Related
I generate 4 parts of big data: cluster1(10000 points), cluster2(15000 points), cluster3(15000 points) and throws(500 points). Here is the code:
library('MASS')
library('fpc')
#library("dbscan")
library("factoextra")
library("clustertend")
library("boot")
library("stream")
set.seed(123)
mu1<-c(-5,-7)
mu1
sigma1<-matrix(c(4,-2,-2,2), nrow=2, ncol=2, byrow = TRUE)
sigma1
n<-10000
cluster1<-mvrnorm(n,mu1,sigma1)
cluster1
#cluster1<-as.data.frame(cluster1)
#cluster1
#c<-runif(10000,1,1000)
#c
phi <- runif(15000, max = 2*pi)
rho <- sqrt(runif(15000))
x <- sqrt(5)*rho*cos(phi) + 6
y <- sqrt(10/3)*rho*sin(phi) + 4
range(2*(x - 6)^2 + 3*(y - 4)^2)
#[1] 0.001536582 9.999425234
plot(x, y)
cluster2<-cbind(x,y)
cluster2
u <- runif(15000, max = 3)
v <- runif(15000, max = 2)
x <- u + v - 10
y <- v - u + 8
range(x + y)
#[1] -1.999774 1.999826
range(x - y + 15)
#[1] -2.999646 2.999692
plot(x, y)
cluster3<-cbind(x,y)
cluster3
#cluster3<-as.data.frame(cluster1)
#cluster3
x <- runif(500, -20, 20)
y <- runif(500, -20, 20)
#u <- runif(500, max = 20)
#v <- runif(500, max = 20)
#x <- u + v - 20
#y <- v - u
range(x)
range(y)
plot(x,y)
throws<-cbind(x,y)
throws
data<-rbind(cluster1,cluster2,cluster3,throws)
data<-as.data.frame(data)
data
plot(data)
Then I try by using the bootstrap method, construct a distribution of H statistics for some
fixed m, which is from 7% of the total number of generated points(m=2835). Here is th code where I do this:
B<-10#number of iterations
H<-NULL#value of Hopkins statistic
for(i in 1:B){
N<-dim(data)[1]
s<-sample(N,0.8*N)
stat<-hopkins(data[s,], n=2835, byrow = TRUE)$H
H[i]<-stat
#print(c(i, stat))
}
It takes very to generate. Then I should to compare this result with beta distribution - B(m,m). Here is the code:
hist(H)
#(density(H), col="red")
#hist(distB)
X<-seq(min(H), max(H), 0.001)
X
lines(X, dbeta(X,2835,2835), type="l", col="red")
The problem is that lined doesn't draw on hist. Can anybody say what is the problem? Here is the image, I see red line, but it's not exactly right.
Your y-axis values plotted by dbeta() are way too low to register on the supplied y-axis (<0.0000001). You need to overlay the second plot:
# sample data
H <- sample(seq(0.455,0.475,0.001), 1000, replace = TRUE)
#plot histogram
hist(H)
# prepare graphics to add second plot
par(new = TRUE)
# sample data for second plot
X <- seq(0.455,0.475, 0.001)
Y <- dbeta(X,2835,2835)
# plot second plot, remove axes
plot(X, dbeta(X,2835,2835), type="l", col="red", axes = FALSE)
axis(4, Y) # add axis on right side
f(x,y)= (1/25)*(20-x)/x 10<x<20, x/2 <y <x
0 o.t
I have to create this image through this expression.
but
x <- seq(10, 20, length=20)
y <- seq(10, 20, length=20)
f <- function(x,y){(1/25)*(20-x)/5}
z <- outer(x,y,f)
persp(x,y,z,theta=30,phi=30, expand=0.5,col=rainbow(19), border=NA)
what is wrong?
You should mask z based on the constraint. As a suggestion, you can use an amazing interactive rgl package in R.
