1. 1 Peter Fox Data Analytics – ITWS-4963/ITWS-6965 Week 5a, February 24, 2015 Weighted kNN, ~ clustering, trees and Bayesian classification

2. Plot tools/ tips http://statmethods.net/advgraphs/layout.html http://flowingdata.com/2014/02/27/how-to-read-histograms-and-use-them-in- r/ pairs, gpairs, scatterplot.matrix, clustergram, etc. data() # precip, presidents, iris, swiss, sunspot.month (!), environmental, ethanol, ionosphere More script fragments in R will be available on the web site (http://escience.rpi.edu/data/DA )http://escience.rpi.edu/data/DA 2

3. Weighted KNN? require(kknn) data(iris) m <- dim(iris)[1] val <- sample(1:m, size = round(m/3), replace = FALSE, prob = rep(1/m, m)) iris.learn <- iris[-val,] iris.valid <- iris[val,] iris.kknn <- kknn(Species~., iris.learn, iris.valid, distance = 1, kernel = "triangular") summary(iris.kknn) fit <- fitted(iris.kknn) table(iris.valid$Species, fit) pcol <- as.character(as.numeric(iris.valid$Species)) pairs(iris.valid[1:4], pch = pcol, col = c("green3", "red”)[(iris.valid$Species != fit)+1]) 3

4. 4 Look at Lab5b_wknn_2015.R

5. Ctree > iris_ctree <- ctree(Species ~ Sepal.Length + Sepal.Width + Petal.Length + Petal.Width, data=iris) > print(iris_ctree) Conditional inference tree with 4 terminal nodes Response: Species Inputs: Sepal.Length, Sepal.Width, Petal.Length, Petal.Width Number of observations: 150 1) Petal.Length <= 1.9; criterion = 1, statistic = 140.264 2)* weights = 50 1) Petal.Length > 1.9 3) Petal.Width <= 1.7; criterion = 1, statistic = 67.894 4) Petal.Length <= 4.8; criterion = 0.999, statistic = 13.865 5)* weights = 46 4) Petal.Length > 4.8 6)* weights = 8 3) Petal.Width > 1.7 7)* weights = 46 5

6. plot(iris_ctree) 6 Try Lab6b_5_2014.R > plot(iris_ctree, type="simple”) # try this

7. Swiss - pairs 7 pairs(~ Fertility + Education + Catholic, data = swiss, subset = Education < 20, main = "Swiss data, Education < 20")

8. New dataset - ionosphere require(kknn) data(ionosphere) ionosphere.learn <- ionosphere[1:200,] ionosphere.valid <- ionosphere[-c(1:200),] fit.kknn <- kknn(class ~., ionosphere.learn, ionosphere.valid) table(ionosphere.valid$class, fit.kknn$fit) # vary kernel (fit.train1 <- train.kknn(class ~., ionosphere.learn, kmax = 15, kernel = c("triangular", "rectangular", "epanechnikov", "optimal"), distance = 1)) table(predict(fit.train1, ionosphere.valid), ionosphere.valid$class) #alter distance (fit.train2 <- train.kknn(class ~., ionosphere.learn, kmax = 15, kernel = c("triangular", "rectangular", "epanechnikov", "optimal"), distance = 2)) table(predict(fit.train2, ionosphere.valid), ionosphere.valid$class) 8

9. Results ionosphere.learn <- ionosphere[1:200,] # convenience samping!!!! ionosphere.valid <- ionosphere[-c(1:200),] fit.kknn <- kknn(class ~., ionosphere.learn, ionosphere.valid) table(ionosphere.valid$class, fit.kknn$fit) b g b 19 8 g 2 122 9

10. (fit.train1 <- train.kknn(class ~., ionosphere.learn, kmax = 15, + kernel = c("triangular", "rectangular", "epanechnikov", "optimal"), distance = 1)) Call: train.kknn(formula = class ~., data = ionosphere.learn, kmax = 15, distance = 1, kernel = c("triangular", "rectangular", "epanechnikov", "optimal")) Type of response variable: nominal Minimal misclassification: 0.12 Best kernel: rectangular Best k: 2 table(predict(fit.train1, ionosphere.valid), ionosphere.valid$class) b g b 25 4 g 2 120 10

11. (fit.train2 <- train.kknn(class ~., ionosphere.learn, kmax = 15, + kernel = c("triangular", "rectangular", "epanechnikov", "optimal"), distance = 2)) Call: train.kknn(formula = class ~., data = ionosphere.learn, kmax = 15, distance = 2, kernel = c("triangular", "rectangular", "epanechnikov", "optimal")) Type of response variable: nominal Minimal misclassification: 0.12 Best kernel: rectangular Best k: 2 table(predict(fit.train2, ionosphere.valid), ionosphere.valid$class) b g b 20 5 g 7 119 11

