I am working on analyzing grade data. As a new way to look at the data I am using a decision tree, for the first time. I believe I have the code right and now I am trying to interpret it. The features are grades gotten for a series of quizzes, and the classification is the final grade the student received. I have a few questions:
If my understanding is correct, each node has a test and a left branch representing the test being true, and the other for false. And when the tree seems to have asked enough questions, it says what the "class" is. If that is the case, how come there's a class= on boxes well before the leaves? I would have thought that just leaves have a class=
How do I "tune" the overall tree? It seems to have too many boxes. Is this an example of "overfitting"? How can I tune that better?
For example, the use of FINAL_GRADE_PA01 seems arbitrary to be based on the ordering of the data. Is that true or did the analysis actually conclude that that feature was the best discriminator?
If I'm not mistaken, those class values indicate what the model would have predicted, had it stopped branching on that node. It still stores those values, but it doesn't use them if there's a branching from that node.
About the number of nodes, as you see in the docs:
The default values for the parameters controlling the size of the
trees (e.g. max_depth, min_samples_leaf, etc.) lead to fully grown and
unpruned trees which can potentially be very large on some data sets.
To reduce memory consumption, the complexity and size of the trees
should be controlled by setting those parameter values.
There are several parameters which you can use to reduce the complexity of your model. The following two parameters are just an example:
max_leaf_nodes : int or None, optional (default=None)
Grow a tree with max_leaf_nodes in best-first fashion. Best nodes are defined as relative reduction in impurity. If None then unlimited
number of leaf nodes.
min_impurity_decrease : float, optional (default=0.)
A node will be split if this split induces a decrease of the impurity greater than or equal to this value.
Related
I was wondering if anyone knows how SPSS builds the interaction terms/calculates the significance for predictors behind the scenes in a GLM? From my understanding it dummy codes variables and treats the one that comes alphabetically last as the reference group.
The reason I'm asking is I have a GLM model which has 3 continuous predictors and two categorical predictors (dummy coded). When I build all the 2-way and 3-way interactions with syntax ie:
Age_Centred Age_CentredDx Age_Centredgender Age_CentredDxgender BMI_Centred BMI_CentredDx BMI_Centredgender BMI_CentredDxgender BPS_Centred BPS_CentredDx BPS_Centredgender BPS_CentredDxgender Dx Dxgender DxICV_Centred DxICV_Centredgender gender ICV_Centred ICV_Centred*gender.
vs manually creating all the variables by hand ie:
Age_Centred Age_Centred_Dx Age_Centred_gender Age_Centred_gender_Dx BMI_Centred BMI_Centred_Dx BMI_Centred_gender BMI_Centred_gender_Dx BPS_Centred BPS_Centred_Dx BPS_Centred_gender BPS_Centred_gender_Dx Dx gender_Dx ICV_Dx ICV_Centred_Dx_gender gender ICV_Centred ICV_gender.
I end up with a model which has the same intercept, overall significance, and R squared however the individual significance of the predictors changes. Refer to output below. To troubleshoot I've tried to flip the references groups when manually creating the variables but it still does not replicate the results. I've had another statistician try the same thing and ended up reaching the same point as what I did. Does it have to do with some of the parameters being redundant?
Building the terms via syntax:
Physically creating the variables by multiplying them together
All the details one might reasonably want about how GLM (and UNIANOVA, which is the same underlying code) parameterizes models, estimates parameters, and conducts hypothesis tests are available in the IBM SPSS Statistics Algorithms manual, available for download as a pdf at ftp://public.dhe.ibm.com/software/analytics/spss/documentation/statistics/26.0/en/client/Manuals/IBM_SPSS_Statistics_Algorithms.pdf. (Note that this is a large file, about 78 MB; clicking on the link starts a download.) In addition to the information in the GLM chapter, appendices F (Indicator Method) and H (Sums of Squares) are relevant, respectively, for building the design matrix and specifying linear combinations of model parameters for computing sums of squares for testing hypotheses.
