How to create a 3D barplot with fitted curve in R - r

I'm trying to make a figure similar to this in R. Basically X and Y represent coordinates (e.g. lon/lat) and Z represents elevation. I want to plot a 3D bar chart with a elevation as the bar height, and then show a smooth curve of fitted Z values going through the bars. I am able to get a 3d barplot with barplot3d but I'm not sure if it's possible to add the fitted curve on top of that. Does anyone know how to do this? I have some example code below demonstrating what I've tried so far.
library(rgl)
library(barplot3d)
library(tidyverse)
x_mat<- matrix(rep(-1:1, each=3),nrow = 3) #x coordinates
y_mat<- matrix(rev(rep(-1:1, each=3)),nrow = 3, byrow = TRUE) #y coordinates
df<- data.frame(x = as.vector(x_mat), y = as.vector(y_mat)) #dataframe
set.seed(5)
df<- df %>% mutate(z= x^2+ y^2 + rnorm(n = 9, mean = 0, sd = 0.1)) #add elevation values
m<- lm(z ~ I(x^2)+I(y^2)+I(x*y)+x+y, data = df) #fitted curve
rgl.open()
rgl::plot3d(m)
barplot3d(rows=3,cols=3, z=df$z,scalexy=1, gap=0, alpha=0.4,theta=30,phi=50,
topcolors = "gray", gridlines = TRUE)
Update: More complicated example
The curve doesn't seem to intersect the bars properly with current solution for a non-symmetrical curve (fixed in most updated answer).
library(rgl)
library(barplot3d)
library(tidyverse)
x_mat<- matrix(rep(-1:1, each=3),nrow = 3) #x coordinates
y_mat<- matrix(rev(rep(-1:1, each=3)),nrow = 3, byrow = TRUE) #y coordinates
A<- 0.2
B<- 0.2
C<- 0.4
D<- 0.4
E<- 0
df<- data.frame(x = as.vector(x_mat), y = as.vector(y_mat)) #dataframe
df<- df %>% mutate(z= A*x^2 + B*y^2 + C*x*y + D*x + E*y) #add elevation values
z_mat<- matrix(data = df$z, nrow=3)
m<- lm(z ~ I(x^2)+I(y^2)+I(x*y)+x+y, data = df) #fitted curve
df$zpred<- predict(m, data.frame(x=df$x, y=df$y))
round(df$z-df$zpred,10) #Predictions should fit observations almost exactly (i.e. intersect exactly with bars)
#0 0 0 0 0 0 0 0 0
n<- 10
xvals <- seq(-1, 1, len = n)
xmat <- replicate(n, seq(1.5, 3.5, len = n))
ymat <- t(xmat)
pred <- expand.grid(x = xvals, y = xvals)
zmat <- matrix(predict(m, pred), nrow = n, ncol = n)
barplot3d(rows=3,cols=3, z=df$z, gap=0, alpha=0.4, phi = 45,
topcolors = "gray", sidecolors = "cyan", linecolors= "blue", gridlines = FALSE, zlabels = FALSE)
surface3d(x = xmat, y = zmat, z = ymat-5, color = "purple", alpha = 0.7)
axes3d()

As you can probably tell, your 3d surface is rotated 90 degrees relative to where it should be. This is not your fault; it is just a difference between the way barplot3d is drawn compared to the other rgl shapes. You also need a bit of shifting and rescaling to get it to fit.
barplot3d(rows=3,cols=3, z=df$z, gap=0, alpha=0.4, phi = 45,
topcolors = "gray", gridlines = TRUE)
xvals <- seq(-1.5, 1.5, len = 10)
xmat <- replicate(10, seq(1, 4, len = 10))
ymat <- t(xmat)
pred <- expand.grid(x = xvals, y = xvals)
zmat <- matrix(predict(m, pred), 10, 10)
surface3d(xmat, zmat, color = "gold", alpha = 0.5, ymat - 5)
Update
To remove the points above the highest bar, just set them to NA. You probably want to increase the resolution when you do this though:
xvals <- seq(-1.5, 1.5, len = 100)
xmat <- replicate(100, seq(1, 4, len = 100))
ymat <- t(xmat) - 5
pred <- expand.grid(x = xvals, y = xvals)
zmat <- matrix(predict(m, pred), 100, 100)
zmat[zmat > 2] <- NA
barplot3d(rows=3,cols=3, z=df$z, gap=0, alpha=0.4, phi = 45,
topcolors = "gray", gridlines = TRUE)
surface3d(xmat, zmat, color = "gold", alpha = 0.5, ymat)
Note that this gives an unavoidably ragged edge to the graphic
An alternative is to shrink the x, y grids at which z is calculated:
xvals <- seq(-1, 1, len = 100)
xmat <- replicate(100, seq(1.5, 3.5, len = 100))
ymat <- t(xmat) - 5
pred <- expand.grid(x = xvals, y = xvals)
zmat <- matrix(predict(m, pred), 100, 100)
barplot3d(rows=3,cols=3, z=df$z, gap=0, alpha=0.4, phi = 45,
topcolors = "gray", gridlines = TRUE)
surface3d(xmat, zmat, ymat, color = "gray20", alpha = 0.5)
Further update
It looks as though we need a double flip to get the z values correct:
barplot3d(rows=3,cols=3, z=df$z, gap=0, alpha=0.4, phi = 45,
topcolors = "gray", sidecolors = "cyan", linecolors= "blue",
gridlines = FALSE, zlabels = FALSE)
surface3d(x = xmat, y = t(apply(t(apply(zmat, 1, rev)), 2, rev)),
z = ymat-5, color = "purple", alpha = 0.7)
axes3d()

Related

Combined persp and ggplot graph on the same window, persp too small

I am combining a persp graph and a ggplot graph in the same window using plot_grid. However, the persp graph is too small, how can I make it bigger?
