on variance due to the size of test data, perform the same procedure in reverse. Train on all of your training data, then randomly sample a percentage of your validation data a number of times, and observe performance. You should now find that the mean performance on small samples of your validation data is roughly the same as the performance on all the validation data, but the variance is much higher with smaller numbers of test samples
You'd be surprised to find out that 80/20 is quite a commonly occurring ratio, often referred to as the Pareto principle. It's usually a safe bet if you use that ratio.
However, depending on the training/validation methodology you employ, the ratio may change. For example: if you use 10-fold cross validation, then you would end up with a validation set of 10% at each fold.
There has been some research into what is the proper ratio between the training set and the validation set:
The fraction of patterns reserved for the validation set should be
inversely proportional to the square root of the number of free
adjustable parameters.
In their conclusion they specify a formula:
Validation set (v) to training set (t) size ratio, v/t, scales like
ln(N/h-max), where N is the number of families of recognizers and
h-max is the largest complexity of those families.
What they mean by complexity is:
Each family of recognizer is characterized by its complexity, which
may or may not be related to the VC-dimension, the description
length, the number of adjustable parameters, or other measures of
complexity.
Taking the first rule of thumb (i.e.validation set should be inversely proportional to the square root of the number of free adjustable parameters), you can conclude that if you have 32 adjustable parameters, the square root of 32 is ~5.65, the fraction should be 1/5.65 or 0.177 (v/t). Roughly 17.7% should be reserved for validation and 82.3% for training.
Last year, I took Prof: Andrew Ng’s online machine learning course. His recommendation was:
Training: 60%
Cross-validation: 20%
Testing: 20%
Well, you should think about one more thing.
If you have a really big dataset, like 1,000,000 examples, split 80/10/10 may be unnecessary, because 10% = 100,000 examples may be just too much for just saying that model works fine.
Maybe 99/0.5/0.5 is enough because 5,000 examples can represent most of the variance in your data and you can easily tell that model works good based on these 5,000 examples in test and dev.
Don't use 80/20 just because you've heard it's ok. Think about the purpose of the test set.
Perhaps a 63.2% / 36.8% is a reasonable choice. The reason would be that if you had a total sample size n and wanted to randomly sample with replacement (a.k.a. re-sample, as in the statistical bootstrap) n cases out of the initial n, the probability of an individual case being selected in the re-sample would be approximately 0.632, provided that n is not too small, as explained here: https://stats.stackexchange.com/a/88993/16263
For a sample of n=250, the probability of an individual case being selected for a re-sample to 4 digits is 0.6329.
For a sample of n=20000, the probability is 0.6321.
It all depends on the data at hand. If you have considerable amount of data then 80/20 is a good choice as mentioned above. But if you do not Cross-Validation with a 50/50 split might help you a lot more and prevent you from creating a model over-fitting your training data.
Suppose you have less data, I suggest to try 70%, 80% and 90% and test which is giving better result. In case of 90% there are chances that for 10% test you get poor accuracy.
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I am using LibSVM to classify some documents. The documents seem to be a bit difficult to classify as the final results show. However, I have noticed something while training my models. and that is: If my training set is for example 1000 around 800 of them are selected as support vectors.
I have looked everywhere to find if this is a good thing or bad. I mean is there a relation between the number of support vectors and the classifiers performance?
I have read this previous post but I am performing a parameter selection and also I am sure that the attributes in the feature vectors are all ordered.
I just need to know the relation.
Thanks.
p.s: I use a linear kernel.
Support Vector Machines are an optimization problem. They are attempting to find a hyperplane that divides the two classes with the largest margin. The support vectors are the points which fall within this margin. It's easiest to understand if you build it up from simple to more complex.
Hard Margin Linear SVM
In a training set where the data is linearly separable, and you are using a hard margin (no slack allowed), the support vectors are the points which lie along the supporting hyperplanes (the hyperplanes parallel to the dividing hyperplane at the edges of the margin)
All of the support vectors lie exactly on the margin. Regardless of the number of dimensions or size of data set, the number of support vectors could be as little as 2.
