I have some data in the folowing format:
y <- c(2637732, 2622262, 2637466, 2630985, 2620929, 2630888, 2625944, 2650034, 2645318, 2636731, 2629001, 2650776, 2648531, 2633905, 2654874, 2637571, 2650994, 2641130, 2652417, 2654005)
x <- c(756994.5, 760190.9, 760898.7, 761690.2, 763064.1, 763089.3, 765942.4, 767058.1, 768265.0, 768471.8, 771393.8, 771394.1, 775332.6, 778324.8, 780480.9, 780961.0,781001.5, 783904.7, 786200.6, 788007.5 )
z <- c(0.008849558,0.260162602,0.115044248,0.109243697,0.066666667,0.000000000,0.022556391,0.157894737,0.045045045,0.378151261,0.028776978,0.128571429,0.064220183,0.148760331,0.514851485,0.173913043,0.019417476,0.037383178,0.041237113,0.150537634)
Here is my code to interpolate the data
df <- data.frame(x=x,y=y,z=z);
gridint <- 500;
xmin <- signif(min(df$x),4) - 1000;
xmax <- signif(max(df$x),4) + 1000;
ymin <- signif(min(df$y),5) - 1000;
ymax <- signif(max(df$y),5) + 1000;
yo <- seq(ymin, ymax, length=gridint);
xo <- seq(xmin, xmax, length=gridint);
library(akima);
fld<- with(df, interp(x = x, y = y, z = z, linear = FALSE, extrap = TRUE, xo = xo, yo= yo));
fld2 <- as.data.frame(interp2xyz(fld));
I would like to create a structure like data(volcano) example to run the bellow script:
library(rgl);
data(volcano)
dim(volcano)
peak.height <- volcano;
ppm.index <- (1:nrow(volcano));
sample.index <- (1:ncol(volcano));
zlim <- range(peak.height)
zlen <- zlim[2] - zlim[1] + 1
colorlut <- terrain.colors(zlen) # height color lookup table
col <- colorlut[(peak.height-zlim[1]+1)] # assign colors to heights for each point
open3d()
ppm.index1 <- ppm.index*zlim[2]/max(ppm.index);
sample.index1 <- sample.index*zlim[2]/max(sample.index)
title.name <- paste("volcano plot3d", sep = "");
surface3d(ppm.index1, sample.index1, peak.height, color=col, back="lines", main = title.name);
grid3d(c("x", "y+", "z"), n =20)
sample.name <- paste("col.", 1:ncol(volcano), sep="");
sample.label <- as.integer(seq(1, length(sample.name), length = 5));
axis3d('y+',at = sample.index1[sample.label], sample.name[sample.label], cex = 0.3);
axis3d('y',at = sample.index1[sample.label], sample.name[sample.label], cex = 0.3)
axis3d('z',pos=c(0, 0, NA))
ppm.label <- as.integer(seq(1, length(ppm.index), length = 10));
axes3d('x', at=c(ppm.index1[ppm.label], 0, 0), abs(round(ppm.index[ppm.label], 2)), cex = 0.3);
title3d(main = title.name, sub = "test", xlab = "ppm", ylab = "samples", zlab = "peak")
rgl.bringtotop();
Can anyone help me? some advice?
Thanks in advance
You can use the deldir package to show a surface based on the raw data. The help page to read is ?persp3d.deldir.
library(rgl)
open3d()
plot3d(x, y, z) # Establish the axes, set labels etc. ...
library(deldir)
dxyz <- deldir(x, y, z = z, suppressMsge = TRUE)
col <- cm.colors(20)[1 + round(19*(z - min(z))/diff(range(z)))]
persp3d(dxyz, col = col, add = TRUE)
That produces this result:
If you don't want to show the points, you can use type = 'n' in the plot3d call, or skip it entirely and drop the add = TRUE argument to persp3d. The latter makes it a little harder to set the aspect ratio and titles.
First let's do it with the x,y,z values you started with:
str(fld)
List of 3
$ x: num [1:50] 756000 756673 757347 758020 758694 ...
$ y: num [1:50] 2619900 2620635 2621369 2622104 2622839 ...
$ z: num [1:50, 1:50] 0.255 0.256 0.257 0.258 0.259 ...
