in a textbook, there was an example i recreated. first step was to create a random scattering of two populations in such a way that there would be some chance of correlation. the second step was to use two methods of classifying the population to predict the color of other members: least squares (the line separates red and blue populations); N-n…

fake_population.R

# create random scattering of points, but cluster them together so
# that it at least looks like there are something to seperate or
# correlate. could have used rmvnorm probably, but did not know before
# i already finished.

nCenters <- 2 
nPointsToGen <- 100
nNearNeighbor <- 5

# create 2 sets of 10 points (x,y). these will be the centers of the
# clusters generated.
m <- rnorm(4 * nCenters)
dim(m) <- c(nCenters,4)

# need to choose the centers at random. we want to generate 100
# points for each color.
sb <- sample(nrow(m), nPointsToGen, replace=T)
sr <- sample(nrow(m), nPointsToGen, replace=T)

d <- matrix(0, nPointsToGen, 4)

# loop through the selected centers and generate a new random point
# close to it.
for(i in seq(sb)){
  for(j in c(1,2)){
    d[i,j] <- rnorm(1,mean=m[sb[i],j],sd=.2)
  }
  for(j in c(3,4)){
    d[i,j] <- rnorm(1,mean=m[sr[i],j],sd=.2)
  }
}

# create data frames with appropriate values (X, Y, 1, Z, color)
# for plane equation Z = aX + bY + c
t <- data.frame(d[,1],d[,2], rep(1,nrow(d)), rep(0,nrow(d)), rep("blue",nrow(d)))
s <- data.frame(d[,3],d[,4], rep(1,nrow(d)), rep(1,nrow(d)), rep("red",nrow(d)))

colnames(t) <- colnames(s) <- c("x", "y", "1", "value", "color")

# combine into one data.frame
points <- merge(t,s,all=T)

# plot them
plot(points$x, points$y, col=as.character(points$color))

# blue circles represent Z = 0 and red Z = 1
# so then we fit a plane to these points, then draw a line where
# Z = .5. this should approximately "separate" the red and blue
# circles.
X <- as.matrix(points[,1:3])
Y <- as.matrix(points[,4])
B <- solve(t(X)%*%X) %*% t(X) %*% Y

# need the equation of the line Y = bX + a
#   .5 = X B_1 + Y B_2 + B_3, solve Y
#   Y  = (-B_1/B_2) X + ((.5 - B_3)/B_2)
abline(a = (.5 - B[3,1])/B[2,1], b = -B[1,1]/B[2,1])

# nearest neighbor classifier: color the background of the plot with
# squares that identify the region.
ys <- seq(min(points$y),max(points$y),.1)
xs <- seq(min(points$x),max(points$x),.1)
hm <- matrix(0, length(ys), length(xs))

# for each (x,y), find N nearest points
N <- nNearNeighbor

for(x in xs){ for(y in ys){
  # find distance to all points
  points$tmp <- apply(points,1,function(row) ((as.numeric(row[1])-x)^2+(as.numeric(row[2])-y)^2))

  # select top N
  t <- head(points[with(points,order(tmp,x,y)),],n=N)

  # color it
  if(mean(t$value)>.5){
    points(x,y, col=rgb(255,0,0,50,maxColorValue=255), pch=15)
  }else{
    points(x,y, col=rgb(0,0,255,50,maxColorValue=255), pch=15)
  }
}}