library(pacman)
p_load(tidyverse)
p_load(mvtnorm)
p_load(cowplot)
p_load(gridGraphics)
p_load(GA)
my_mean<-c(25,65)
mycors<-seq(-1,1,by=.25)
sd_vec<-c(5,7)
i<-3
temp_cor<-matrix(c(1,mycors[i],
mycors[i],1),
byrow = T,ncol=2)
V<-sd_vec %*% t(sd_vec) *temp_cor
my_x<-seq(my_mean[1]-3*sd_vec[1], my_mean[1]+3*sd_vec[1], length.out=20)
my_y<-seq(my_mean[2]-3*sd_vec[2], my_mean[2]+3*sd_vec[2], length.out=20)
temp_f<-function(a,b){dmvnorm(cbind(a,b), my_mean,V)}
my_z<-outer(my_x, my_y,temp_f)
nlevels<-20
my_zlim <- range(my_z, finite = TRUE)
my_levels <- pretty(my_zlim, nlevels)
zz <- (my_z[-1, -1] + my_z[-1, -ncol(my_z)] + my_z[-nrow(my_z), -1] + my_z[-nrow(my_z),
-ncol(my_z)])/4
cols <- jet.colors(length(my_levels) - 1)
zzz <- cut(zz, breaks = my_levels, labels = cols)
persp(my_x, my_y, my_z, theta = -25, phi = 45, expand = 0.5,xlab="x",ylab="y",zlab="f(x,y)",col = as.character(zzz))
p1 <- recordPlot()
data.grid <- expand.grid(x = seq(my_mean[1]-3*sd_vec[1], my_mean[1]+3*sd_vec[1], length.out=200),
y = seq(my_mean[2]-3*sd_vec[2], my_mean[2]+3*sd_vec[2], length.out=200))
q.samp <- cbind(data.grid, prob = dmvnorm(data.grid, mean = my_mean, sigma = V))
p2<-ggplot(q.samp, aes(x, y, z = prob)) +
geom_contour(aes(color = ..level..), bins = 11, size = 1) +
scale_color_gradientn(colours = jet.colors(11)) +
theme_bw()
plot_grid(p1, p2)
Created on 2020-10-31 by the reprex package (v0.3.0)
I think there are two things you need to do:
Set par(mar = c(0, 0, 0, 0)) before calling persp. Ensure you save your default parameters before and reset them afterwards.
Resize your plotting window to give it a wider aspect ratio
So basically you can change your persp call to:
par_store <- par()
par(mar = c(0, 0, 0, 0))
persp(my_x, my_y, my_z, theta = -25, phi = 45, expand = 0.5,
xlab = "x", ylab = "y", zlab = "f(x,y)", col = as.character(zzz))
p1 <- recordPlot()
par(par_store)
And after resizing the plotting window you get:

Draw line through 2d density plot

I have a large dataset of gene expression from ~10,000 patient samples (TCGA), and I'm plotting a predicted expression value (x) and the actual observed value (y) of a certain gene signature. For my downstream analysis, I need to draw a precise line through the plot and calculate different parameters in samples above/below the line.
No matter how I draw a line through the data (geom_smooth(method = 'lm', 'glm', 'gam', or 'loess')), the line always seems imperfect - it doesn't cut through the data to my liking (red line is lm in figure).
After playing around for a while, I realized that the 2d kernel density lines (geom_density2d) actually do a good job of showing the slope/trends of my data, so I manually drew a line that kind of cuts through the density lines (black line in figure).
My question: how can I automatically draw a line that cuts through the kernel density lines, as for the black line in the figure? (Rather than manually playing with different intercepts and slopes till something looks good).
The best approach I can think of is to somehow calculate intercept and slope of the longest diameter for each of the kernel lines, take an average of all those intercepts and slopes and plot that line, but that's a bit out of my league. Maybe someone here has experience with this and can help?
A more hacky approach may be getting the x,y coords of each kernel density line from ggplot_build, and going from there, but it feels too hacky (and is also out of my league).
Thanks!
EDIT: Changed a few details to make the figure/analysis easier. (Density lines are smoother now).
Reprex:
library(MASS)
set.seed(123)
samples <- 10000
r <- 0.9
data <- mvrnorm(n=samples, mu=c(0, 0), Sigma=matrix(c(2, r, r, 2), nrow=2))
x <- data[, 1] # standard normal (mu=0, sd=1)
y <- data[, 2] # standard normal (mu=0, sd=1)
test.df <- data.frame(x = x, y = y)
lm(y ~ x, test.df)
ggplot(test.df, aes(x, y)) +
geom_point(color = 'grey') +
geom_density2d(color = 'red', lwd = 0.5, contour = T, h = c(2,2)) + ### EDIT: h = c(2,2)
geom_smooth(method = "glm", se = F, lwd = 1, color = 'red') +
geom_abline(intercept = 0, slope = 0.7, lwd = 1, col = 'black') ## EDIT: slope to 0.7
Figure:
I generally agree with #Hack-R.
However, it was kind of a fun problem and looking into ggplot_build is not such a big deal.
require(dplyr)
require(ggplot2)
p <- ggplot(test.df, aes(x, y)) +
geom_density2d(color = 'red', lwd = 0.5, contour = T, h = c(2,2))
#basic version of your plot
p_built <- ggplot_build(p)
p_data <- p_built$data[[1]]
p_maxring <- p_data[p_data[['level']] == min(p_data[['level']]),] %>%
select(x,y) # extracts the x/y coordinates of the points on the largest ellipse from your 2d-density contour
Now this answer helped me to find the points on this ellipse which are furthest apart.
coord_mean <- c(x = mean(p_maxring$x), y = mean(p_maxring$y))
p_maxring <- p_maxring %>%
mutate (mean_dev = sqrt((x - mean(x))^2 + (y - mean(y))^2)) #extra column specifying the distance of each point to the mean of those points
coord_farthest <- c('x' = p_maxring$x[which.max(p_maxring$mean_dev)], 'y' = p_maxring$y[which.max(p_maxring$mean_dev)])
# gives the coordinates of the point farthest away from the mean point
farthest_from_farthest <- sqrt((p_maxring$x - coord_farthest['x'])^2 + (p_maxring$y - coord_farthest['y'])^2)
#now this looks which of the points is the farthest from the point farthest from the mean point :D
coord_fff <- c('x' = p_maxring$x[which.max(farthest_from_farthest)], 'y' = p_maxring$y[which.max(farthest_from_farthest)])
ggplot(test.df, aes(x, y)) +
geom_density2d(color = 'red', lwd = 0.5, contour = T, h = c(2,2)) +
# geom_segment using the coordinates of the points farthest apart
geom_segment((aes(x = coord_farthest['x'], y = coord_farthest['y'],
xend = coord_fff['x'], yend = coord_fff['y']))) +
geom_smooth(method = "glm", se = F, lwd = 1, color = 'red') +
# as per your request with your geom_smooth line
coord_equal()
coord_equal is super important, because otherwise you will get super weird results - it messed up my brain too. Because if the coordinates are not set equal, the line will seemingly not pass through the point furthest apart from the mean...