#source: https://stackoverflow.com/questions/50079316/plot3d-how-to-change-z-axis-surface-color-to-heat-map-color
map2color <- function(x, pal, limits = range(x,na.rm=T)){
pal[findInterval(x, seq(limits[1], limits[2], length.out = length(pal) + 1),
all.inside=TRUE)]
}
x <- seq(10, 20, length=20)
y <- seq(10, 20, length=20)
mask <- sapply(x,function(m) sapply(y,function(n) if((n>m/2)&(n<m)){(1/25)*(20-m)/5}else{ NA }))
z <- outer(x,y,f)
z <- z * mask
#persp(x,y,z, col= map2color(z, rainbow(100)),border = NA)
library(rgl)
persp3d(x,y,z,col = map2color(z, rainbow(100)),theta=30,phi=30)
#SRhm's answer is probably the best choice, but if you want to live on the bleeding edge, you can get rid of the jagged diagonal edge using a development version of rgl (from R-forge), at least version 0.100.8.
This version supports triangulations with boundaries using the tripack package. So you set up a grid of values over the x-y range, then define the boundaries of the region using the equations, and you get smooth edges. For example:
library(tripack)
library(rgl)
g <- expand.grid(x=10:20, y=5:20)
keep <- with(g, 10 < x & x < 20 & x/2 < y & y < x)
g2 <- g[keep,]
tri <- tri.mesh(g2)
# Set up boundary constraints
cx <- c(10:20, 20: 10)
cy <- c(seq(5, 10, len=11), 20:10)
tri2 <- add.constraint(tri, cx, cy, reverse = TRUE)
# This isn't necessary, but shows where the formula will be evaluated
plot(tri2)
It might be better to fill in some of the left and right edges with more points to avoid those big triangles,
but skip that for now.
z <- with(tri2, (1/25)*(20-x)/x)
# Now plot it, using the map2color function #SRhm found:
#source: https://stackoverflow.com/questions/50079316/plot3d-how-to-change-z-axis-surface-color-to-heat-map-color
map2color <- function(x, pal, limits = range(x,na.rm=T)){
pal[findInterval(x, seq(limits[1], limits[2], length.out = length(pal) + 1),
all.inside=TRUE)]
}
persp3d(tri2, z, col = map2color(z, rainbow(100)))
After rotation, you get this view:
I have a question about fitting ellipses to data with the ellipse center at the origin. I have explored two methods that fit ellipses but generate an arbitrary center unless I manipulate the data with some imaginary mirror points.
Method#01
This portion of the script directly comes from this useful post. I'm copying the codes directly here for ease.
