Georgia Tech ISYE6501 Week 2 HW, 100% pass rate,

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ISYE6501 Week 2 HW Question 4.1 At my job we use clustering in an image system to understand the minerals that are coming into our processing facility. Comparing the images to known mineral images, the system can quantify the mineral content and monitor the colors of the minerals to compare to centroids of the clusters. Question 4.2 library(kernlab) library(ggplot2) ## Warning: package 'ggplot2' was built under R version 3.6.3 ## ## Attaching package: 'ggplot2' ## The following object is masked from 'package:kernlab': ## ## alpha library(kknn) ## Warning: package 'kknn' was built under R version 3.6.3 data<-read.table("C:\\Users\\Colin Shumaker\\Desktop\\ISYE6501\\Week 2\\data 4.2\\iri s.txt", header = TRUE) data1 <-data[,1:4] head(data1) ## Sepal.Length Sepal.Width Petal.Length Petal.Width ## 1 5.1 3.5 1.4 0.2 ## 2 4.9 3.0 1.4 0.2 ## 3 4.7 3.2 1.3 0.2 ## 4 4.6 3.1 1.5 0.2 ## 5 5.0 3.6 1.4 0.2 ## 6 5.4 3.9 1.7 0.4species <-data[,5] kvalues <- rep(0,10) kx <- 1:10 set.seed(1000) for (r in 1:10){ ktest <- kmeans(data1,r,nstart=5,iter.max = 10) kx[r]<-r kvalues[r]<-ktest$tot.withins } plot(kx,kvalues,type="b",xlab="Number of Clusters", ylab="Total Distance") set.seed(4) k <- kmeans(data1,3) cat("Predictors 1,2,3,4","\n") ## Predictors 1,2,3,4 table(k$cluster,species)## species ## setosa versicolor virginica ## 1 0 48 14 ## 2 0 2 36 ## 3 50 0 0 set.seed(5) k1 <- kmeans(data1[,1:3],3) cat("Predictors 1,2,3","\n") ## Predictors 1,2,3 table(k1$cluster,species) ## species ## setosa versicolor virginica ## 1 0 45 13 ## 2 50 0 0 ## 3 0 5 37 set.seed(6) k2 <- kmeans(data1[,2:4],3) cat("Predictors 2,3,4","\n") ## Predictors 2,3,4 table(k2$cluster,species) ## species ## setosa versicolor virginica ## 1 0 48 5 ## 2 50 0 0 ## 3 0 2 45 set.seed(7) k3 <- kmeans(data1[,c(1,2,4)],3) cat("Predictors 1,2,4","\n") ## Predictors 1,2,4table(k3$cluster,species) ## species ## setosa versicolor virginica ## 1 0 39 15 ## 2 0 11 35 ## 3 50 0 0 set.seed(8) k4 <- kmeans(data1[,c(1,3,4)],3) cat("Predictors 1,3,4","\n") ## Predictors 1,3,4 table(k4$cluster,species) ## species ## setosa versicolor virginica ## 1 50 0 0 ## 2 0 2 36 ## 3 0 48 14 set.seed(9) k5 <- kmeans(data1[,3:4],3) cat("Predictors 3,4","\n") ## Predictors 3,4 table(k5$cluster,species) ## species ## setosa versicolor virginica ## 1 0 2 46 ## 2 50 0 0 ## 3 0 48 4 ggplot(data,aes(x=Petal.Length,y=Petal.Width, col=Species))+ geom_point()# best k value is 3 based on the plot to find the best combination of predictors. The best accuracy is 96% showing that the best combination of predictors is 3 and 4. Evalu ating the clusters using petal.length and petal.width shows the best response for spec ies Question 5.1 library(outliers) data1<-read.table("C:\\Users\\Colin Shumaker\\Desktop\\ISYE6501\\Week 2\\data 5.1\\usc rime.txt", header = TRUE) crimesort <- t(data1$Crime[order(data1$Crime)]) plot(seq(1:length(crimesort)),crimesort,xlab="Sort", ylab="Crime Rates")grubbs.test(crimesort, type=10) ## ## Grubbs test for one outlier ## ## data: crimesort ## G = 2.81287, U = 0.82426, p-value = 0.07887 ## alternative hypothesis: highest value 1993 is an outlier

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