1. 1 Peter Fox Data Analytics – ITWS-4963/ITWS-6965 Week 4b, February 20, 2015 Lab: regression, kNN and K- means results, interpreting and evaluating models

2. Classification (2) Retrieve the abalone.csv dataset Predicting the age of abalone from physical measurements. The age of abalone is determined by cutting the shell through the cone, staining it, and counting the number of rings through a microscope: a boring and time-consuming task. Other measurements, which are easier to obtain, are used to predict the age. Perform knn classification to get predictors for Age (Rings). Interpretation not required. 2

3. What did you get? See pdf 3

4. Clustering (3) The Iris dataset (in R use data(“iris”) to load it) The 5 th column is the species and you want to find how many clusters without using that information Create a new data frame and remove the fifth column Apply kmeans (you choose k) with 1000 iterations Use table(iris[,5], ) to assess results 4

5. Return object clusterA vector of integers (from 1:k) indicating the cluster to which each point is allocated. centersA matrix of cluster centres. totssThe total sum of squares. withinssVector of within-cluster sum of squares, one component per cluster. tot.withinssTotal within-cluster sum of squares, i.e., sum(withinss). betweenssThe between-cluster sum of squares, i.e. totss-tot.withinss. sizeThe number of points in each cluster. 5

6. Contingency tables See pdf file 6

7. Contingency tables > table(nyt1$Impressions,nyt1$Gender) # 0 1 1 69 85 2 389 395 3 975 937 4 1496 1572 5 1897 2012 6 1822 1927 7 1525 1696 8 1142 1203 9 722 711 10 366 400 11 214 200 12 86 101 13 41 43 14 10 9 15 5 7 16 0 4 17 0 1 7 Contingency table - displays the (multivariate) frequency distribution of the variable. Tests for significance (not now) > table(nyt1$Clicks,nyt1$Gender) 0 1 1 10335 10846 2 415 440 3 9 17

8. Regression Exercises Using the EPI dataset find the single most important factor in increasing the EPI in a given region Examine distributions down to the leaf nodes and build up an EPI “model” 8

9. Linear and least-squares > EPI_data<- read.csv(”EPI_data.csv") > attach(EPI_data) > boxplot(ENVHEALTH,DALY,AIR_H,WATER_H) > lmENVH<- lm(ENVHEALTH~DALY+AIR_H+WATER_H) > lmENVH … (what should you get?) > summary(lmENVH) … > cENVH<-coef(lmENVH) 9

10. Linear and least-squares > lmENVH<-lm(ENVHEALTH~DALY+AIR_H+WATER_H) > lmENVH Call: lm(formula = ENVHEALTH ~ DALY + AIR_H + WATER_H) Coefficients: (Intercept) DALY AIR_H WATER_H -2.673e-05 5.000e-01 2.500e-01 2.500e-01 > summary(lmENVH) … > cENVH<-coef(lmENVH) 10

11. Read the documentation! 11

12. Linear and least-squares > summary(lmENVH) Call: lm(formula = ENVHEALTH ~ DALY + AIR_H + WATER_H) Residuals: Min 1Q Median 3Q Max -0.0072734 -0.0027299 0.0001145 0.0021423 0.0055205 Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) -2.673e-05 6.377e-04 -0.042 0.967 DALY 5.000e-01 1.922e-05 26020.669 <2e-16 *** AIR_H 2.500e-01 1.273e-05 19645.297 <2e-16 *** WATER_H 2.500e-01 1.751e-05 14279.903 <2e-16 *** --- 12 p < 0.01 : very strong presumption against null hypothesis vs. this fit 0.01 < p < 0.05 : strong presumption against null hypothesis 0.05 < p < 0.1 : low presumption against null hypothesis p > 0.1 : no presumption against the null hypothesis