12. However… there is more 12

13. Bayes > cl <- kmeans(iris[,1:4], 3) > table(cl$cluster, iris[,5]) setosa versicolor virginica 2 0 2 36 1 0 48 14 3 50 0 0 # > m <- naiveBayes(iris[,1:4], iris[,5]) > table(predict(m, iris[,1:4]), iris[,5]) setosa versicolor virginica setosa 50 0 0 versicolor 0 47 3 virginica 0 3 47 13 pairs(iris[1:4],main="Iris Data (red=setosa,green=versicolor,blue=virginica)", pch=21, bg=c("red","green3","blue")[u nclass(iris$Species)])

14. Using a contingency table > data(Titanic) > mdl <- naiveBayes(Survived ~., data = Titanic) > mdl 14 Naive Bayes Classifier for Discrete Predictors Call: naiveBayes.formula(formula = Survived ~., data = Titanic) A-priori probabilities: Survived No Yes 0.676965 0.323035 Conditional probabilities: Class Survived 1st 2nd 3rd Crew No 0.08187919 0.11208054 0.35436242 0.45167785 Yes 0.28551336 0.16596343 0.25035162 0.29817159 Sex Survived Male Female No 0.91543624 0.08456376 Yes 0.51617440 0.48382560 Age Survived Child Adult No 0.03489933 0.96510067 Yes 0.08016878 0.91983122

15. Using a contingency table > predict(mdl, as.data.frame(Titanic)[,1:3]) [1] Yes No No No Yes Yes Yes Yes No No No No Yes Yes Yes Yes Yes No No No Yes Yes Yes Yes No [26] No No No Yes Yes Yes Yes Levels: No Yes 15

16. Naïve Bayes – what is it? Example: testing for a specific item of knowledge that 1% of the population has been informed of (don’t ask how). An imperfect test: –99% of knowledgeable people test positive –99% of ignorant people test negative If a person tests positive – what is the probability that they know the fact? 16

17. Naïve approach… We have 10,000 representative people 100 know the fact/item, 9,900 do not We test them all: –Get 99 knowing people testing knowing –Get 99 not knowing people testing not knowing –But 99 not knowing people testing as knowing Testing positive (knowing) – equally likely to know or not = 50% 17

18. Tree diagram 10000 ppl 1% know (100ppl) 99% test to know (99ppl) 1% test not to know (1per) 99% do not know (9900ppl) 1% test to know (99ppl) 99% test not to know (9801ppl) 18

19. Relation between probabilities For outcomes x and y there are probabilities of p(x) and p (y) that either happened If there’s a connection then the joint probability - both happen = p(x,y) Or x happens given y happens = p(x|y) or vice versa then: –p(x|y)*p(y)=p(x,y)=p(y|x)*p(x) So p(y|x)=p(x|y)*p(y)/p(x) (Bayes’ Law) E.g. p(know|+ve)=p(+ve|know)*p(know)/p(+ve)= (.99*.01)/(.99*.01+.01*.99) = 0.5 19

20. How do you use it? If the population contains x what is the chance that y is true? p(SPAM|word)=p(word|SPAM)*p(SPAM)/p(w ord) Base this on data: –p(spam) counts proportion of spam versus not –p(word|spam) counts prevalence of spam containing the ‘word’ –p(word|!spam) counts prevalence of non-spam containing the ‘word’ 20

21. Or.. What is the probability that you are in one class (i) over another class (j) given another factor (X)? Invoke Bayes: Maximize p(X|Ci)p(Ci)/p(X) (p(X)~constant and p(Ci) are equal if not known) So: conditional indep - 21

22. P(x k | C i ) is estimated from the training samples – Categorical: Estimate P(x k | C i ) as percentage of samples of class i with value x k Training involves counting percentage of occurrence of each possible value for each class –Numeric: Actual form of density function is generally not known, so “normal” density is often assumed 22

23. Digging into iris classifier<-naiveBayes(iris[,1:4], iris[,5]) table(predict(classifier, iris[,-5]), iris[,5], dnn=list('predicted','actual')) classifier$apriori classifier$tables$Petal.Length plot(function(x) dnorm(x, 1.462, 0.1736640), 0, 8, col="red", main="Petal length distribution for the 3 different species") curve(dnorm(x, 4.260, 0.4699110), add=TRUE, col="blue") curve(dnorm(x, 5.552, 0.5518947 ), add=TRUE, col = "green") 23