I leave it to you to build this into a function in order to automate it. Also, I'll leave it to you to calculate the y-intercept and slope from the two points
Tjebo's approach was kind of good initially, but after a close look, I found that it found the longest distance between two points on an ellipse. While this is close to what I wanted, it failed with either an irregular shape of the ellipse, or the sparsity of points in the ellipse. This is because it measured the longest distance between two points; whereas what I really wanted is the longest diameter of an ellipse; i.e.: the semi-major axis. See image below for examples/details.
Briefly:
To find/draw density contours of specific density/percentage:
R - How to find points within specific Contour
To get the longest diameter ("semi-major axis") of an ellipse:
https://stackoverflow.com/a/18278767/3579613
For function that returns intercept and slope (as in OP), see last piece of code.
The two pieces of code and images below compare two Tjebo's approach vs. my new approach based on the above posts.
#### Reprex from OP
require(dplyr)
require(ggplot2)
require(MASS)
set.seed(123)
samples <- 10000
r <- 0.9
data <- mvrnorm(n=samples, mu=c(0, 0), Sigma=matrix(c(2, r, r, 2), nrow=2))
x <- data[, 1] # standard normal (mu=0, sd=1)
y <- data[, 2] # standard normal (mu=0, sd=1)
test.df <- data.frame(x = x, y = y)
#### From Tjebo
p <- ggplot(test.df, aes(x, y)) +
geom_density2d(color = 'red', lwd = 0.5, contour = T, h = 2)
p_built <- ggplot_build(p)
p_data <- p_built$data[[1]]
p_maxring <- p_data[p_data[['level']] == min(p_data[['level']]),][,2:3]
coord_mean <- c(x = mean(p_maxring$x), y = mean(p_maxring$y))
p_maxring <- p_maxring %>%
mutate (mean_dev = sqrt((x - mean(x))^2 + (y - mean(y))^2)) #extra column specifying the distance of each point to the mean of those points
p_maxring = p_maxring[round(seq(1, nrow(p_maxring), nrow(p_maxring)/23)),] #### Make a small ellipse to illustrate flaws of approach
coord_farthest <- c('x' = p_maxring$x[which.max(p_maxring$mean_dev)], 'y' = p_maxring$y[which.max(p_maxring$mean_dev)])
# gives the coordinates of the point farthest away from the mean point
farthest_from_farthest <- sqrt((p_maxring$x - coord_farthest['x'])^2 + (p_maxring$y - coord_farthest['y'])^2)
#now this looks which of the points is the farthest from the point farthest from the mean point :D
coord_fff <- c('x' = p_maxring$x[which.max(farthest_from_farthest)], 'y' = p_maxring$y[which.max(farthest_from_farthest)])
farthest_2_points = data.frame(t(cbind(coord_farthest, coord_fff)))
plot(p_maxring[,1:2], asp=1)
lines(farthest_2_points, col = 'blue', lwd = 2)
#### From answer in another post
d = cbind(p_maxring[,1], p_maxring[,2])
r = ellipsoidhull(d)
exy = predict(r) ## the ellipsoid boundary
lines(exy)
me = colMeans((exy))
dist2center = sqrt(rowSums((t(t(exy)-me))^2))
max(dist2center) ## major axis
lines(exy[dist2center == max(dist2center),], col = 'red', lwd = 2)
#### The plot here is made from the data in the reprex in OP, but with h = 0.5
library(MASS)
set.seed(123)
samples <- 10000
r <- 0.9
data <- mvrnorm(n=samples, mu=c(0, 0), Sigma=matrix(c(2, r, r, 2), nrow=2))
x <- data[, 1] # standard normal (mu=0, sd=1)
y <- data[, 2] # standard normal (mu=0, sd=1)
test.df <- data.frame(x = x, y = y)
## MAKE BLUE LINE
p <- ggplot(test.df, aes(x, y)) +
geom_density2d(color = 'red', lwd = 0.5, contour = T, h = 0.5) ## NOTE h = 0.5
p_built <- ggplot_build(p)
p_data <- p_built$data[[1]]
p_maxring <- p_data[p_data[['level']] == min(p_data[['level']]),][,2:3]
coord_mean <- c(x = mean(p_maxring$x), y = mean(p_maxring$y))
p_maxring <- p_maxring %>%
mutate (mean_dev = sqrt((x - mean(x))^2 + (y - mean(y))^2))
coord_farthest <- c('x' = p_maxring$x[which.max(p_maxring$mean_dev)], 'y' = p_maxring$y[which.max(p_maxring$mean_dev)])
farthest_from_farthest <- sqrt((p_maxring$x - coord_farthest['x'])^2 + (p_maxring$y - coord_farthest['y'])^2)
coord_fff <- c('x' = p_maxring$x[which.max(farthest_from_farthest)], 'y' = p_maxring$y[which.max(farthest_from_farthest)])
## MAKE RED LINE
## h = 0.5
## Given the highly irregular shape of the contours, I will use only the largest contour line (0.95) for draing the line.
## Thus, average = 1. See function below for details.
ln = long.diam("x", "y", test.df, h = 0.5, average = 1) ## NOTE h = 0.5
## PLOT
ggplot(test.df, aes(x, y)) +
geom_density2d(color = 'red', lwd = 0.5, contour = T, h = 0.5) + ## NOTE h = 0.5
geom_segment((aes(x = coord_farthest['x'], y = coord_farthest['y'],
xend = coord_fff['x'], yend = coord_fff['y'])), col = 'blue', lwd = 2) +
geom_abline(intercept = ln[1], slope = ln[2], color = 'red', lwd = 2) +
coord_equal()
Finally, I came up with the following function to deal with all this. Sorry for the lack of comments/clarity
#### This will return the intercept and slope of the longest diameter (semi-major axis).
####If Average = TRUE, it will average the int and slope across different density contours.