13. Linear and least-squares Continued: --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 Residual standard error: 0.003097 on 178 degrees of freedom (49 observations deleted due to missingness) Multiple R-squared: 1,Adjusted R-squared: 1 F-statistic: 3.983e+09 on 3 and 178 DF, p-value: < 2.2e-16 > names(lmENVH) [1] "coefficients" "residuals" "effects" "rank" "fitted.values" "assign" [7] "qr" "df.residual" "na.action" "xlevels" "call" "terms" [13] "model" 13

14. Object of class lm: An object of class "lm" is a list containing at least the following components: coefficientsa named vector of coefficients residualsthe residuals, that is response minus fitted values. fitted.valuesthe fitted mean values. rankthe numeric rank of the fitted linear model. weights(only for weighted fits) the specified weights. df.residualthe residual degrees of freedom. callthe matched call. termsthe terms object used.terms object used. contrasts(only where relevant) the contrasts used. xlevels(only where relevant) a record of the levels of the factors used in fitting. offsetthe offset used (missing if none were used). yif requested, the response used. xif requested, the model matrix used. modelif requested (the default), the model frame used. 14

15. > plot(ENVHEALTH,c ol="red") > points(lmENVH$fitte d.values,col="blue") > Huh? 15 Plot original versus fitted

16. Try again! 16 > plot(ENVHEALTH[!is.na(ENVHEALTH)], col="red") > points(lmENVH$fitted.values,col="blue")

17. Predict > cENVH<- coef(lmENVH) > DALYNEW<- c(seq(5,95,5)) #2 > AIR_HNEW<- c(seq(5,95,5)) #3 > WATER_HNEW<- c(seq(5,95,5)) #4 17

18. Predict > NEW<- data.frame(DALYNEW,AIR_HNEW,WATER_H NEW) > pENV<- predict(lmENVH,NEW,interval=“prediction”) > cENV<- predict(lmENVH,NEW,interval=“confidence”) # look up what this does 18

19. Predict object returns predict.lm produces a vector of predictions or a matrix of predictions and bounds with column names fit, lwr, and upr if interval is set. Access via [,1] etc. If se.fit is TRUE, a list with the following components is returned: fitvector or matrix as above se.fitstandard error of predicted means residual.scaleresidual standard deviations dfdegrees of freedom for residual 19

20. Output from predict > head(pENV) fit lwr upr 1 NA NA NA 2 11.55213 11.54591 11.55834 3 18.29168 18.28546 18.29791 4 NA NA NA 5 69.92533 69.91915 69.93151 6 90.20589 90.19974 90.21204 … 20

21. > tail(pENV) fit lwr upr 226 NA NA NA 227 NA NA NA 228 34.95256 34.94641 34.95871 229 59.00213 58.99593 59.00834 230 24.20951 24.20334 24.21569 231 38.03701 38.03084 38.04319 21

22. Read the documentation! 22

23. Classification Exercises (Lab3b_knn1.R) > nyt1<-read.csv(“nyt1.csv") > nyt1 0 & nyt1$Clicks>0 & nyt1$Age>0),] > nnyt1<-dim(nyt1)[1]# shrink it down! > sampling.rate=0.9 > num.test.set.labels=nnyt1*(1.-sampling.rate) > training <-sample(1:nnyt1,sampling.rate*nnyt1, replace=FALSE) > train<-subset(nyt1[training,],select=c(Age,Impressions)) > testing<-setdiff(1:nnyt1,training) > test<-subset(nyt1[testing,],select=c(Age,Impressions)) > cg<-nyt1$Gender[training] > true.labels<-nyt1$Gender[testing] > classif<-knn(train,test,cg,k=5) # > classif > attributes(.Last.value) # interpretation to come! 23

24. K Nearest Neighbors (classification) Script – Lab3b_knn1_2015.R > nyt1<-read.csv(“nyt1.csv") … from week 3b slides or script > classif<-knn(train,test,cg,k=5) # > head(true.labels) [1] 1 0 0 1 1 0 > head(classif) [1] 1 1 1 1 0 0 Levels: 0 1 > ncorrect<-true.labels==classif > table(ncorrect)["TRUE"]# or > length(which(ncorrect)) > What do you conclude? 24

