I need some help rewriting my code. I have some code that I wrote in R for a contour plot that I want. The code works, however, it is very inefficient in that it has to plot thousands of points to get exactly what I want (the grey area), and so I would like to see if there is a simpler way to do what my code is doing.
I basically want to plot the function
f = x1 + x2
subject to the following constraint functions
c1 = 3/2 - x1 - 2x2 - 1/2*sin(2*pi(x1^2 - 2x2)) < 0
c2 = x1^2+x2^2-3/2 < 0
And so where c1 and c2 are both greater than zero, I would like to grey out those areas, and only show the function f where c1 and c2 are less than 0. And the domain of x1 and x2 are between 0 and 1.
Here is my current R code:
x1 = seq(0,1,.001)
x2 = seq(0,1,.001)
f = function(x1,x2){
ans = x1 + x2
return(ans) }
h = function(x1,x2){
ans1 = 1.5-x1-2*x2-.5*sin(2*pi*(x1^2-2*x2))
ans2 = x1^2+x2^2-1.5
ans1 = sapply(ans1,function(x){max(x,0)})
ans2 = sapply(ans2,function(x){max(x,0)})
ans = ans1 + ans2
return(ans) }
z = outer(x1,x2,f)
w = outer(x1,x2,h)
image(x1,x2,z,xlab=expression(x[1]),ylab=expression(x[2]))
contour(x1,x2,z,add=TRUE)
X = cbind(expand.grid(x1,x2),c(w))
points(X[X[,3]!=0,1],X[X[,3]!=0,2],pch=19,col="lightgrey")
You can do it with .filled.contour, the working function behind filled.contour. (The full filled.contour makes it hard to annotate your plot.) For example
x1 = seq(0,1,.001)
x2 = seq(0,1,.001)
f = function(x1,x2){
ans = x1 + x2
return(ans) }
h = function(x1,x2){
ans1 = 1.5-x1-2*x2-.5*sin(2*pi*(x1^2-2*x2))
ans2 = x1^2+x2^2-1.5
ans1 = pmax(ans1,0)
ans2 = pmax(ans2,0)
ans = ans1 + ans2
return(ans) }
z = outer(x1,x2,f)
w = outer(x1,x2,h)
# Set up the plot, .filled.contour doesn't do that
plot(x1, x2, type="n", xlab=expression(x[1]),ylab=expression(x[2]))
# Set a gray background
rect(min(x1), min(x2), max(x1), max(x2), col = "gray")
# Make parts transparent
z[ w != 0 ] <- NA
# Choose the contour levels
levels <- pretty(z)
# Plot them
.filled.contour(x1,x2,z,levels,
hcl.colors(length(levels)-1, "ylOrRd", rev = TRUE))
# If your device does antialiasing, plot the filled contours twice to avoid
# ugly effects
.filled.contour(x1,x2,z,levels,
hcl.colors(length(levels)-1, "ylOrRd", rev = TRUE))
contour(x1,x2,z, levels = levels, add=TRUE)
This gives me
Related
I'm trying to create a code such that Y = 5g1(X1) + 3g2(X2) + 4g3(X3) + 6g4(X4) + sqrt(1.74)*eps (the functions g, are defined in the code).
X = (X1,...,Xp) should be an nxp dimensional design matrix, however I'm not sure about how to generate that based on this information where Xj = W+U is simulated according to a random effects model. I tried using X = do.call(cbind, replicate(p, X, simplify=FALSE)) but this just replicates each Xj, i'm not sure that's what should be done, they should be different.
Any advice on what i have missed would be appreciated and any improvements on the code too to make it more concise.
n<- 400
p<- 1000
W = runif(n)
U = runif(n)
eps = rnorm(n)
for (j in 1:p){
X = W+U
X = as.matrix(X)
return(X)} #This is a nx1 matrix...