long.diam = function(x, y, df, probs = c(0.95, 0.5, 0.1), average = T, h = 2) {
fun.df = data.frame(cbind(df[,x], df[,y]))
colnames(fun.df) = c("x", "y")
dens = kde2d(fun.df$x, fun.df$y, n = 200, h = h)
dx <- diff(dens$x[1:2])
dy <- diff(dens$y[1:2])
sz <- sort(dens$z)
c1 <- cumsum(sz) * dx * dy
levels <- sapply(probs, function(x) {
approx(c1, sz, xout = 1 - x)$y
})
names(levels) = paste0("L", str_sub(formatC(probs, 2, format = 'f'), -2))
#plot(fun.df$x,fun.df$y, asp = 1)
#contour(dens, levels = levels, labels=probs, add=T, col = c('red', 'blue', 'green'), lwd = 2)
#contour(dens, add = T, col = 'red', lwd = 2)
#abline(lm(fun.df$y~fun.df$x))
ls <- contourLines(dens, levels = levels)
names(ls) = names(levels)
lines.info = list()
for (i in 1:length(ls)) {
d = cbind(ls[[i]]$x, ls[[i]]$y)
exy = predict(ellipsoidhull(d))## the ellipsoid boundary
colnames(exy) = c("x", "y")
me = colMeans((exy)) ## center of the ellipse
dist2center = sqrt(rowSums((t(t(exy)-me))^2))
#plot(exy,type='l',asp=1)
#points(d,col='blue')
#lines(exy[order(dist2center)[1:2],])
#lines(exy[rev(order(dist2center))[1:2],])
max.dist = data.frame(exy[rev(order(dist2center))[1:2],])
line.fit = lm(max.dist$y ~ max.dist$x)
lines.info[[i]] = c(as.numeric(line.fit$coefficients[1]), as.numeric(line.fit$coefficients[2]))
}
names(lines.info) = names(ls)
#plot(fun.df$x,fun.df$y, asp = 1)
#contour(dens, levels = levels, labels=probs, add=T, col = c('red', 'blue', 'green'), lwd = 2)
#abline(lines.info[[1]], col = 'red', lwd = 2)
#abline(lines.info[[2]], col = 'blue', lwd = 2)
#abline(lines.info[[3]], col = 'green', lwd = 2)
#abline(apply(simplify2array(lines.info), 1, mean), col = 'black', lwd = 4)
if (isTRUE(average)) {
apply(simplify2array(lines.info), 1, mean)
} else {
lines.info[[average]]
}
}
Finally, here's the final implementation of the different answers:
library(MASS)
set.seed(123)
samples = 10000
r = 0.9
data = mvrnorm(n=samples, mu=c(0, 0), Sigma=matrix(c(2, r, r, 2), nrow=2))
x = data[, 1] # standard normal (mu=0, sd=1)
y = data[, 2] # standard normal (mu=0, sd=1)
#plot(x, y)
test.df = data.frame(x = x, y = y)
#### Find furthest two points of contour
## BLUE
p <- ggplot(test.df, aes(x, y)) +
geom_density2d(color = 'red', lwd = 2, contour = T, h = 2)
p_built <- ggplot_build(p)
p_data <- p_built$data[[1]]
p_maxring <- p_data[p_data[['level']] == min(p_data[['level']]),][,2:3]
coord_mean <- c(x = mean(p_maxring$x), y = mean(p_maxring$y))
p_maxring <- p_maxring %>%
mutate (mean_dev = sqrt((x - mean(x))^2 + (y - mean(y))^2))
coord_farthest <- c('x' = p_maxring$x[which.max(p_maxring$mean_dev)], 'y' = p_maxring$y[which.max(p_maxring$mean_dev)])
farthest_from_farthest <- sqrt((p_maxring$x - coord_farthest['x'])^2 + (p_maxring$y - coord_farthest['y'])^2)
coord_fff <- c('x' = p_maxring$x[which.max(farthest_from_farthest)], 'y' = p_maxring$y[which.max(farthest_from_farthest)])
#### Find the average intercept and slope of 3 contour lines (0.95, 0.5, 0.1), as in my long.diam function above.
## RED
ln = long.diam("x", "y", test.df)
#### Plot everything. Black line is GLM
ggplot(test.df, aes(x, y)) +
geom_point(color = 'grey') +
geom_density2d(color = 'red', lwd = 1, contour = T, h = 2) +
geom_smooth(method = "glm", se = F, lwd = 1, color = 'black') +
geom_abline(intercept = ln[1], slope = ln[2], col = 'red', lwd = 1) +
geom_segment((aes(x = coord_farthest['x'], y = coord_farthest['y'],
xend = coord_fff['x'], yend = coord_fff['y'])), col = 'blue', lwd = 1) +
coord_equal()

Fix interpolated polar contour plot function to works with current R and (possibly) use ggplot

The question R interpolated polar contour plot shows an excellent way to produce interpolated polar plots in R. I include the very slightly modified version I'm using:
PolarImageInterpolate <- function(
### Plotting data (in cartesian) - will be converted to polar space.
x, y, z,
### Plot component flags
contours=TRUE, # Add contours to the plotted surface
legend=TRUE, # Plot a surface data legend?
axes=TRUE, # Plot axes?
points=TRUE, # Plot individual data points
extrapolate=FALSE, # Should we extrapolate outside data points?