fit.ellipse <- function (x, y = NULL) {
# from:
# http://r.789695.n4.nabble.com/Fitting-a-half-ellipse-curve-tp2719037p2720560.html
#
# Least squares fitting of an ellipse to point data
# using the algorithm described in:
# Radim Halir & Jan Flusser. 1998.
# Numerically stable direct least squares fitting of ellipses.
# Proceedings of the 6th International Conference in Central Europe
# on Computer Graphics and Visualization. WSCG '98, p. 125-132
#
# Adapted from the original Matlab code by Michael Bedward (2010)
# michael.bedward#gmail.com
#
# Subsequently improved by John Minter (2012)
#
# Arguments:
# x, y - x and y coordinates of the data points.
# If a single arg is provided it is assumed to be a
# two column matrix.
#
# Returns a list with the following elements:
#
# coef - coefficients of the ellipse as described by the general
# quadratic: ax^2 + bxy + cy^2 + dx + ey + f = 0
#
# center - center x and y
#
# major - major semi-axis length
#
# minor - minor semi-axis length
#
EPS <- 1.0e-8
dat <- xy.coords(x, y)
D1 <- cbind(dat$x * dat$x, dat$x * dat$y, dat$y * dat$y)
D2 <- cbind(dat$x, dat$y, 1)
S1 <- t(D1) %*% D1
S2 <- t(D1) %*% D2
S3 <- t(D2) %*% D2
T <- -solve(S3) %*% t(S2)
M <- S1 + S2 %*% T
M <- rbind(M[3,] / 2, -M[2,], M[1,] / 2)
evec <- eigen(M)$vec
cond <- 4 * evec[1,] * evec[3,] - evec[2,]^2
a1 <- evec[, which(cond > 0)]
f <- c(a1, T %*% a1)
names(f) <- letters[1:6]
# calculate the center and lengths of the semi-axes
#
# see http://www.ncbi.nlm.nih.gov/pmc/articles/PMC2288654/
# J. R. Minter
# for the center, linear algebra to the rescue
# center is the solution to the pair of equations
# 2ax + by + d = 0
# bx + 2cy + e = 0
# or
# | 2a b | |x| |-d|
# | b 2c | * |y| = |-e|
# or
# A x = b
# or
# x = Ainv b
# or
# x = solve(A) %*% b
A <- matrix(c(2*f[1], f[2], f[2], 2*f[3]), nrow=2, ncol=2, byrow=T )
b <- matrix(c(-f[4], -f[5]), nrow=2, ncol=1, byrow=T)
soln <- solve(A) %*% b
b2 <- f[2]^2 / 4
center <- c(soln[1], soln[2])
names(center) <- c("x", "y")
num <- 2 * (f[1] * f[5]^2 / 4 + f[3] * f[4]^2 / 4 + f[6] * b2 - f[2]*f[4]*f[5]/4 - f[1]*f[3]*f[6])
den1 <- (b2 - f[1]*f[3])
den2 <- sqrt((f[1] - f[3])^2 + 4*b2)
den3 <- f[1] + f[3]
semi.axes <- sqrt(c( num / (den1 * (den2 - den3)), num / (den1 * (-den2 - den3)) ))
# calculate the angle of rotation
term <- (f[1] - f[3]) / f[2]
angle <- atan(1 / term) / 2
list(coef=f, center = center, major = max(semi.axes), minor = min(semi.axes), angle = unname(angle))
}
Let's take a example distribution of polar points for illustration purpose
X<-structure(list(x_polar = c(0, 229.777200000011, 246.746099999989,
-10.8621999999741, -60.8808999999892, 75.8904999999795, -83.938199999975,
-62.9770000000135, 49.1650999999838, 52.3093000000226, 49.6891000000178,
-66.4248999999836, 34.3671999999788, 242.386400000018, 343.60619999998
), y_polar = c(0, 214.868299999973, 161.063599999994, -68.8972000000067,
-77.0230000000447, 93.2863000000361, -16.2356000000145, 27.7828000000445,
-17.8077000000048, 2.10540000000037, 25.6866000000155, -84.6034999999683,
-31.1800000000512, 192.010800000047, 222.003700000001)), .Names = c("x_polar",
"y_polar"), row.names = c(NA, -15L), class = "data.frame")
efit <- fit.ellipse(X)
e <- get.ellipse(efit)
#plot
par(bg=NA)
plot(X, pch=3, col='gray', lwd=2, axes=F, xlab="", ylab="", type='n',
ylim=c(min(X$y_polar)-150, max(X$y_polar)), xlim=c(min(X$x_polar)-150, max(X$x_polar))) #blank plot
points(X$x_polar, X$y_polar, pch=3, col='gray', lwd=2, axes=F, xlab="", ylab="") #observations
lines(e, col="red", lwd=3, lty=2) #plotting the ellipse
points(0,0,col=2, lwd=2, cex=2) #center/origin
To bring the origin of the ellipse at the center we could modify as follows (surely not the best way of doing it)
#generate mirror coordinates
X$x_polar_mirror<- -X$x_polar
X$y_polar_mirror<- -X$y_polar
mydata<-as.matrix(data.frame(c(X$x_polar, X$x_polar_mirror), c(X$y_polar, X$y_polar_mirror)))
#fit the data
efit <- fit.ellipse(mydata)
e <- get.ellipse(efit)
par(bg=NA)
plot(mydata, pch=3, col='gray', lwd=2, axes=F, xlab="", ylab="", type='n',
ylim=c(min(X$y_polar)-150, max(X$y_polar)), xlim=c(min(X$x_polar)-150, max(X$x_polar)))
points(X$x_polar, X$y_polar, pch=3, col='gray', lwd=2, axes=F, xlab="", ylab="")
lines(e, col="red", lwd=3, lty=2)
points(0,0,col=2, lwd=2, cex=2) #center
Well ... it sort of does the job but none would be happy with all those imaginary points considered in the calculation.