That can be plotted with the base graphics function persp:
png(); with(fld, persp(x,y,z) ) ; dev.off()
Now to build a proper rgl plot, if .... I can. Turns out you need to scale the range of the coordinates to [0-1] for rgl to have the correct aspect ratio to see anything. (Could also fiddle with aspect3d() but I've stumbled with that one
open3d()
with(fld, surface3d( (x -min(x))/(max(x) -min(x)),
(y -min(y))/(max(y) -min(y)),
(z -min(z))/(max(z) -min(z) )))
rgl.snapshot("test.png")
Related
I'm trying to print a vector field based on a matrix multiplication. The problem is that the function that will print values to make the matrix multiplication can only take a single number. When a range of number is put into the all.p function, the output is not usable to do the matrix multiplication. Is there a way to change all.p so that with multiple inputs, the matrix multiplication can still be valid, and the vector field can be computed? The code fails at the vectorfield function as this function with put the values into the range 0 to 1, but the all.p can't take multiple inputs.
geno.fit = matrix(c(0.791,1.000,0.834,
0.670,1.006,0.901,
0.657,0.657,1.067),
nrow = 3,
ncol = 3,
byrow = T)
all.p <- function(p) {
if (length(p)>1) {
stop("More numbers in input than expected")
}
P = p^2
PQ = 2*p*(1-p)
Q = (1-p)^2
return(list=c(P=P,PQ=PQ,Q=Q))
}
library(pracma)
f <- function(x, y) all.p(x) %*% geno.fit %*% all.p(y)
xx <- c(0, 1); yy <- c(0, 1)
vectorfield(fun = f, xlim = xx, ylim = yy, scale = 0.1)
for (xs in seq(0, 1, by = 0.25)) {
sol <- rk4(f, 0, 1, xs, 100)
lines(sol$x, sol$y, col="darkgreen")
}
grid()
I also tried to use a for loop.
f <- function(x, y, n = 16) {
space3 = matrix(NA,nrow = n,ncol = n)
for (i in 1:(length(x))) {
for (j in 1:(length(y))) {
# Calculate mean fitness
space3[i,j] = all.p(x[i]) %*% geno.fit %*% all.p(y[j])
}
}
return(space3)
}
xx <- c(0, 1); yy <- c(0, 1)
f(seq(0,1,length.out = 16), seq(0,1,length.out = 16))
vectorfield(fun = f, xlim = xx, ylim = yy, scale = 0.1)
Below is the code to make the gradient ascend (without the vectors).
library(fields) # for image.plot
res = 0.01
seq.x = seq(0,1,by = res)
space = outer(seq.x,seq.x,"*")
pace2 = space
for (i in 1:length(seq.x)) {
for (j in 1:length(seq.x)) {
space[i,j] = all.p(1-seq.x[i]) %*% geno.fit %*% all.p(1-seq.x[j])
}
}
round(t(space),3)
new.space = t(space)
image.plot(new.space)
by.text = 8
for (i in seq(1,length(seq.x),by = by.text)) {
for (j in seq(1,length(seq.x),by = by.text)) {
text(seq.x[i],seq.x[j],
labels = round(new.space[i,j],4),
cex = new.space[i,j]/2,
col = "black")
}
}
contour(new.space,ylim=c(1,0),add = T, nlevels = 50)
I was able to make the vector field function work, but it's not showing what I was expecting from the previous gradient ascend vector field:
How can the 2 be reconciled? (i.e., plotting the vectors on the gradient ascend image which would show the proper direction of the vectors in the steepest ascend)
Here is my solution:
library(fields) # for image.plot
library(plotly)
library(raster)
# Genotype fitness matrix -------------------------------------------------
geno.fit = matrix(c(0.791,1.000,0.834,
0.670,1.006,0.901,
0.657,0.657,1.067),
nrow = 3,
ncol = 3,
byrow = T)
# Resolution
res = 0.01
# Sequence of X
seq.x = seq(0,1,by = res)
# Make a matrix
space = outer(seq.x,seq.x,"*")
# Function to calculate the AVERAGE fitness for a given frequency of an allele to get the expected frequency of genotypes in a population
all.p <- function(p) { # Takes frequency of an allele in the population