In building the design matrix, categorical predictors (factors) are indeed represented by sets of indicator (0-1) variables. For a factor with k levels, k indicator variables are created, one for each observed level of the factor. The procedure does not explicitly treat the last category (sorted in ascending order, alphabetical for strings) as a reference category, though in simpler models the effect of what's done is essentially the same. If there is an intercept in the model, then the kth indicator will be redundant (linearly dependent) on the intercept and the preceding k-1 indicators. The estimation algorithm used in GLM/UNIANOVA will set the row and column in the cross-product matrix representing the redundant column in the design matrix to 0s, alias the corresponding parameter estimate to 0, and the results are similar to a reparameterization approach treating the last category as a reference category, except that you have to remember that it's there if you want to specify a linear combination of the parameters to estimate.
If you suppress the intercept, then for the first factor entered into the model the kth indicator would not be redundant (unless the factor is preceded by an unusual covariate or set of covariates). Any subsequent factors included in the model would involve redundant parameters, as would any interactions among factors, whether or not an intercept is included. Interactions among factors are created by multiplying the 0s and 1s for each level of the factors by those for each level of the other factor. So for an interaction of two two-level factors, there are four columns generated, of which typically the last three are redundant.
Covariates are entered simply by copying the values of the variables into the design matrix. Interactions involving covariates and other covariates multiply values for the columns involved within each row, and interactions involving covariates and factors multiply covariates (or products of them) by the indicator variables for the factor(s). Usually covariate-by-covariate terms do not involve redundancies, but factor-by-covariate terms do.
To get to the specifics of what's going on with your data, I can't replicate your exact results without your data, but I am able to replicate the patterns shown if I assume you've used the binary Dx variable as a covariate and the binary gender variable as a factor in each analysis. (There seem to actually be four continuous predictors in your model rather than three, but that doesn't affect anything of importance for understanding what's going on.)
There are two aspects of the situation to be considered. One is the parameterization and how the two ways of entering the variables into the model treat the variables and whether or not they produce the same estimates of parameters. The second is how the model specification results in the Type III tests shown in the ANOVA tables.
If I'm understanding things correctly based on what you've posted here, you should find if you compare parameter estimates for the two analyses that the parameter estimates for the intercepts and the non-redundant estimates for gender ([gender=0]) are the same, and have the same standard errors. For the terms involving just covariates or products of covariates, I expect that you will find the parameter estimates to differ between the two analyses and produce different t statistics. For interactions involving gender and covariates (which is all the other variables or products created outside the procedure), I expect the estimates will be the same in magnitude and opposite in sign, with the same standard errors.
None of the estimates or tests here are wrong. The models fitted involve interaction effects. An interaction means that effect of one variable varies by the levels of the other variable(s) in the interaction, and in order to estimate the same simple effects you have to parameterize the model in the same way, at least as far as the non-redundant parameters are concerned. However, to get the Type III tests for all terms to be identical, it's not always enough to have the same parameter estimates and standard errors. Type III tests involve a concept called containment that must also be considered.
For two effects in a model, effect A is contained in effect B if:
A and B contain the same covariate terms, if any.
B contains all factor effects in A, and at least one more (with the intercept being contained in all factor-only effects).
In your original model, the intercept is included in the gender effect, gender is not included in any effects, and all the covariate main effects and two-way interactions among covariates are contained within the interactions between those terms and gender, while the three-way interactions (which include gender) are not contained within any other effects.
Type III sums of squares (not invented by SPSS, but by our friends at SAS) are based on linear combinations of parameters where a given effect is adjusted for any effects that do not contain it, and made orthogonal to any effects that contain it. The practical application of these rules is complicated (see Appendix H of the algorithms).
If you recode the gender variable to swap the 0 and 1 values, specify it as a covariate along with all the other variables, and fit the same model, you should be able to match all the non-redundant parameter estimates from the original model, along with their standard errors and t statistics. However, because the containment relationships in the original model are no longer there, the Type III tests for the terms not involving gender (which were previously contained in terms involving gender) will not match up.
The bottom line is that all results are translatable and all correct for what's being done, and that in order to make much sense out of individual terms you have to carefully focus on what's being estimated in a given parameterization, as well as the containment relationships. The difficult part gets simpler when you take seriously the fact that when variable X is involved in interaction terms, there is no single estimate of the effect of X. Any estimates are conditional one where you fix the value(s) of the terms with which X interacts.