Soft-Margin Linear SVM
But what if our dataset isn't linearly separable? We introduce soft margin SVM. We no longer require that our datapoints lie outside the margin, we allow some amount of them to stray over the line into the margin. We use the slack parameter C to control this. (nu in nu-SVM) This gives us a wider margin and greater error on the training dataset, but improves generalization and/or allows us to find a linear separation of data that is not linearly separable.
Now, the number of support vectors depends on how much slack we allow and the distribution of the data. If we allow a large amount of slack, we will have a large number of support vectors. If we allow very little slack, we will have very few support vectors. The accuracy depends on finding the right level of slack for the data being analyzed. Some data it will not be possible to get a high level of accuracy, we must simply find the best fit we can.
Non-Linear SVM
This brings us to non-linear SVM. We are still trying to linearly divide the data, but we are now trying to do it in a higher dimensional space. This is done via a kernel function, which of course has its own set of parameters. When we translate this back to the original feature space, the result is non-linear:
Now, the number of support vectors still depends on how much slack we allow, but it also depends on the complexity of our model. Each twist and turn in the final model in our input space requires one or more support vectors to define. Ultimately, the output of an SVM is the support vectors and an alpha, which in essence is defining how much influence that specific support vector has on the final decision.
Here, accuracy depends on the trade-off between a high-complexity model which may over-fit the data and a large-margin which will incorrectly classify some of the training data in the interest of better generalization. The number of support vectors can range from very few to every single data point if you completely over-fit your data. This tradeoff is controlled via C and through the choice of kernel and kernel parameters.
I assume when you said performance you were referring to accuracy, but I thought I would also speak to performance in terms of computational complexity. In order to test a data point using an SVM model, you need to compute the dot product of each support vector with the test point. Therefore the computational complexity of the model is linear in the number of support vectors. Fewer support vectors means faster classification of test points.
A good resource:
A Tutorial on Support Vector Machines for Pattern Recognition
800 out of 1000 basically tells you that the SVM needs to use almost every single training sample to encode the training set. That basically tells you that there isn't much regularity in your data.
Sounds like you have major issues with not enough training data. Also, maybe think about some specific features that separate this data better.
Both number of samples and number of attributes may influence the number of support vectors, making model more complex. I believe you use words or even ngrams as attributes, so there are quite many of them, and natural language models are very complex themselves. So, 800 support vectors of 1000 samples seem to be ok. (Also pay attention to #karenu's comments about C/nu parameters that also have large effect on SVs number).
To get intuition about this recall SVM main idea. SVM works in a multidimensional feature space and tries to find hyperplane that separates all given samples. If you have a lot of samples and only 2 features (2 dimensions), the data and hyperplane may look like this:
Here there are only 3 support vectors, all the others are behind them and thus don't play any role. Note, that these support vectors are defined by only 2 coordinates.
Now imagine that you have 3 dimensional space and thus support vectors are defined by 3 coordinates.
This means that there's one more parameter (coordinate) to be adjusted, and this adjustment may need more samples to find optimal hyperplane. In other words, in worst case SVM finds only 1 hyperplane coordinate per sample.
When the data is well-structured (i.e. holds patterns quite well) only several support vectors may be needed - all the others will stay behind those. But text is very, very bad structured data. SVM does its best, trying to fit sample as well as possible, and thus takes as support vectors even more samples than drops. With increasing number of samples this "anomaly" is reduced (more insignificant samples appear), but absolute number of support vectors stays very high.
SVM classification is linear in the number of support vectors (SVs). The number of SVs is in the worst case equal to the number of training samples, so 800/1000 is not yet the worst case, but it's still pretty bad.
Then again, 1000 training documents is a small training set. You should check what happens when you scale up to 10000s or more documents. If things don't improve, consider using linear SVMs, trained with LibLinear, for document classification; those scale up much better (model size and classification time are linear in the number of features and independent of the number of training samples).
There is some confusion between sources. In the textbook ISLR 6th Ed, for instance, C is described as a "boundary violation budget" from where it follows that higher C will allow for more boundary violations and more support vectors.
But in svm implementations in R and python the parameter C is implemented as "violation penalty" which is the opposite and then you will observe that for higher values of C there are fewer support vectors.