Method#02
This is another indirect way of fitting the data but again the ellipse center is not at the origin. Any workaround?
require(car)
dataEllipse(X$x_polar, X$y_polar, levels=c(0.15, 0.7),
xlim=c(-150, 400), ylim=c(-200,300))
My questions: (a) is there a robust alternative way of fitting these points with the ellipse center at the origin (0,0)? (b) is there a measure of the goodness of ellipse fit? Thank you in advance.
I'm not really happy with aproach I've concieved, there should be a closed form solution, but still:
# Ellipse equasion with center in (0, 0) with semiaxis pars[1] and pars[2] rotated by pars[3].
# t and pars[3] in radians
ellipsePoints <- function(t, pars) {
data.frame(x = cos(pars[3]) * pars[1] * cos(t) - sin(pars[3]) * pars[2] * sin(t),
y = sin(pars[3]) * pars[1] * cos(t) + cos(pars[3]) * pars[2] * sin(t))
}
# Way to fit an ellipse through minimising distance to data points.
# If weighted then points which are most remote from center will have bigger impact.
ellipseBrute <- function(x, y, pars, weighted = FALSE) {
d <- sqrt(x**2 + y**2)
t <- asin(y/d)
w <- (d/sum(d))**weighted
t[x == 0 & y == 0] <- 0
ep <- ellipsePoints(t, pars)
sum(w*(sqrt(ep$x**2 + ep$y**2) - d)**2)
}
# Fit through optim.
opt_res <- optim(c(diff(range(X$x_polar)),
diff(range(X$y_polar)),
2*pi)/2,
ellipseBrute,
x = X$x_polar, y = X$y_polar,
weighted = TRUE
)
# Check resulting ellipse throuh plot
df <- ellipsePoints(seq(0, 2*pi, length.out = 1e3), opt_res$par)
plot(y ~ x, df, col = 'blue', t = 'l',
xlim = range(c(X$x_polar, df$x)),
ylim = range(c(X$y_polar, df$y)))
points(0, 0, pch = 3, col = 'blue')
points(y_polar ~ x_polar, X)
Could you give me an example on how to use rgl to plot 3 variables at the axes x, y and z and a fourth one with different colours?
thanks
You use a combination of persp and colour according to a separate function. Here's some example code:
## Create a simple surface f(x,y) = -x^2 - y^2
## Colour the surface according to x^2 only
nx = 31; ny = 31
x = seq(-1, 1, length = nx)
y = seq(-1, 1, length = ny)
z = outer(x, y, function(x,y) -x^2 -y^2)
## Fourth dim
z_col = outer(x, y, function(x,y) x^2)
## Average the values at the corner of each facet
## and scale to a value in [0, 1]. We will use this
## to select a gray for colouring the facet.
hgt = 0.25 * (z_col[-nx,-ny] + z_col[-1,-ny] + z_col[-nx,-1] + z_col[-1,-1])
hgt = (hgt - min(hgt))/ (max(hgt) - min(hgt))
## Plot the surface with the specified facet colours.
persp(x, y, z, col = gray(1 - hgt))
persp(x, y, z, col=cm.colors(32)[floor(31*hgt+1)], theta=-35, phi=10)
This gives:
RGL
It's fairly straightforward to use the above technique with the rgl library:
library(rgl)
## Generate the data using the above commands
## New window
open3d()
## clear scene:
clear3d("all")
## setup env:
bg3d(color="#887777")
light3d()
surface3d(x, y, z, color=cm.colors(32)[floor(31*hgt+1)], alpha=0.5)
There is an example in ?plot3d if you are talking about plotting points in a 3d space and colouring them:
x <- sort(rnorm(1000))
y <- rnorm(1000)
z <- rnorm(1000) + atan2(x,y)
plot3d(x, y, z, col=rainbow(1000))
But if you mean to colour the points by a 4th variable, say a grouping variable, then we can modify the example above to do this by creating a grouping variable
grp <- gl(5, 200) ## 5 groups 200 members each
## now select the colours we want
cols <- 1:5
## Now plot
plot3d(x, y, z, col=cols[grp])
OK, is this more what you want?
