Below is a hand coded (in R) K-Nearest Neighbor algorithm. The algorithm is built to accept any 2dim dataset and will output a label vector. I really just put this together as a way to show just how intuitive a lot of machine learning methods can be. The R code is reasonably documented, but most readers will be able to read through without documentation as everything used is base R and the implementation is very straightforward. ..just for fun. 🙂

set.seed(111) #create a basic 2 dim sample data set with four apparent cluster centers a1<-rnorm(100,.5,.2);a2<-rnorm(100,.5,.3) b1<-rnorm(100,1.5,.2);b2<-rnorm(100,.5,.3) c1<-rnorm(100,.5,.3);c2<-rnorm(100,1.5,.1) d1<-rnorm(100,1.5,.1);d2<-rnorm(100,1.5,.3) X1<-cbind(a1,a2);X2<-cbind(b1,b2);X3<-cbind(c1,c2);X4<-cbind(d1,d2) data_<-rbind(X1,X2,X3,X4) plot(data_) #add a labels column label<-rep(0,400) for (i in 1:400){ label[i]<-floor((i-1)/100) } label<-as.matrix(label) data<-cbind(data_,label) colnames(data)<-c("x","y","label") write.csv(data,file="data.csv")

Above is just code that can be used to generate a makeshift dataset with 4 apparent data centers

#Import our dataset data <- read.csv("...") set.seed(111) #create a distance matrix function dmatrix<-function(d){ n=nrow(d) dmat<-matrix(rep(0,n^2),nrow=n,ncol=n) for(i in 1:n){ for(j in 1:n){ dmat[i,j]=sqrt((data[i,2]-data[j,2])^2+(data[i,3]-data[j,3])^2) } } return(dmat) } #create a nearest neighbor ID function kn<-function(i,dmat,k=5){ x<-dmat[i,] #return the row of interest x<-order(x) #order the row return(x[2:k+1]) #return the first k entries (excluding the first) } #create a function to output predictions based on new data knn<-function(data,k=5){ n<-nrow(data) dmat<-dmatrix(data) pred<-rep(0,n) for(i in 1:n){ index<-kn(i,dmat,k=k) #extract the k nearest indices using our kn function pred[i]<-names(sort(table(label[index])))[1] } return(pred) } #run the function and assign the output to the variable x x<-knn(data) cbind(data$label,x) t<-table(data$label,x);t # x # 0 1 2 3 # 0 98 2 0 0 # 1 4 92 0 4 # 2 0 0 99 1 # 3 0 13 1 86 cat("the proportion of correct classifications is: ",(t[1,1]+t[2,2]+t[3,3]+t[4,4])/sum(t),"\n") #the proportion of correct classifications is: 0.9225

As can be seen in the preceding table, the algorithm correctly classifies most of the data points in our data set (the values on the diagonal).

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