Data Analytics CMIS Short Course part II Day 1 Part 2: Trees Sam Buttrey December 2015

Regression Trees The usual set-up: numeric responses y 1, …, y n ; predictors X i for each y i –These might be numeric, categorical, logical… Start with all the y’s in one node; measure the “impurity” of that node (the extent to which the y’s are spread out) Object will be to divide the observations into sub-nodes of high purity (that is, with similar y’s)

Impurity Measure Any measure of impurity should be 0 when all y’s are the same; otherwise > 0 Natural choice for continuous y’s: compute predictions y-hat and take impurity as D =  i (y i – y-hat) 2 –RSS (deviance for Normal model) is preferable to SD; impurity should relate to sample size –This is “just right” if the y’s are Normal and unweighted (we care about everything equally)

Reducing Impurity R implementations in tree(), rpart() Both measure impurity by RSS by default Now consider each X column in turn 1.If X j is numeric, divide the Y’s into two pieces, one where X j  a and one where X j > a (a “split”) –Try every a; there are at most n – 1 of these E.g. Alcohol < 4; Alcohol < 4.5, etc.; Price < 3; Price < 3.5, etc; Sodium < 10, …

Reducing Impurity, cont’d In left and right “child” nodes, compute separate means y L -hat, y R -hat and separate deviances D L and D R The decrease in deviance (i.e. increase in purity) for this split is D – (D R + D L ) Our goal is to find the split for which this decrease is largest, or the sum of D L and D R is smallest (i.e. for which the two resulting nodes are purest)

Beer Example “Root” impurity is calories 2 6 n = 35 D = Alc  4 Alc > 4 n = 3 D = 1241 n = 32 D = 8995 Sum: n = 35 D = Alc  4.5 Alc > 4.5 n = 10 D = 7308 n = 25 D = 1584 Sum: 8892 n = 35 D = Alc  4.9 Alc > 4.9 n = 27 D = n = 8 D = 929 Sum: n = 35 D = Price  2.75 Price > 2.75 n = 21 D = n = 14 D = 3647 Sum: n = 35 D = Cost  0.46 Cost > 0.46 n = 21 D = n = 14 D = 3647 Sum: n = 35 D = Sod  10 Sod > 10 n = 11 D = 9965 n = 24 D = 9431 Sum: Best among these

Reducing Impurity 2.If X is categorical, split the data into two pieces, one with one subset of categories and one with the rest (a “split”) –For k categories, there are 2 k–1 – 1 of these E.g. divide men from women, (sub/air) from (surface/supply), (old/young) from (med.) Measure decrease in deviance exactly as before Select the best split among all possibilities –Subject to rules on minimum node size etc.

The Recursion Now split the two “child” nodes (say, #2 and #3) separately #2’s split will normally be different from #3’s; it could be on the same variable used at the root, but usually won’t be Split #2 into 4 and 5, #3 into 6 and 7, so as to decrease the impurity as much as possible; then split the resulting children –Node q’s children are numbered 2q, 2q+1

Some Practical Considerations 1.When do we stop splitting? –Clearly too much splitting  over-fitting –Stay tuned for this one 2.It feels bad to create child nodes with one observation – maybe even < 10 3.The addition of deviances implies we think they’re on the same scale – that is, we assume homoscedasticity when we use RSS as our impurity measure

Prediction For a new case, find the terminal node it falls into (based on its X’s) Predict the average of the y’s there –SDs are harder to come by! Diagnostic: residuals, fitted values, within- node variance vs. mean Instead of fitting a plane in the X space, we’re dividing it up into oblong pieces and assuming constant y in each piece

R Example R has tree and rpart libraries for trees Example 1: beer –There’s a tie for best split plot() plus text() draws pictures –Or use rpart.plot() from that library In this case, the linear model is better… –Unless you include both Price and Cost… –Tree model unaffected by “multi-collinearity” Trees are easy to understand; it’s easy to make nice pictures; almost no theoretical results 11

Tree Example (1985 cps wage) Training set (n = 427), test set (107) Step 1: produce tree for Wage –Notice heteroscedasticity, as reported by meanvar() –We hope for a roughly flat picture, indicating constant variance across leaves –In this case let’s take the log of Wage –Let tree stop splitting according to its defaults We will discuss these shortly!