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25. Decision tree (example) > require(party) # don’t get me started! > str(iris) 'data.frame':150 obs. of 5 variables: $ Sepal.Length: num 5.1 4.9 4.7 4.6 5 5.4 4.6 5 4.4 4.9... $ Sepal.Width : num 3.5 3 3.2 3.1 3.6 3.9 3.4 3.4 2.9 3.1... $ Petal.Length: num 1.4 1.4 1.3 1.5 1.4 1.7 1.4 1.5 1.4 1.5... $ Petal.Width : num 0.2 0.2 0.2 0.2 0.2 0.4 0.3 0.2 0.2 0.1... $ Species : Factor w/ 3 levels "setosa","versicolor",..: 1 1 1 1 1 1 1 1 1 1... > iris_ctree <- ctree(Species ~ Sepal.Length + Sepal.Width + Petal.Length + Petal.Width, data=iris) 25

26. plot(iris_ctree) 26 Try Lab6b_5_2014.R > plot(iris_ctree, type="simple”) # try this

27. Beyond plot: pairs pairs(iris[1:4], main = "Anderson's Iris Data -- 3 species”, pch = 21, bg = c("red", "green3", "blue")[unclass(iris$Species)]) 27 Try Lab6b_2_2014.R - USJudgeRatings

28. Try hclust for iris 28

29. gpairs(iris) 29 Try Lab6b_3_2014.R

30. Better scatterplots 30 install.packages("car") require(car) scatterplotMatrix(iris) Try Lab6b_4_2014.R

31. splom(iris) # default 31 Try Lab6b_7_2014.R

32. splom extra! require(lattice) super.sym <- trellis.par.get("superpose.symbol") splom(~iris[1:4], groups = Species, data = iris, panel = panel.superpose, key = list(title = "Three Varieties of Iris", columns = 3, points = list(pch = super.sym$pch[1:3], col = super.sym$col[1:3]), text = list(c("Setosa", "Versicolor", "Virginica")))) splom(~iris[1:3]|Species, data = iris, layout=c(2,2), pscales = 0, varnames = c("Sepal\nLength", "Sepal\nWidth", "Petal\nLength"), page = function(...) { ltext(x = seq(.6,.8, length.out = 4), y = seq(.9,.6, length.out = 4), labels = c("Three", "Varieties", "of", "Iris"), cex = 2) }) parallelplot(~iris[1:4] | Species, iris) parallelplot(~iris[1:4], iris, groups = Species, horizontal.axis = FALSE, scales = list(x = list(rot = 90))) > Lab6b_7_2014.R 32

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35. Using a contingency table > data(Titanic) > mdl <- naiveBayes(Survived ~., data = Titanic) > mdl 35 Naive Bayes Classifier for Discrete Predictors Call: naiveBayes.formula(formula = Survived ~., data = Titanic) A-priori probabilities: Survived No Yes 0.676965 0.323035 Conditional probabilities: Class Survived 1st 2nd 3rd Crew No 0.08187919 0.11208054 0.35436242 0.45167785 Yes 0.28551336 0.16596343 0.25035162 0.29817159 Sex Survived Male Female No 0.91543624 0.08456376 Yes 0.51617440 0.48382560 Age Survived Child Adult No 0.03489933 0.96510067 Yes 0.08016878 0.91983122 Try Lab6b_9_2014.R

36. http://www.ugrad.stat.ubc.ca/R/library/mlb ench/html/HouseVotes84.html require(mlbench) data(HouseVotes84) model <- naiveBayes(Class ~., data = HouseVotes84) predict(model, HouseVotes84[1:10,-1]) predict(model, HouseVotes84[1:10,-1], type = "raw") pred <- predict(model, HouseVotes84[,-1]) table(pred, HouseVotes84$Class) 36

37. Exercise for you > data(HairEyeColor) > mosaicplot(HairEyeColor) > margin.table(HairEyeColor,3) Sex Male Female 279 313 > margin.table(HairEyeColor,c(1,3)) Sex Hair Male Female Black 56 52 Brown 143 143 Red 34 37 Blond 46 81 How would you construct a naïve Bayes classifier and test it? 37

38. Hierarchical clustering > d <- dist(as.matrix(mtcars)) > hc <- hclust(d) > plot(hc) 38

39. ctree 39 require(party) swiss_ctree <- ctree(Fertility ~ Agriculture + Education + Catholic, data = swiss) plot(swiss_ctree)

40. Hierarchical clustering 40 > dswiss <- dist(as.matrix(swiss)) > hs <- hclust(dswiss) > plot(hs)

41. scatterplotMatrix 41

42. require(lattice); splom(swiss) 42

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45. At this point… You may realize the inter-relation among classification at an absolute and relative level (i.e. hierarchical -> trees…) –Trees are interesting from a decision perspective: if this or that, then this…. Beyond just distance measures (kmeans) to probabilities (Bayesian) So many ways to visualize them… 45