#alternatively write: X = do.call(cbind, replicate(p, X, simplify=FALSE))
g1 = X
g2 = (2*X-1)^2
g3 = sin(2*pi*X)/(2-sin(2*pi*X))
g4 = 0.1*sin(2*pi*X) + 0.2*cos(2*pi*X) + 0.3*sin(2*pi*X)^2 + 0.4*cos(2*pi*X)^3 + 0.5*sin(2*pi*X)^3
Y = 5*g1 + 3*g2 + 4*g3 + 6*g4 + sqrt(1.74)*eps
return(Y)
}
I am not sure to capture the logic of your calculation, eventually it is something like this:
n <- 40 # 400
p <- 100 # 1000
X <- replicate(p, runif(n) + runif(n)) ## W+U
y <- function(X) {
g1 <- X
g2 <- (2*X-1)^2
g3 <- sin(2*pi*X)/(2-sin(2*pi*X))
g4 <- 0.1*sin(2*pi*X) + 0.2*cos(2*pi*X) + 0.3*sin(2*pi*X)^2 + 0.4*cos(2*pi*X)^3 + 0.5*sin(2*pi*X)^3
eps <- rnorm(length(X))
Y <- 5*g1 + 3*g2 + 4*g3 + 6*g4 + sqrt(1.74)*eps
return(Y)
}
Y <- apply(X, 2, FUN=y)
Also the variant without apply() works:
Y <- y(X)
To compare both variants:
set.seed(42)
Y1 <- apply(X, 2, FUN=y)
set.seed(42)
Y2 <- y(X)
identical(Y1, Y2)
Many books illustrate the idea of Fisher linear discriminant analysis using the following figure (this particular is from Pattern Recognition and Machine Learning, p. 188)
I wonder how to reproduce this figure in R (or in any other language). Pasted below is my initial effort in R. I simulate two groups of data and draw linear discriminant using abline() function. Any suggestions are welcome.
set.seed(2014)
library(MASS)
library(DiscriMiner) # For scatter matrices
# Simulate bivariate normal distribution with 2 classes
mu1 <- c(2, -4)
mu2 <- c(2, 6)
rho <- 0.8
s1 <- 1
s2 <- 3
Sigma <- matrix(c(s1^2, rho * s1 * s2, rho * s1 * s2, s2^2), byrow = TRUE, nrow = 2)
n <- 50
X1 <- mvrnorm(n, mu = mu1, Sigma = Sigma)
X2 <- mvrnorm(n, mu = mu2, Sigma = Sigma)
y <- rep(c(0, 1), each = n)
X <- rbind(x1 = X1, x2 = X2)
X <- scale(X)
# Scatter matrices
B <- betweenCov(variables = X, group = y)
W <- withinCov(variables = X, group = y)
# Eigenvectors
ev <- eigen(solve(W) %*% B)$vectors
slope <- - ev[1,1] / ev[2,1]
intercept <- ev[2,1]
par(pty = "s")
plot(X, col = y + 1, pch = 16)
abline(a = slope, b = intercept, lwd = 2, lty = 2)
MY (UNFINISHED) WORK
I pasted my current solution below. The main question is how to rotate (and move) the density plot according to decision boundary. Any suggestions are still welcome.
require(ggplot2)
library(grid)
library(MASS)
# Simulation parameters
mu1 <- c(5, -9)
mu2 <- c(4, 9)
rho <- 0.5
s1 <- 1
s2 <- 3
Sigma <- matrix(c(s1^2, rho * s1 * s2, rho * s1 * s2, s2^2), byrow = TRUE, nrow = 2)
n <- 50
# Multivariate normal sampling
X1 <- mvrnorm(n, mu = mu1, Sigma = Sigma)
X2 <- mvrnorm(n, mu = mu2, Sigma = Sigma)
# Combine into data frame
y <- rep(c(0, 1), each = n)
X <- rbind(x1 = X1, x2 = X2)
X <- scale(X)
X <- data.frame(X, class = y)
# Apply lda()
m1 <- lda(class ~ X1 + X2, data = X)
m1.pred <- predict(m1)
# Compute intercept and slope for abline
gmean <- m1$prior %*% m1$means
const <- as.numeric(gmean %*% m1$scaling)
z <- as.matrix(X[, 1:2]) %*% m1$scaling - const
slope <- - m1$scaling[1] / m1$scaling[2]
intercept <- const / m1$scaling[2]
# Projected values
LD <- data.frame(predict(m1)$x, class = y)
# Scatterplot
p1 <- ggplot(X, aes(X1, X2, color=as.factor(class))) +
geom_point() +
theme_bw() +
theme(legend.position = "none") +
scale_x_continuous(limits=c(-5, 5)) +
scale_y_continuous(limits=c(-5, 5)) +
geom_abline(intecept = intercept, slope = slope)
# Density plot
p2 <- ggplot(LD, aes(x = LD1)) +
geom_density(aes(fill = as.factor(class), y = ..scaled..)) +
theme_bw() +
theme(legend.position = "none")
grid.newpage()
print(p1)
vp <- viewport(width = .7, height = 0.6, x = 0.5, y = 0.3, just = c("centre"))
pushViewport(vp)
print(p2, vp = vp)
Basically you need to project the data along the direction of the classifier, plot a histogram for each class, and then rotate the histogram so its x axis is parallel to the classifier. Some trial-and-error with scaling the histogram is needed in order to get a nice result. Here's an example of how to do it in Matlab, for the naive classifier (difference of class' means). For the Fisher classifier it is of course similar, you just use a different classifier w. I changed the parameters from your code so the plot is more similar to the one you gave.