### Data splitting params for color scale and contours
col_breaks_source = 1, # Where to calculate the color brakes from (1=data,2=surface)
# If you know the levels, input directly (i.e. c(0,1))
col_levels = 10, # Number of color levels to use - must match length(col) if
#col specified separately
col = rev(heat.colors(col_levels)), # Colors to plot
# col = rev(heat.colors(col_levels)), # Colors to plot
contour_breaks_source = 1, # 1=z data, 2=calculated surface data
# If you know the levels, input directly (i.e. c(0,1))
contour_levels = col_levels+1, # One more contour break than col_levels (must be
# specified correctly if done manually
### Plotting params
outer.radius = ceiling(max(sqrt(x^2+y^2))),
circle.rads = pretty(c(0,outer.radius)), #Radius lines
spatial_res=1000, #Resolution of fitted surface
single_point_overlay=0, #Overlay "key" data point with square
#(0 = No, Other = number of pt)
### Fitting parameters
interp.type = 1, #1 = linear, 2 = Thin plate spline
lambda=0){ #Used only when interp.type = 2
minitics <- seq(-outer.radius, outer.radius, length.out = spatial_res)
# interpolate the data
if (interp.type ==1 ){
Interp <- akima:::interp(x = x, y = y, z = z,
extrap = extrapolate,
xo = minitics,
yo = minitics,
linear = FALSE)
Mat <- Interp[[3]]
}
else if (interp.type == 2){
library(fields)
grid.list = list(x=minitics,y=minitics)
t = Tps(cbind(x,y),z,lambda=lambda)
tmp = predict.surface(t,grid.list,extrap=extrapolate)
Mat = tmp$z
}
else {stop("interp.type value not valid")}
# mark cells outside circle as NA
markNA <- matrix(minitics, ncol = spatial_res, nrow = spatial_res)
Mat[!sqrt(markNA ^ 2 + t(markNA) ^ 2) < outer.radius] <- NA
### Set contour_breaks based on requested source
if ((length(contour_breaks_source == 1)) & (contour_breaks_source[1] == 1)){
contour_breaks = seq(min(z,na.rm=TRUE),max(z,na.rm=TRUE),
by=(max(z,na.rm=TRUE)-min(z,na.rm=TRUE))/(contour_levels-1))
}
else if ((length(contour_breaks_source == 1)) & (contour_breaks_source[1] == 2)){
contour_breaks = seq(min(Mat,na.rm=TRUE),max(Mat,na.rm=TRUE),
by=(max(Mat,na.rm=TRUE)-min(Mat,na.rm=TRUE))/(contour_levels-1))
}
else if ((length(contour_breaks_source) == 2) & (is.numeric(contour_breaks_source))){
contour_breaks = pretty(contour_breaks_source,n=contour_levels)
contour_breaks = seq(contour_breaks_source[1],contour_breaks_source[2],
by=(contour_breaks_source[2]-contour_breaks_source[1])/(contour_levels-1))
}
else {stop("Invalid selection for \"contour_breaks_source\"")}
### Set color breaks based on requested source
if ((length(col_breaks_source) == 1) & (col_breaks_source[1] == 1))
{zlim=c(min(z,na.rm=TRUE),max(z,na.rm=TRUE))}
else if ((length(col_breaks_source) == 1) & (col_breaks_source[1] == 2))
{zlim=c(min(Mat,na.rm=TRUE),max(Mat,na.rm=TRUE))}
else if ((length(col_breaks_source) == 2) & (is.numeric(col_breaks_source)))
{zlim=col_breaks_source}
else {stop("Invalid selection for \"col_breaks_source\"")}
# begin plot
Mat_plot = Mat
Mat_plot[which(Mat_plot<zlim[1])]=zlim[1]
Mat_plot[which(Mat_plot>zlim[2])]=zlim[2]
image(x = minitics, y = minitics, Mat_plot , useRaster = TRUE, asp = 1, axes = FALSE, xlab = "", ylab = "", zlim = zlim, col = col)
# add contours if desired
if (contours){
CL <- contourLines(x = minitics, y = minitics, Mat, levels = contour_breaks)
A <- lapply(CL, function(xy){
lines(xy$x, xy$y, col = gray(.2), lwd = .5)
})
}
# add interpolated point if desired
if (points){
points(x, y, pch = 21, bg ="blue")
}
# add overlay point (used for trained image marking) if desired
if (single_point_overlay!=0){
points(x[single_point_overlay],y[single_point_overlay],pch=0)
}
# add radial axes if desired
if (axes){
# internals for axis markup
RMat <- function(radians){
matrix(c(cos(radians), sin(radians), -sin(radians), cos(radians)), ncol = 2)
}
circle <- function(x, y, rad = 1, nvert = 500){
rads <- seq(0,2*pi,length.out = nvert)
xcoords <- cos(rads) * rad + x
ycoords <- sin(rads) * rad + y
cbind(xcoords, ycoords)
}
# draw circles
if (missing(circle.rads)){
circle.rads <- pretty(c(0,outer.radius))
}
for (i in circle.rads){
lines(circle(0, 0, i), col = "#66666650")
}
# put on radial spoke axes:
axis.rads <- c(0, pi / 6, pi / 3, pi / 2, 2 * pi / 3, 5 * pi / 6)
r.labs <- c(90, 60, 30, 0, 330, 300)
l.labs <- c(270, 240, 210, 180, 150, 120)
for (i in 1:length(axis.rads)){
endpoints <- zapsmall(c(RMat(axis.rads[i]) %*% matrix(c(1, 0, -1, 0) * outer.radius,ncol = 2)))
segments(endpoints[1], endpoints[2], endpoints[3], endpoints[4], col = "#66666650")
endpoints <- c(RMat(axis.rads[i]) %*% matrix(c(1.1, 0, -1.1, 0) * outer.radius, ncol = 2))
lab1 <- bquote(.(r.labs[i]) * degree)
lab2 <- bquote(.(l.labs[i]) * degree)
text(endpoints[1], endpoints[2], lab1, xpd = TRUE)
text(endpoints[3], endpoints[4], lab2, xpd = TRUE)
}
axis(2, pos = -1.25 * outer.radius, at = sort(union(circle.rads,-circle.rads)), labels = NA)
text( -1.26 * outer.radius, sort(union(circle.rads, -circle.rads)),sort(union(circle.rads, -circle.rads)), xpd = TRUE, pos = 2)
}
# add legend if desired
# this could be sloppy if there are lots of breaks, and that's why it's optional.
# another option would be to use fields:::image.plot(), using only the legend.
# There's an example for how to do so in its documentation
if (legend){
library(fields)
image.plot(legend.only=TRUE, smallplot=c(.78,.82,.1,.8), col=col, zlim=zlim)
# ylevs <- seq(-outer.radius, outer.radius, length = contour_levels+ 1)
# #ylevs <- seq(-outer.radius, outer.radius, length = length(contour_breaks))
# rect(1.2 * outer.radius, ylevs[1:(length(ylevs) - 1)], 1.3 * outer.radius, ylevs[2:length(ylevs)], col = col, border = NA, xpd = TRUE)
# rect(1.2 * outer.radius, min(ylevs), 1.3 * outer.radius, max(ylevs), border = "#66666650", xpd = TRUE)
# text(1.3 * outer.radius, ylevs[seq(1,length(ylevs),length.out=length(contour_breaks))],round(contour_breaks, 1), pos = 4, xpd = TRUE)
}
}
Unfortunately, this function has a few bugs:
a) Even with a purely radial pattern, the produced plot has a distortion whose origin I don't understand:
#example
r <- rep(seq(0.1, 0.9, len = 8), each = 8)
theta <- rep(seq(0, 7/4*pi, by = pi/4), times = 8)
x <- r*sin(theta)
y <- r*cos(theta)
z <- z <- rep(seq(0, 1, len = 8), each = 8)
PolarImageInterpolate(x, y, z)
why the wiggles between 300° and 360°? The z function is constant in theta, so there's no reason why there should be wiggles.