X <- 1:10
Y <- 1:10
## Z is now a 100 row object of X,Y combinations
Z <- expand.grid(X = X, Y = Y)
## Add in Z1, which is the 3rd variable
## X,Y,Z1 define the surface, which we colour according to
## 4th variable Z2
Z <- within(Z, {
Z1 <- 1.2 + (1.4 * X) + (-1.9 * Y)
Z2 <- 1.2 + (1.4 * X) - (1.2 * X^2) + (1.9 * Y) + (-1.3 * Y^2)
Z3 <- 1.2 + (1.4 * X) + (-1.9 * Y) + (-X^2) + (-Y^2)})
## show the data
head(Z)
## Set-up the rgl device
with(Z, plot3d(X, Y, Z1, type = "n"))
## Need a scale for Z2 to display as colours
## Here I choose 10 equally spaced colours from a palette
cols <- heat.colors(10)
## Break Z2 into 10 equal regions
cuts <- with(Z, cut(Z2, breaks = 10))
## Add in the surface, colouring by Z2
with(Z, surface3d(1:10,1:10, matrix(Z1, ncol = 10),
color = cols[cuts], back = "fill"))
with(Z, points3d(X, Y, Z1, size = 5)) ## show grid X,Y,Z1
Here's a modification where the plane surface Z1 is curved (Z3).
## Set-up the rgl device plotting Z3, a curved surface
with(Z, plot3d(X, Y, Z3, type = "n"))
with(Z, surface3d(1:10,1:10, matrix(Z3, ncol = 10),
color = cols[cuts], back = "fill"))
The detail of what I did to get Z2 probably doesn't matter, but I tried to get something like the graph you linked to.
If I've still not got what you want, can you edit your Q with some example data and give us a better idea of what you want?
HTH
Take a look at example(points3d).
The r3d help page shows you how to draw axes.
x <- c(0, 10, 0, 0)
y <- c(0, 0, 100, 0)
z <- c(0, 0, 0, 1)
i <- c(1,2,1,3,1,4)
labels <- c("Origin", "X", "Y", "Z")
text3d(x,y,z,labels)
segments3d(x[i],y[i],z[i])
Now you add some points
dfr <- data.frame(x = 1:10, y = (1:10)^2, z = runif(10), col = rainbow(10))
with(dfr, points3d(x, y, z, col = col))
I'm trying to perfect a method for comparing regression and PCA, inspired by the blog Cerebral Mastication which has also has been discussed from a different angle on SO. Before I forget, many thanks to JD Long and Josh Ulrich for much of the core of this. I'm going to use this in a course next semester. Sorry this is long!
UPDATE: I found a different approach which almost works (please fix it if you can!). I posted it at the bottom. A much smarter and shorter approach than I was able to come up with!
I basically followed the previous schemes up to a point: Generate random data, figure out the line of best fit, draw the residuals. This is shown in the second code chunk below. But I also dug around and wrote some functions to draw lines normal to a line through a random point (the data points in this case). I think these work fine, and they are shown in First Code Chunk along with proof they work.
Now, the Second Code Chunk shows the whole thing in action using the same flow as #JDLong and I'm adding an image of the resulting plot. Data in black, red is the regression with residuals pink, blue is the 1st PC and the light blue should be the normals, but obviously they are not. The functions in First Code Chunk that draw these normals seem fine, but something is not right with the demonstration: I think I must be misunderstanding something or passing the wrong values. My normals come in horizontal, which seems like a useful clue (but so far, not to me). Can anyone see what's wrong here?
Thanks, this has been vexing me for a while...