rng('default')
n = 1000;
mu1 = [1,3]';
mu2 = [4,1]';
rho = 0.3;
s1 = .8;
s2 = .5;
Sigma = [s1^2,rho*s1*s1;rho*s1*s1, s2^2];
X1 = mvnrnd(mu1,Sigma,n);
X2 = mvnrnd(mu2,Sigma,n);
X = [X1; X2];
Y = [zeros(n,1);ones(n,1)];
scatter(X1(:,1), X1(:,2), [], 'b' );
hold on
scatter(X2(:,1), X2(:,2), [], 'r' );
axis equal
m1 = mean(X(1:n,:))';
m2 = mean(X(n+1:end,:))';
plot(m1(1),m1(2),'bx','markersize',18)
plot(m2(1),m2(2),'rx','markersize',18)
plot([m1(1),m2(1)], [m1(2),m2(2)],'g')
%% classifier taking only means into account
w = m2 - m1;
w = w / norm(w);
% project data onto w
X1_projected = X1 * w;
X2_projected = X2 * w;
% plot histogram and rotate it
angle = 180/pi * atan(w(2)/w(1));
[hy1, hx1] = hist(X1_projected);
[hy2, hx2] = hist(X2_projected);
hy1 = hy1 / sum(hy1); % normalize
hy2 = hy2 / sum(hy2); % normalize
scale = 4; % set manually
h1 = bar(hx1, scale*hy1,'b');
h2 = bar(hx2, scale*hy2,'r');
set([h1, h2],'ShowBaseLine','off')
% rotate around the origin
rotate(get(h1,'children'),[0,0,1], angle, [0,0,0])
rotate(get(h2,'children'),[0,0,1], angle, [0,0,0])
Using car::scatter3d(), I am trying to create a 3D figure with a regression surface indicating an interaction between a categorical and a continuous variable. Partly following the code here, I obtained a figure below.
The figure is obviously wrong in that the regression surface does not reach one of the values of the categorical variable. The problem perhaps lies in the use of the rgl::persp3d() (the last block of the code below), but I have not been able to identify what exactly I'm doing wrongly. Could someone let me know what I'm missing and how to fix the problem?
library(rgl)
library(car)
n <- 100
set.seed(1)
x <- runif(n, 0, 10)
set.seed(1)
z <- sample(c(0, 1), n, replace = TRUE)
set.seed(1)
y <- 0.5 * x + 0.1 * z + 0.3 * x * z + rnorm(n, sd = 1.5)
d <- data.frame(x, z, y)
scatter3d(y ~ x + z, data = d,
xlab = "continuous", zlab = "categorical", ylab = "outcome",
residuals = FALSE, surface = FALSE
)
d2 <- d
d2$x <- d$x / (max(d$x) - min(d$x))
d2$y <- d$y / (max(d$y) - min(d$y))
mod <- lm(y ~ x * z, data = d2)
grd <- expand.grid(x = unique(d2$x), z = unique(d2$z))
grd$pred <- predict(mod, newdata = grd)
grd <- grd[order(grd$z, grd$x), ]
# The problem is likely to lie somewhere below.
persp3d(x = unique(grd$x), y = unique(grd$z),
z = matrix(grd$pred, length(unique(grd$z)), length(unique(grd$x))),
alpha = 0.5,
col = "blue",
add = TRUE,
xlab = "", ylab = "", zlab = ""
)
I prefer sticking to car::scatter3d() in drawing the original graph because I already made several figures with car::scatter3d() and want to make this figure consistent with them as well.
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.