b) After 4 years, some of the packages loaded have been modified and at least one functionality of the function is broken. Setting interp.type = 2 should use thin plate splines for interpolation instead than a basic linear interpolation, but it doesn't work:
> PolarImageInterpolate(x, y, z, interp.type = 2)
Warning:
Grid searches over lambda (nugget and sill variances) with minima at the endpoints:
(GCV) Generalized Cross-Validation
minimum at right endpoint lambda = 9.493563e-06 (eff. df= 60.80002 )
predict.surface is now the function predictSurface
Error in image.default(x = minitics, y = minitics, Mat_plot, useRaster = TRUE, :
'z' must be a matrix
the first message is a warning and doesn't worry me, but the second one is actually an error and prevents me from using thin plate splines. Can you help me solve these two problems?
Also, I'd like to "upgrade" to using ggplot2, so if you can give an answer which does that, it would be great. Otherwise, after the bugs are fixed, I'll try asking a specific question which only asks to modify the function so that it uses ggplot2.
For the ggplot2 solution, here is a start. geom_raster allows interpolation, but does not work for polar coordinates. Instead, you can use geom_tile, though then you may need to do the interpolation yourself before passing the values to ggplot.
Of important note: the example data you gave gives an error when working with geom_raster that I believe is caused by the spacing of the values. Here is an example set that works (note, used this blog as a guide, though it is now outdated):
dat_grid <-
expand.grid(x = seq(0,350,10), y = 0:10)
dat_grid$density <- runif(nrow(dat_grid))
ggplot(dat_grid
, aes(x = x, y = y, fill = density)) +
geom_tile() +
coord_polar() +
scale_x_continuous(breaks = seq(0,360,90)) +
scale_fill_gradient2(low = "white"
, mid = "yellow"
, high = "red3"
, midpoint = 0.5)
If you are working from raw data, you might be able to get ggplot to do the work for you. Here is an example working from raw data. There are a lot of manual tinkering things to do, but it is at least an optional starting point:
polarData <-
data.frame(
theta = runif(10000, 0, 2*pi)
, r = log(abs(rnorm(10000, 0, 10)))
)
toCart <-
data.frame(
x = polarData$r * cos(polarData$theta)
, y = polarData$r * sin(polarData$theta)
)
axisLines <-
data.frame(
x = 0
, y = 0
, xend = max(polarData$r)*cos(seq(0, 2*pi, pi/4))
, yend = max(polarData$r)*sin(seq(0, 2*pi, pi/4))
, angle = paste(seq(0, 2, 1/4), "pi") )
ticks <-
data.frame(
label = pretty(c(0, max(polarData$r)) )[-1]
)
ggplot(toCart) +
# geom_point(aes(x = x, y = y)) +
stat_density_2d(aes(x = x, y = y
, fill = ..level..)
, geom = "polygon") +
scale_fill_gradient(low = "white"
, high = "red3") +
theme(axis.text = element_blank()
, axis.title = element_blank()
, axis.line = element_blank()
, axis.ticks = element_blank()) +
geom_segment(data = axisLines
, aes(x = x, y = y
, xend = xend
, yend = yend)) +
geom_label(data = axisLines
, aes(x = xend, y = yend, label = angle)) +
geom_label(data = ticks
, aes(x = 0, y = label, label = label))
From an another post, I came to know that the fucnction predict.surface from package fields is deprecated whic is used for interp.type = 2 in PolarImageInterpolate. Instead, a new predictSurface function is introduced, which can be used here.
Example:
r <- rep(seq(0.1, 0.9, len = 8), each = 8)
theta <- rep(seq(0, 7/4*pi, by = pi/4), times = 8)
x <- r*sin(theta)
y <- r*cos(theta)
z <- z <- rep(seq(0, 1, len = 8), each = 8)
PolarImageInterpolate(x, y, z, interp.type = 2)

Plot vectors of gradient descent in R

I've code gradient descent algorithm in R and now I'm trying to "draw" the path of the vectors.
I've got draw points in my contour plot, but it's not correct because nobody knows what happened first.
In my algorith always I have an previous state P=(Xi,Yi) and a later state L=(Xi+1,Yi+1), so, How can I draw the vector PL in a contour or a persp plot?
I only got this with contour, where the red point is the convergence:
The same for persp:
Thanks all!
EDIT:
Graphics can be obtanined respectively:
f<-function(u,v){
u*u*exp(2*v)+4*v*v*exp(-2*u)-4*u*v*exp(v-u)
}
x = seq(-2, 2, by = 0.5)
y = seq(-2, 2, by = 0.5)
z <- outer(x,y,f)
#Contour plot
contour(x,y,z)
#Persp plot
persp(x, y, z, phi = 25, theta = 55, xlim=c(-2,2), ylim=c(-2,2),
xlab = "U", ylab = "V",
main = "F(u,v)", col="yellow", ticktype = "detailed"
) -> res
Taking Himmelblau's function as a test example:
f <- function(x, y) { (x^2+y-11)^2 + (x+y^2-7)^2 }
Its partial derivatives:
dx <- function(x,y) {4*x**3-4*x*y-42*x+4*x*y-14}
dy <- function(x,y) {4*y**3+2*x**2-26*y+4*x*y-22}
Running the gradient descent:
# gradient descent parameters
num_iter <- 100
learning_rate <- 0.001
x_val <- 6
y_val <- 6
updates_x <- vector("numeric", length = num_iter)
updates_y <- vector("numeric", length = num_iter)
updates_z <- vector("numeric", length = num_iter)
# parameter updates
for (i in 1:num_iter) {
dx_val = dx(x_val,y_val)
dy_val = dy(x_val,y_val)
x_val <- x_val-learning_rate*dx_val
y_val <- y_val-learning_rate*dx_val
z_val <- f(x_val, y_val)
updates_x[i] <- x_val
updates_y[i] <- y_val
updates_z[i] <- z_val
}
Plotting:
x <- seq(-6, 6, length = 100)
y <- x
z <- outer(x, y, f)
plt <- persp(x, y, z,
theta = -50-log(i), phi = 20+log(i),
expand = 0.5,
col = "lightblue", border = 'lightblue',
axes = FALSE, box = FALSE,
ltheta = 60, shade = 0.90
)
points(trans3d(updates_x[1:i], updates_y[1:i], updates_z[1:i],pmat = plt),
col = c(rep('white', num_iter-1), 'blue'),
pch = 16,
cex = c(rep(0.5, num_iter-1), 1))
There's a trick to plotting points using persp, as mentioned in ?persp. By employing the power of trans3d, you can successfully put points and lines on a perspective plot.