25. Classification Exercises (Lab3b_knn2_2015.R) 2 examples in the script 25

26. Clustering Exercises Lab3b_kmeans1.R Lab3b_kmeans2.R – plotting up results from the iris clustering 26

27. Regression > bronx<- read.xlsx(”sales/rollingsales_bronx.xls",pattern ="BOROUGH",stringsAsFactors=FALSE,sheetI ndex=1,startRow=5,header=TRUE) > plot(log(bronx$GROSS.SQUARE.FEET), log(bronx$SALE.PRICE) ) > m1<- lm(log(bronx$SALE.PRICE)~log(bronx$GROS S.SQUARE.FEET),data=bronx)  What’s wrong? 27

28. Clean up… > bronx 0 & bronx$LAND.SQUARE.FEET>0 & bronx$SALE.PRICE>0),] > m1<- lm(log(bronx$SALE.PRICE)~log(bronx$GROS S.SQUARE.FEET),data=bronx) # > summary(m1) 28

29. Call: lm(formula = log(SALE.PRICE) ~ log(GROSS.SQUARE.FEET), data = bronx) Residuals: Min 1Q Median 3Q Max -14.4529 0.0377 0.4160 0.6572 3.8159 Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 7.0271 0.3088 22.75 <2e-16 *** log(GROSS.SQUARE.FEET) 0.7013 0.0379 18.50 <2e-16 *** --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 Residual standard error: 1.95 on 2435 degrees of freedom Multiple R-squared: 0.1233, Adjusted R-squared: 0.1229 F-statistic: 342.4 on 1 and 2435 DF, p-value: < 2.2e-16 29

30. Plot > plot(log(bronx$GROSS.SQUARE.FEET), log(bronx$SALE.PRICE)) > abline(m1,col="red",lwd=2) # then > plot(resid(m1)) 30

31. Another model (2)? Add two more variables to the linear model LAND.SQUARE.FEET and NEIGHBORHOOD Repeat but suppress the intercept (2a) 31

32. Model 3/4 Model 3 Log(SALE.PRICE) vs. no intercept Log(GROSS.SQUARE.FEET), Log(LAND.SQUARE.FEET), NEIGHBORHOOD, BUILDING.CLASS.CATEGORY Model 4 Log(SALE.PRICE) vs. no intercept Log(GROSS.SQUARE.FEET), Log(LAND.SQUARE.FEET), NEIGHBORHOOD*BUILDING.CLASS.CATEG ORY 32

33. Solution model 2 > m2<- lm(log(bronx$SALE.PRICE)~log(bronx$GROSS.SQUARE.FEE T)+log(bronx$LAND.SQUARE.FEET)+factor(bronx$NEIGHBO RHOOD),data=bronx) > summary(m2) > plot(resid(m2)) # > m2a<- lm(log(bronx$SALE.PRICE)~0+log(bronx$GROSS.SQUARE.F EET)+log(bronx$LAND.SQUARE.FEET)+factor(bronx$NEIGH BORHOOD),data=bronx) > summary(m2a) > plot(resid(m2a)) 33

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35. Solution model 3 and 4 > m3<- lm(log(bronx$SALE.PRICE)~0+log(bronx$GROSS.SQUARE.F EET)+log(bronx$LAND.SQUARE.FEET)+factor(bronx$NEIGH BORHOOD)+factor(bronx$BUILDING.CLASS.CATEGORY),dat a=bronx) > summary(m3) > plot(resid(m3)) # > m4<- lm(log(bronx$SALE.PRICE)~0+log(bronx$GROSS.SQUARE.F EET)+log(bronx$LAND.SQUARE.FEET)+factor(bronx$NEIGH BORHOOD)*factor(bronx$BUILDING.CLASS.CATEGORY),dat a=bronx) > summary(m4) > plot(resid(m4)) 35

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37. Assignment 3 Preliminary and Statistical Analysis. Due ~ March 6. 15% (written) –Distribution analysis and comparison, visual ‘analysis’, statistical model fitting and testing of some of the nyt1…31 datasets. 37