Lets say I have 100 independent features - 90 are binary (e.g. 0/1) and 10 are continuous variables (e.g. age, height, weight, etc). I use the 100 features to predict a classifier problem with an adequate amount of samples.
When I set a XGBClassifier function and fit it, then the 10 most important features from the standpoint of gain are always the 10 continuous variable. For now I am not interested in cover or frequency. The 10 continuous variables take up like .8 to .9 of space in gain list ( sum(gain) = 1).
I tried tuning the gamma, reg_alpha , reg_lambda , max_depth, colsample. Still top 10 features by gain are always the 10 continuous features.
Any suggestions?
small update -- someone asked why I think this is happening. I believe it's because a continuous variable can be split on multiple times per decision tree. A binary variable can only be split on once. Hence, the higher prevalence of continuous variables in trees and thus a higher gain score
Yes, it's well-known that a tree(/forest) algorithm (xgboost/rpart/etc.) will generally 'prefer' continuous variables over binary categorical ones in its variable selection, since it can choose the continuous split-point wherever it wants to maximize the information gain (and can freely choose different split-points for that same variable at other nodes, or in other trees). If that's the optimal tree (for those particular variables), well then it's the optimal tree. See Why do Decision Trees/rpart prefer to choose continuous over categorical variables? on sister site CrossValidated.
When you say "any suggestions", depends what exactly do you want, it could be one of the following:
a) To find which of the other 90 binary categorical features give the most information gain
b) To train a suboptimal tree just to find out which features those are
c) To engineer some "compound" features by combining the binary features into n-bit categorical features which have more information gain (while being sure to remove the individual binary features from the input)
d) You could look into association rules : What is the practical difference between association rules and decision trees in data mining?
If you want to explore a)...c), suggest something vaguely like this:
exclude various subsets of the 10 continuous variables, then see which binary features show up as having the most gain. Let's say that gives you N candidate features. N will be << 90, let's assume N < 20 to make the following more computationally efficient.
then compute the pairwise measure of association or correlation (Spearman or Kendall) between each of the N features. Look at a corrplot. Pick the clusters of variables which are most associated with each other. Create compound n-bit variables which combine those individual binary features. Then retrain the tree, including the compound variables, and excluding the individual binary variables (to avoid changing the total variance in the input).
iterate for excluding various subsets of the 10 continuous variables. See which patterns emerge in your compound variables. I'm sure there's an algorithm for doing this (compound feature-engineering of n-bit categoricals) more formally and methodically, I just don't know it.
Anyway, for hacking a tree-based method for better performance, I imagine the most naive way is "at every step, pick the two most highly-correlated/associated categorical features and combine them". Then retrain the tree (include new feature, exclude its constituent features) and use the revised gain numbers.
perhaps a more robust way might be:
Pick some threshold T for correlation/association, say start at a high level T = 0.9 or 0.95
At each step, merge any features whose absolute correlation/association to each other >= T
If there were no merges at this step, reduce T by some value (like T -= 0.05) or ratio (e.g. T *= 0.9 . If still no merges, keep reducing T until there are merges, or until you hit some termination value (e.g. T = 0.03)
Retrain the tree including the compound variables, excluding their constituent subvariables.
Now go back and retrain what should be an improved tree with all 10 continuous variables, and your compound categorical features.
Or you could early-terminate the compound feature selection to see what the full retrained tree looks like.
This issue arose in the 2014 Kaggle Allstate Purchase Prediction Challenge, where the policy coverage options A,B,C,D,E,F,G were each categoricals with between 2-4 values, and very highly correlated with each other. (The current option of C, "C_previous", is one of the input features). See that competitions's forums and published solutions for more. Be aware that policy = (A,B,C,D,E,F,G) is the output. But C_previous is an input variable.
Some general fast-and-dirty rules-of-thumb on feature selection from Kaggle are:
throw out any near-constant/ very-low-variance variables (because they have near-zero information content)
throw out any very-high-cardinality categorical variables (cardinality >~ training-set-size/2), (because they will also tend to have low information content, but cause lots of spurious overfitting and blow up training time). This can include customer IDs, row IDs, transaction IDs, sequence IDs, and other variables which shouldn't be trained on in the first place but accidentally ended up in the training set.