f<-function(u,v){
u*u*exp(2*v)+4*v*v*exp(-2*u)-4*u*v*exp(v-u)
}
x = seq(-2, 2, by = 0.5)
y = seq(-2, 2, by = 0.5)
z <- scale(outer(x,y,f))
view <- persp(x, y, z, phi = 30, theta = 30, xlim=c(-2,2), ylim=c(-2,2),
xlab = "X", ylab = "Y", zlab = "Z", scale = FALSE,
main = "F(u,v)", col="yellow", ticktype = "detailed")
set.seed(2)
pts <- data.frame(x = sample(x, 3),
y = sample(y, 3),
z = sample(z, 3))
points(trans3d(x = pts$x, y = pts$y, z = pts$z, pmat = view), pch = 16)
lines(trans3d(x = pts$x, y = pts$y, z = pts$z, pmat = view))

How can I plot data with confidence intervals?

If I have 10 values, each of which has a fitted value F, and an upper and lower confidence interval U and L:
set.seed(0815)
F <- runif(10, 1, 2)
L <- runif(10, 0, 1)
U <- runif(10, 2, 3)
How can I show these 10 fitted values and their confidence intervals in the same plot like the one below in R?
Here is a plotrix solution:
set.seed(0815)
x <- 1:10
F <- runif(10,1,2)
L <- runif(10,0,1)
U <- runif(10,2,3)
require(plotrix)
plotCI(x, F, ui=U, li=L)
And here is a ggplot solution:
set.seed(0815)
df <- data.frame(x =1:10,
F =runif(10,1,2),
L =runif(10,0,1),
U =runif(10,2,3))
require(ggplot2)
ggplot(df, aes(x = x, y = F)) +
geom_point(size = 4) +
geom_errorbar(aes(ymax = U, ymin = L))
UPDATE:
Here is a base solution to your edits:
set.seed(1234)
x <- rnorm(20)
df <- data.frame(x = x,
y = x + rnorm(20))
plot(y ~ x, data = df)
# model
mod <- lm(y ~ x, data = df)
# predicts + interval
newx <- seq(min(df$x), max(df$x), length.out=100)
preds <- predict(mod, newdata = data.frame(x=newx),
interval = 'confidence')
# plot
plot(y ~ x, data = df, type = 'n')
# add fill
polygon(c(rev(newx), newx), c(rev(preds[ ,3]), preds[ ,2]), col = 'grey80', border = NA)
# model
abline(mod)
# intervals
lines(newx, preds[ ,3], lty = 'dashed', col = 'red')
lines(newx, preds[ ,2], lty = 'dashed', col = 'red')
Here is a solution using functions plot(), polygon() and lines().
set.seed(1234)
df <- data.frame(x =1:10,
F =runif(10,1,2),
L =runif(10,0,1),
U =runif(10,2,3))
plot(df$x, df$F, ylim = c(0,4), type = "l")
#make polygon where coordinates start with lower limit and
# then upper limit in reverse order
polygon(c(df$x,rev(df$x)),c(df$L,rev(df$U)),col = "grey75", border = FALSE)
lines(df$x, df$F, lwd = 2)
#add red lines on borders of polygon
lines(df$x, df$U, col="red",lty=2)
lines(df$x, df$L, col="red",lty=2)
Now use example data provided by OP in another question:
Lower <- c(0.418116841, 0.391011834, 0.393297710,
0.366144073,0.569956636,0.224775521,0.599166016,0.512269587,
0.531378573, 0.311448219, 0.392045751,0.153614913, 0.366684097,
0.161100849,0.700274810,0.629714150, 0.661641288, 0.533404093,
0.412427559, 0.432905333, 0.525306427,0.224292061,
0.28893064,0.099543648, 0.342995605,0.086973739,0.289030388,
0.081230826,0.164505624, -0.031290586,0.148383474,0.070517523,0.009686605,
-0.052703529,0.475924192,0.253382210, 0.354011010,0.130295355,0.102253218,
0.446598823,0.548330752,0.393985810,0.481691632,0.111811248,0.339626541,
0.267831909,0.133460254,0.347996621,0.412472322,0.133671128,0.178969601,0.484070587,
0.335833224,0.037258467, 0.141312363,0.361392799,0.129791998,
0.283759439,0.333893418,0.569533076,0.385258093,0.356201955,0.481816148,
0.531282473,0.273126565,0.267815691,0.138127486,0.008865700,0.018118398,0.080143484,
0.117861634,0.073697418,0.230002398,0.105855042,0.262367348,0.217799352,0.289108011,
0.161271889,0.219663224,0.306117717,0.538088622,0.320711912,0.264395149,0.396061543,
0.397350946,0.151726970,0.048650180,0.131914718,0.076629840,0.425849394,
0.068692279,0.155144797,0.137939059,0.301912657,-0.071415593,-0.030141781,0.119450922,
0.312927614,0.231345972)
Upper.limit <- c(0.6446223,0.6177311, 0.6034427, 0.5726503,
0.7644718, 0.4585430, 0.8205418, 0.7154043,0.7370033,
0.5285199, 0.5973728, 0.3764209, 0.5818298,
0.3960867,0.8972357, 0.8370151, 0.8359921, 0.7449118,
0.6152879, 0.6200704, 0.7041068, 0.4541011, 0.5222653,
0.3472364, 0.5956551, 0.3068065, 0.5112895, 0.3081448,
0.3745473, 0.1931089, 0.3890704, 0.3031025, 0.2472591,
0.1976092, 0.6906118, 0.4736644, 0.5770463, 0.3528607,
0.3307651, 0.6681629, 0.7476231, 0.5959025, 0.7128883,
0.3451623, 0.5609742, 0.4739216, 0.3694883, 0.5609220,