First Code Chunk (Functions to Draw Normals and Proof They Work):
##### The functions below are based very loosely on the citation at the end
pointOnLineNearPoint <- function(Px, Py, slope, intercept) {
# Px, Py is the point to test, can be a vector.
# slope, intercept is the line to check distance.
Ax <- Px-10*diff(range(Px))
Bx <- Px+10*diff(range(Px))
Ay <- Ax * slope + intercept
By <- Bx * slope + intercept
pointOnLine(Px, Py, Ax, Ay, Bx, By)
}
pointOnLine <- function(Px, Py, Ax, Ay, Bx, By) {
# This approach based upon comingstorm's answer on
# stackoverflow.com/questions/3120357/get-closest-point-to-a-line
# Vectorized by Bryan
PB <- data.frame(x = Px - Bx, y = Py - By)
AB <- data.frame(x = Ax - Bx, y = Ay - By)
PB <- as.matrix(PB)
AB <- as.matrix(AB)
k_raw <- k <- c()
for (n in 1:nrow(PB)) {
k_raw[n] <- (PB[n,] %*% AB[n,])/(AB[n,] %*% AB[n,])
if (k_raw[n] < 0) { k[n] <- 0
} else { if (k_raw[n] > 1) k[n] <- 1
else k[n] <- k_raw[n] }
}
x = (k * Ax + (1 - k)* Bx)
y = (k * Ay + (1 - k)* By)
ans <- data.frame(x, y)
ans
}
# The following proves that pointOnLineNearPoint
# and pointOnLine work properly and accept vectors
par(mar = c(4, 4, 4, 4)) # otherwise the plot is slightly distorted
# and right angles don't appear as right angles
m <- runif(1, -5, 5)
b <- runif(1, -20, 20)
plot(-20:20, -20:20, type = "n", xlab = "x values", ylab = "y values")
abline(b, m )
Px <- rnorm(10, 0, 4)
Py <- rnorm(10, 0, 4)
res <- pointOnLineNearPoint(Px, Py, m, b)
points(Px, Py, col = "red")
segments(Px, Py, res[,1], res[,2], col = "blue")
##========================================================
##
## Credits:
## Theory by Paul Bourke http://local.wasp.uwa.edu.au/~pbourke/geometry/pointline/
## Based in part on C code by Damian Coventry Tuesday, 16 July 2002
## Based on VBA code by Brandon Crosby 9-6-05 (2 dimensions)
## With grateful thanks for answering our needs!
## This is an R (http://www.r-project.org) implementation by Gregoire Thomas 7/11/08
##
##========================================================
Second Code Chunk (Plots the Demonstration):
set.seed(55)
np <- 10 # number of data points
x <- 1:np
e <- rnorm(np, 0, 60)
y <- 12 + 5 * x + e
par(mar = c(4, 4, 4, 4)) # otherwise the plot is slightly distorted
plot(x, y, main = "Regression minimizes the y-residuals & PCA the normals")
yx.lm <- lm(y ~ x)
lines(x, predict(yx.lm), col = "red", lwd = 2)
segments(x, y, x, fitted(yx.lm), col = "pink")
# pca "by hand"
xyNorm <- cbind(x = x - mean(x), y = y - mean(y)) # mean centers
xyCov <- cov(xyNorm)
eigenValues <- eigen(xyCov)$values
eigenVectors <- eigen(xyCov)$vectors
# Add the first PC by denormalizing back to original coords:
new.y <- (eigenVectors[2,1]/eigenVectors[1,1] * xyNorm[x]) + mean(y)
lines(x, new.y, col = "blue", lwd = 2)
# Now add the normals
yx2.lm <- lm(new.y ~ x) # zero residuals: already a line
res <- pointOnLineNearPoint(x, y, yx2.lm$coef[2], yx2.lm$coef[1])
points(res[,1], res[,2], col = "blue", pch = 20) # segments should end here
segments(x, y, res[,1], res[,2], col = "lightblue1") # the normals
############ UPDATE
Over at Vincent Zoonekynd's Page I found almost exactly what I wanted. But, it doesn't quite work (obviously used to work). Here is a code excerpt from that site which plots normals to the first PC reflected through a vertical axis:
set.seed(1)
x <- rnorm(20)
y <- x + rnorm(20)
plot(y~x, asp = 1)
r <- lm(y~x)
abline(r, col='red')
r <- princomp(cbind(x,y))
b <- r$loadings[2,1] / r$loadings[1,1]
a <- r$center[2] - b * r$center[1]
abline(a, b, col = "blue")
title(main='Appears to use the reflection of PC1')
u <- r$loadings
# Projection onto the first axis
p <- matrix( c(1,0,0,0), nrow=2 )
X <- rbind(x,y)
X <- r$center + solve(u, p %*% u %*% (X - r$center))
segments( x, y, X[1,], X[2,] , col = "lightblue1")
And here is the result:
Alright, I'll have to answer my own question! After further reading and comparison of methods that people have put on the internet, I have solved the problem. I'm not sure I can clearly state what I "fixed" because I went through quite a few iterations. Anyway, here is the plot and the code (MWE). The helper functions are at the end for clarity.