I can suggest few things for you to try.
Test your model without this data (only 90 features) and evaluate the decrease in your score. If it's insignificant you might want to remove those features.
Turn them into groups.
For example, age can be categorized into groups, 0 : 0-7, 1 : 8-16, 2 : 17-25 and so on.
Turn them into binary. Out of the box idea on how to chose the best value to split them into binary is: Build 1 tree with 1 node (max depth = 1) and use only 1 feature. (1 out of the continuous features). then, dump the model to a .txt file and see the value it chose to split on. using this value, you can transform all that feature column into binary
I'm dealing myself with very similar problems right now, So i'll be happy to hear your results and the paths you chose to try.
I learned a lot from the answer by #smci, so I would recommend to follow his suggestions.
In the case, when your binary categorical features are in fact OHE representations of several categorical features with several classes in each, you can follow two more approaches:
Convert OHE into label encoding. Yes, this has the caveat that one introduces an order into a categorical features, which might be meaningless, for example green=3 > red=2 > blue=1. But in practice is seems that trees handle label=encoded categorical variables (even with meaningless order) reasonably well.
Convert OHE into target-/mean-/likelihood encoding. This is tricky, because you need to apply regularisation to avoid data leakage.
Both of those ideas are meant to group together several binary features into a single one based on prior knowledge about feature meaning. If you do not have that luxury, you can also try to deduce such groups by doing scalar product of columns and finding those giving zero product.
I have a set of 3-5 black box scoring functions that assign positive real value scores to candidates.
Each is decent at ranking the best candidate highest, but they don't always agree--I'd like to find how to combine the scores together for an optimal meta-score such that, among a pool of candidates, the one with the highest meta-score is usually the actual correct candidate.
So they are plain R^n vectors, but each dimension individually tends to have higher value for correct candidates. Naively I could just multiply the components, but I hope there's something more subtle to benefit from.
If the highest score is too low (or perhaps the two highest are too close), I just give up and say 'none'.
So for each trial, my input is a set of these score-vectors, and the output is which vector corresponds to the actual right answer, or 'none'. This is kind of like tech interviewing where a pool of candidates are interviewed by a few people who might have differing opinions but in general each tend to prefer the best candidate. My own application has an objective best candidate.
I'd like to maximize correct answers and minimize false positives.
More concretely, my training data might look like many instances of
{[0.2, 0.45, 1.37], [5.9, 0.02, 2], ...} -> i
where i is the ith candidate vector in the input set.
So I'd like to learn a function that tends to maximize the actual best candidate's score vector from the input. There are no degrees of bestness. It's binary right or wrong. However, it doesn't seem like traditional binary classification because among an input set of vectors, there can be at most 1 "classified" as right, the rest are wrong.
Thanks
Your problem doesn't exactly belong in the machine learning category. The multiplication method might work better. You can also try different statistical models for your output function.
ML, and more specifically classification, problems need training data from which your network can learn any existing patterns in the data and use them to assign a particular class to an input vector.
If you really want to use classification then I think your problem can fit into the category of OnevsAll classification. You will need a network (or just a single output layer) with number of cells/sigmoid units equal to your number of candidates (each representing one). Note, here your number of candidates will be fixed.
You can use your entire candidate vector as input to all the cells of your network. The output can be specified using one-hot encoding i.e. 00100 if your candidate no. 3 was the actual correct candidate and in case of no correct candidate output will be 00000.
For this to work, you will need a big data set containing your candidate vectors and corresponding actual correct candidate. For this data you will either need a function (again like multiplication) or you can assign the outputs yourself, in which case the system will learn how you classify the output given different inputs and will classify new data in the same way as you did. This way, it will maximize the number of correct outputs but the definition of correct here will be how you classify the training data.
You can also use a different type of output where each cell of output layer corresponds to your scoring functions and 00001 means that the candidate your 5th scoring function selected was the right one. This way your candidates will not have to be fixed. But again, you will have to manually set the outputs of the training data for your network to learn it.
OnevsAll is a classification technique where there are multiple cells in the output layer and each perform binary classification in between one of the classes vs all others. At the end the sigmoid with the highest probability is assigned 1 and rest zero.