R Interlude Wage tree using log of wage Compare performance to several linear models In this example the tree is not as strong as a reasonable linear model… …But sometimes it’s better… …And stay tuned for ensembles of models 13

Tree Benefits Trees are interpretable, easy to understand Extend naturally to classification, including the case with more than two classes Insensitive to monotonic transformation of X’s variables (unlike linear model) –Reduces impact of outliers Interactions included automatically Smart missing value handling –Both building the tree and predicting 14

Tree Benefits and Drawbacks (Almost) Entirely automatic model generation –Model is local, rather than global like regression No problem when columns overlap (e.g. beer) or # colums > # rows On the other hand… Abrupt decision boundaries look weird Some problems in life are approximately linear, at least after lots of analyst input No inference – just test set performance 15

Bias vs Variance 16 All Relationships Linear Linear Plus Best Linear Model Best LM with transformations, interactions… Best Tree True Model

Bias vs Variance 17 All Relationships Linear Linear Plus Best Linear Model Best LM with transformations, interactions… Bias Best Tree True Model

Stopping Rules Defaults –Do not split a node with  < 20 observations –Do not create a node with < 7 –R 2 must increase by.01 each step (this value is the complexity parameter cp) –Maximum depth = 30 (!) The plan is to intentionally grow a tree that’s too big, then “prune” it back to the optimal size using… Cross-validation!

Cross-validation in Rpart Cross-validation is done automatically –Results in the cp table of the rpart object 19 CPnsplitrel errorxerrorxstd Impurity as a fraction of the impurity at the root – always goes down for larger trees Cross-validated error: often goes down and then back up Find minimum xerror value… …And use corresponding cp (rounding upwards)

Pruning Recap In wage example: wage.rp <- rpart (LW ~., data = wage) plotcp (wage.rp) # show minimum prune (wage.rp, cp =.02) “Optimal” tree size is random – it depends on the cross-validation Or prune to one-SE rule by selecting the smallest size whose xerror < (min xerror + 1  corresponding SE) –Less variable? 20

Rpart Features rpart.control() function to set things like minimum leaf size, minimum within- leaf deviance For y’s like Poisson counts, compute Poisson deviance Methods also exist for exponential data (e.g. component lifetimes) with censoring Case weights can be applied Most importantly, rpart can also handle binary or multinomial categorical y’s

Missing Value Handling Missing values handling: Building: surrogate splits –Splits that “look like” the best split, to be used at prediction time if the “best” has NAs –Cases with missing values deleted, but… –(tree) na.tree.replace() for categoricals Predicting: use avg. y at stopping point We really want to avoid this because in real life lots of observations are missing at least a little data

Classification Two of the big problems in statistics: –Regression: estimate E(y i | X i ) when y is continuous (numeric) –Classification: predict y i | X i when y is categorical Example 1: two classes (“0” and “1”) One method: logistic regression Choose prediction threshold c, say 0.5 Predicted p > c  classify object into class 1; otherwise classify into class 0 For > 2 categories, logit can be extended

Classification Object: produce a rule that classifies new observations accurately –Or that assigns a good probability estimate –…Or at least rank-orders observations well Measure of quality: Area under the ROC curve, or misclassification rate, or deviance, or something else Issues: rare (or common) events are hard to classify, since the “null model” is so good –E.g. “No airplane will ever crash”

Other Classifiers

Classification Tree Same machinery as regression tree Still need to find optimal size of tree –As before, with plotcp/prune Example: Fisher iris data (3 classes); predict species based on measurements plot() + text(), rpart.plot() as before Example: spam data –Logistic regression tricky to build here

More on classification trees predict() works with a classification tree, just as with regression ones Follow each observation to its leaf; plurality “vote” determines classification By default, predict() produces a vector of class proportions in each obs.’s chosen node Use type="class" to choose the most probable class As always, there is more to a good model than just raw misclassification rate

Splitting Rules Several kinds of splits for classification trees in R It’s not a good idea to split so as to minimized misclassifcation rate Example: 151/200 75% “Yes” 52/100 52% “Yes” 99/100 99% “Yes” Misclass rate: 49/200 = Misclass rate: (1 + 48)/( ) = 0.245

Misclass Rate Minimizing misclass rate deprives us of splits that produce more homogeneous children, if all children continue to be majority 1 (or 0)… …So misclassification rate is particularly weak for very rare or very common events In our example, deviance starts at -2 [151 log (151/200) + 49 log (49/200)] = 222 and goes to -2 {[99 log (99/100) + 1 log (1/100)] + [52 log (52/100) + 48 log (48/100)]} = 150 Decrease !

Other Choices of Criterion Gini index within node t : 1 –  j p( j) 2 Smallest when one p is 1, the others 0 Gini split is then the weighted average of the two child-node Gini indices Information (also called “entropy”) E (t) = 1 –  p (j) log p (j) Compare to deviance, which is D (t)  –  n j log p (j) – not much difference Values not displayed in rpart printout

Multinomial Example Exactly the same setup when there are multiple classes –Whereas logistic regression gets lots more complicated –Other natural choice here: neural networks? Example: Optical digit data –No obvious stopping rule here (?) 31