# Comparison of Linear Regression & PCA
# Generate sample data
set.seed(39) # gives a decent-looking example
np <- 10 # number of data points
x <- -np:np
e <- rnorm(length(x), 0, 10)
y <- rnorm(1, 0, 2) * x + 3*rnorm(1, 0, 2) + e
# Plot the main data & residuals
plot(x, y, main = "Regression minimizes the y-residuals & PCA the normals", asp = 1)
yx.lm <- lm(y ~ x)
lines(x, predict(yx.lm), col = "red", lwd = 2)
segments(x, y, x, fitted(yx.lm), col = "pink")
# Now the PCA using built-in functions
# rotation = loadings = eigenvectors
r <- prcomp(cbind(x,y), retx = TRUE)
b <- r$rotation[2,1] / r$rotation[1,1] # gets slope of loading/eigenvector 1
a <- r$center[2] - b * r$center[1]
abline(a, b, col = "blue") # Plot 1st PC
# Plot normals to 1st PC
X <- pointOnLineNearPoint(x, y, b, a)
segments( x, y, X[,1], X[,2], col = "lightblue1")
###### Needed Functions
pointOnLineNearPoint <- function(Px, Py, slope, intercept) {
# Px, Py is the point to test, can be a vector.
# slope, intercept is the line to check distance.
Ax <- Px-10*diff(range(Px))
Bx <- Px+10*diff(range(Px))
Ay <- Ax * slope + intercept
By <- Bx * slope + intercept
pointOnLine(Px, Py, Ax, Ay, Bx, By)
}
pointOnLine <- function(Px, Py, Ax, Ay, Bx, By) {
# This approach based upon comingstorm's answer on
# stackoverflow.com/questions/3120357/get-closest-point-to-a-line
# Vectorized by Bryan
PB <- data.frame(x = Px - Bx, y = Py - By)
AB <- data.frame(x = Ax - Bx, y = Ay - By)
PB <- as.matrix(PB)
AB <- as.matrix(AB)
k_raw <- k <- c()
for (n in 1:nrow(PB)) {
k_raw[n] <- (PB[n,] %*% AB[n,])/(AB[n,] %*% AB[n,])
if (k_raw[n] < 0) { k[n] <- 0
} else { if (k_raw[n] > 1) k[n] <- 1
else k[n] <- k_raw[n] }
}
x = (k * Ax + (1 - k)* Bx)
y = (k * Ay + (1 - k)* By)
ans <- data.frame(x, y)
ans
}
Try changing this line of your code:
res <- pointOnLineNearPoint(x, y, yx2.lm$coef[2], yx2.lm$coef[1])
to
res <- pointOnLineNearPoint(x, new.y, yx2.lm$coef[2], yx2.lm$coef[1])
So you're calling the correct y values.
In Vincent Zoonekynd's code, change the line u <- r$loadings to u <- solve(r$loadings). In the second instance of solve(), the predicted component scores along the first principal axis (i.e., the matrix of predicted scores with the second predicted components scores set to zero) need to be multiplied by the inverse of the loadings/eigenvectors. Multiplying data by the loadings gives predicted scores; dividing predicted scores by the loadings give data. Hope that helps.