Once your system has learned how you classify data through your training data, you can feed your new data in and it will give you output in the same way i.e. 01000 etc.
I hope my answer was able to help you.:)
I am reading about n-grams and I am wondering whether there is a case in practice when uni-grams would are preferred to be used over bi-grams (or higher N-grams). As I understand, the bigger N, the bigger complexity to calculate the probabilities and establish the vector space. But apart from that, are there other reasons (e.g. related to type of data)?
This boils down to data sparsity: As your n-gram length increases, the amount of times you will see any given n-gram will decrease: In the most extreme example, if you have a corpus where the maximum document length is n tokens and you are looking for an m-gram where m=n+1, you will, of course, have no data points at all because it's simply not possible to have a sequence of that length in your data set. The more sparse your data set, the worse you can model it. For this reason, despite that a higher-order n-gram model, in theory, contains more information about a word's context, it cannot easily generalize to other data sets (known as overfitting) because the number of events (i.e. n-grams) it has seen during training becomes progressively less as n increases. On the other hand, a lower-order model lacks contextual information and so may underfit your data.
For this reason, if you have a very relatively large amount of token types (i.e. the vocabulary of your text is very rich) but each of these types has a very low frequency, you may get better results with a lower-order n-gram model. Similarly, if your training data set is very small, you may do better with a lower-order n-gram model. However, assuming that you have enough data to avoid over-fitting, you then get better separability of your data with a higher-order model.
Usually, n-grams more than 1 is better as it carries more information about the context in general. However, sometimes unigrams are also calculated besides bigram and trigrams and used as fallback for them. This is usefull also, if you want high recall than precision to search unigrams, for instance, you are searching for all possible uses of verb "make".
Lets use Statistical Machine Translation as an Example:
Intuitively, the best scenario is that your model has seen the full sentence (lets say 6-grams) before and knows its translation as a whole. If this is not the case you try to divide it to smaller n-grams, keeping into consideration that the more information you know about the word surroundings, the better the translation. For example, if you want to translate "Tom Green" to German, if you have seen the bi-gram you will know it is a person name and should remain as it is but if your model never saw it, you would fall back to unigrams and translate "Tom" and "Green" separately. Thus "Green" will be translated as a color to "Grün" and so on.
Also, in search knowing more about the surrounding context makes the results more accurate.
How to convert or visualize decision table to decision tree graph,
is there an algorithm to solve it, or a software to visualize it?
For example, I want to visualize my decision table below:
http://i.stack.imgur.com/Qe2Pw.jpg
Gotta say that is an interesting question.
I don't know the definitive answer, but I'd propose such a method:
use Karnaugh map to turn your decision table to minimized boolean function
turn your function into a tree
Lets simplyify an example, and assume that using Karnaugh got you function (a and b) or c or d. You can turn that into a tree as:
Source: my own
It certainly is easier to generate a decision table from a decision tree, not the other way around.
But the way I see it you could convert your decision table to a data set. Let the 'Disease' be the class attribute and treat the evidence as simple binary instance attributes. From that you can easily generate a decision tree using one of available decision tree induction algorithms, for example C4.5. Just remember to disable pruning and lower the minimum number of objects parameter.
During that process you would lose a bit of information, but the accuracy would remain the same. Take a look at both rows describing disease D04 - the second row is in fact more general than the first. Decision tree generated from this data would recognize the mentioned disease only from E11, 12 and 13 attributes, since it's enough to correctly label the instance.
I've spent few hours looking for a good algorithm. But I'm happy with my results.
My code is too dirty now to paste here (I can share privately on request, on your discretion) but the general idea is as the following.
Assume you have a data set with some decision criteria and outcome.
Define a tree structure (e.g. data.tree in R) and create "Start" root node.
Calculate outcome entropy of your data set. If entropy is zero you are done.
Using each criterion, one by one, as tree node calculate entropy for all branches created with this criterion. Take the minimum one entropy of all branches.
Branches created with the criterion with the smallest (minimum) entropy are your next tree node. Add them as child nodes.
Split your data according to decision point/tree node found in step 4 and remove the criterion used.
Repeat step 2-4 for each branch until your all branches have entropy = 0.
Enjoy your ideal decision tree :)