0.6343219, 0.3647751, 0.4247147, 0.6996334, 0.5562876,
0.2586490, 0.3750040, 0.5922248, 0.3626322, 0.5243285,
0.5548211, 0.7409648, 0.5820070, 0.5530232, 0.6863703,
0.7206998, 0.4952387, 0.4993264, 0.3527727, 0.2203694,
0.2583149, 0.3035342, 0.3462009, 0.3003602, 0.4506054,
0.3359478, 0.4834151, 0.4391330, 0.5273411, 0.3947622,
0.4133769, 0.5288060, 0.7492071, 0.5381701, 0.4825456,
0.6121942, 0.6192227, 0.3784870, 0.2574025, 0.3704140,
0.2945623, 0.6532694, 0.2697202, 0.3652230, 0.3696383,
0.5268808, 0.1545602, 0.2221450, 0.3553377, 0.5204076,
0.3550094)
Fitted.values<- c(0.53136955, 0.50437146, 0.49837019,
0.46939721, 0.66721423, 0.34165926, 0.70985388, 0.61383696,
0.63419092, 0.41998407, 0.49470927, 0.26501789, 0.47425695,
0.27859380, 0.79875525, 0.73336461, 0.74881668, 0.63915795,
0.51385774, 0.52648789, 0.61470661, 0.33919656, 0.40559797,
0.22339000, 0.46932536, 0.19689011, 0.40015996, 0.19468781,
0.26952645, 0.08090917, 0.26872696, 0.18680999, 0.12847285,
0.07245286, 0.58326799, 0.36352329, 0.46552867, 0.24157804,
0.21650915, 0.55738088, 0.64797691, 0.49494416, 0.59728999,
0.22848680, 0.45030036, 0.37087676, 0.25147426, 0.45445930,
0.52339711, 0.24922310, 0.30184215, 0.59185198, 0.44606040,
0.14795374, 0.25815819, 0.47680880, 0.24621212, 0.40404398,
0.44435727, 0.65524894, 0.48363255, 0.45461258, 0.58409323,
0.62599114, 0.38418264, 0.38357103, 0.24545011, 0.11461756,
0.13821664, 0.19183886, 0.23203127, 0.18702881, 0.34030391,
0.22090140, 0.37289121, 0.32846615, 0.40822456, 0.27801706,
0.31652008, 0.41746184, 0.64364785, 0.42944100, 0.37347037,
0.50412786, 0.50828681, 0.26510696, 0.15302635, 0.25116438,
0.18559609, 0.53955941, 0.16920626, 0.26018389, 0.25378867,
0.41439675, 0.04157232, 0.09600163, 0.23739430, 0.41666762,
0.29317767)
Assemble into a data frame (no x provided, so using indices)
df2 <- data.frame(x=seq(length(Fitted.values)),
fit=Fitted.values,lwr=Lower,upr=Upper.limit)
plot(fit~x,data=df2,ylim=range(c(df2$lwr,df2$upr)))
#make polygon where coordinates start with lower limit and then upper limit in reverse order
with(df2,polygon(c(x,rev(x)),c(lwr,rev(upr)),col = "grey75", border = FALSE))
matlines(df2[,1],df2[,-1],
lwd=c(2,1,1),
lty=1,
col=c("black","red","red"))
Here is part of my program related to plotting confidence interval.
1. Generate the test data
ads = 1
require(stats); require(graphics)
library(splines)
x_raw <- seq(1,10,0.1)
y <- cos(x_raw)+rnorm(len_data,0,0.1)
y[30] <- 1.4 # outlier point
len_data = length(x_raw)
N <- len_data
summary(fm1 <- lm(y~bs(x_raw, df=5), model = TRUE, x =T, y = T))
ht <-seq(1,10,length.out = len_data)
plot(x = x_raw, y = y,type = 'p')
y_e <- predict(fm1, data.frame(height = ht))
lines(x= ht, y = y_e)
Result
2. Fitting the raw data using B-spline smoother method
sigma_e <- sqrt(sum((y-y_e)^2)/N)
print(sigma_e)
H<-fm1$x
A <-solve(t(H) %*% H)
y_e_minus <- rep(0,N)
y_e_plus <- rep(0,N)
y_e_minus[N]
for (i in 1:N)
{
tmp <-t(matrix(H[i,])) %*% A %*% matrix(H[i,])
tmp <- 1.96*sqrt(tmp)
y_e_minus[i] <- y_e[i] - tmp
y_e_plus[i] <- y_e[i] + tmp
}
plot(x = x_raw, y = y,type = 'p')
polygon(c(ht,rev(ht)),c(y_e_minus,rev(y_e_plus)),col = rgb(1, 0, 0,0.5), border = NA)
#plot(x = x_raw, y = y,type = 'p')
lines(x= ht, y = y_e_plus, lty = 'dashed', col = 'red')
lines(x= ht, y = y_e)
lines(x= ht, y = y_e_minus, lty = 'dashed', col = 'red')
Result
Some addition to the previous answers. It is nice to regulate the density of the polygon to avoid obscuring the data points.
library(MASS)
attach(Boston)
lm.fit2 = lm(medv~poly(lstat,2))
plot(lstat,medv)
new.lstat = seq(min(lstat), max(lstat), length.out=100)
preds <- predict(lm.fit2, newdata = data.frame(lstat=new.lstat), interval = 'prediction')
lines(sort(lstat), fitted(lm.fit2)[order(lstat)], col='red', lwd=3)
polygon(c(rev(new.lstat), new.lstat), c(rev(preds[ ,3]), preds[ ,2]), density=10, col = 'blue', border = NA)
lines(new.lstat, preds[ ,3], lty = 'dashed', col = 'red')
lines(new.lstat, preds[ ,2], lty = 'dashed', col = 'red')
Please note that you see the prediction interval on the picture, which is several times wider than the confidence interval. You can read here the detailed explanation of those two types of interval estimates.

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