I have 5 (x,y) data points and I'm trying to find a best fit solution consisting of two lines which intersect at a point (x0,y0), and which follow these equations:
y1 = (m1)(x1 - x0) + y0
y2 = (m2)(x2 - x0) + y0
Specifically, I require that the intersection must occur between x=2 and x=3. Have a look at the code:
#Initialize x1, y1, x2, y2
x1 <- c(1,2)
y1 <- c(10,10)
x2 <- c(3,4,5)
y2 <- c(20,30,40)
g <- c(TRUE, TRUE, FALSE, FALSE, FALSE)
q <- nls(c(y1, y2) ~ ifelse(g == TRUE, m1 * (x1 - x0) + y0, m2 * (x2 - x0) + y0), start = c(m1 = -1, m2 = 1, y0 = 0, x0 = 2), algorithm = "port", lower = c(m1 = -Inf, m2 = -Inf, y0 = -Inf, x0 = 2), upper = c(m1 = Inf, m2 = Inf, y0 = Inf, x0 = 3))
coef <- coef(q)
m1 <- coef[1]
m2 <- coef[2]
y0 <- coef[3]
x0 <- coef[4]
#Plot the original x1, y1, and x2, y2
plot(x1,y1,xlim=c(1,5),ylim=c(0,50))
points(x2,y2)
#Plot the fits
x1 <- c(1,2,3,4,5)
fit1 <- m1 * (x1 - x0) + y0
lines(x1, fit1, col="red")
x2 <- c(1,2,3,4,5)
fit2 <- m2 * (x2 - x0) + y0
lines(x2, fit2, col="blue")
So, you can see the data points listed there. Then, I run it through my nls, get my parameters m1, m2, x0, y0 (the slopes, and the intersection point).
But, take a look at the solution:
Clearly, the red line (which is supposed to only be based on the first 2 points) is not the best line of fit for the first 2 points. This is the same case with the blue line (the 2nd fit), which supposed to be is dependent on the last 3 points). What is wrong here?
This is segmented regression:
# input data
x1 <- c(1,2); y1 <- c(10,10); x2 <- c(3,4,5); y2 <- c(20,30,40)
x <- c(x1, x2); y <- c(y1, y2)
# segmented regression
library(segmented)
fm <- segmented.lm(lm(y ~ x), ~ x, NA, seg.control(stop.if.error = FALSE, K = 2))
summary(fm)
# plot
plot(fm)
points(y ~ x)
See ?lm, ?segmented.lm and ?seg.control for more info.
I'm not exactly sure what's wrong but I can get it to work by rearranging things a bit. Please note the comment in ?nls about "Do not use ‘nls’ on artificial "zero-residual" data."; I added a bit of noise.
## Initialize x1, y1, x2, y2
x1 <- c(1,2)
y1 <- c(10,10)
x2 <- c(3,4,5)
y2 <- c(20,30,40)
## make single x, y vector
x <- c(x1,x2)
set.seed(1001)
## (add a bit of noise to avoid zero-residual artificiality)
y <- c(y1,y2)+rnorm(5,sd=0.01)
g <- c(TRUE,TRUE,FALSE,FALSE,FALSE) ## specify identities of points
## particular changes:
## * you have lower=upper=2 for x0. Did you want 2<x0<3?
## * specified data argument explicitly (allows use of predict() etc.)
## * changed name from 'q' to 'fit1' (avoid R built-in function)
fit1 <- nls(y ~ ifelse(g,m1,m1+delta_m)*(x - x0) + y0,
start = c(m1 = -1, delta_m = 2, y0 = 0, x0 = 2),
algorithm = "port",
lower = c(m1 = -Inf, delta_m = 0, y0 = -Inf, x0 = 2),
upper = c(m1 = Inf, delta_m = Inf, y0 = Inf, x0 = 3),
data=data.frame(x,y))
#Plot the original 'data'
plot(x,y,col=rep(c("red","blue"),c(2,3)),
xlim=c(1,5),ylim=c(0,50))
## add predicted values
xvec <- seq(1,5,length.out=101)
lines(xvec,predict(fit1,newdata=data.frame(x=xvec)))
edit: based ifelse clause on point identity, not x position
edit: changed to require second slope to be > first slope
On a second look, I think the issue above is probably due to the use of separate vectors for x1 and x2 above, rather than a single x vector: I suspect these got replicated by R to match up with the g vector, which would have messed things up pretty badly. For example, this stripped-down example:
g <- c(TRUE, TRUE, FALSE, FALSE, FALSE)
ifelse(g,x1,x2)
## [1] 1 2 5 3 4
shows that x2 gets extended to (3 4 5 3 4) before being used in the ifelse clause. The scariest part is that normally one gets a warning such as this:
> x2 + 1:5
[1] 4 6 8 7 9
Warning message:
In x2 + 1:5 :
longer object length is not a multiple of shorter object length
but in this case there is no warning ...