if (length(p)>1) { # Has to be only 1 number
stop("More numbers in input than expected")
}
P = p^2 # Gets the AA
PQ = 2*p*(1-p) # gets the Aa
Q = (1-p)^2 # Gets the aa
return(list=c(P=P, # Return the values
PQ=PQ,
Q=Q))
}
# Examples
all.p(0)
all.p(1)
# Plot the matrix of all combinations of genotype frequencies
image.plot(space,
ylim=c(1.05,-0.05),
ylab= "Percentage of Chromosome EF of TD form",
xlab= "Percentage of Chromosome CD of BL form")
# Backup the data
space2 = space
# calculate the average fitness for EVERY combination of frequency of 2 genotypes
for (i in 1:length(seq.x)) {
for (j in 1:length(seq.x)) {
# Calculate mean fitness
space[i,j] = all.p(1-seq.x[i]) %*% geno.fit %*% all.p(1-seq.x[j])
}
}
# Show the result
round(t(space),3)
# Transform the space
new.space = t(space)
image.plot(new.space,
# ylim=c( 1.01,-0.01),
ylab= "Percentage of Chromosome EF of TD (Tidbinbilla) form",
xlab= "Percentage of Chromosome CD of BL (Blundell) form")
# Add the numbers to get a better sense of the average fitness values at each point
by.text = 8
for (i in seq(1,length(seq.x),by = by.text)) {
for (j in seq(1,length(seq.x),by = by.text)) {
text(seq.x[i],seq.x[j],
labels = round(new.space[i,j],4),
cex = new.space[i,j]/2,
col = "black") # col = "gray70"
}
}
# Add contour lines
contour(new.space,ylim=c(1,0),add = T, nlevels = 50)
# Plotly 3D graph --------------------------------------------------------
# To get the 3D plane in an INTERACTIVE graph
xyz=cbind(expand.grid(seq.x,
seq.x),
as.vector(new.space))
plot_ly(x = xyz[,1],y = xyz[,2],z = xyz[,3],
color = xyz[,3])
# Vector field on the Adaptive landscape ----------------------------------
library(tidyverse)
library(ggquiver)
raster2quiver <- function(rast, aggregate = 50, colours = terrain.colors(6), contour.breaks = 200)
{
names(rast) <- "z"
quiv <- aggregate(rast, aggregate)
terr <- terrain(quiv, opt = c('slope', 'aspect'))
quiv$u <- -terr$slope[] * sin(terr$aspect[])
quiv$v <- -terr$slope[] * cos(terr$aspect[])
quiv_df <- as.data.frame(quiv, xy = TRUE)
rast_df <- as.data.frame(rast, xy = TRUE)
print(ggplot(mapping = aes(x = x, y = y, fill = z)) +
geom_raster(data = rast_df, na.rm = TRUE) +
geom_contour(data = rast_df,
aes(z=z, color=..level..),
breaks = seq(0,3, length.out = contour.breaks),
size = 1.4)+
scale_color_gradient(low="blue", high="red")+
geom_quiver(data = quiv_df, aes(u = u, v = v), vecsize = 1.5) +
scale_fill_gradientn(colours = colours, na.value = "transparent") +
theme_bw())
return(quiv_df)
}
r <-raster(
space,
xmn=range(seq.x)[1], xmx=range(seq.x)[2],
ymn=range(seq.x)[1], ymx=range(seq.x)[2],
crs=CRS("+proj=utm +zone=11 +datum=NAD83")
)
# Draw the adaptive landscape
raster2quiver(rast = r, aggregate = 2, colours = tim.colors(100))
Not exactly what I wanted, but it does what I was looking for!
I have always been interested in learning about how computers are able to animate the "Mandelbrot Set" (https://en.wikipedia.org/wiki/Mandelbrot_set).
I found this website (https://www.dandelbrot.com/post/the-mandelbrot-set-in-r/) that shows how to create Mandelbrot Set:
mandelbrot_generator <- function(
p = 2,
q = 1,
xmin = -2.1, # minimum x value
xmax = 0.8, # maximum x value
nx = 500,
ymin = -1.3, # minimum y value
ymax = 1.3, # maximum y value
ny = 500,
n = 100,
showplot = TRUE, # If TRUE then display image,
showvals = FALSE,
cols = colorRampPalette(c("black","cyan","cyan3","black"))(11))
{
# variables
x <- seq(xmin, xmax, length.out=nx)
y <- seq(ymin, ymax, length.out=ny)
c <- outer(x,y*1i,FUN="+")
z <- matrix(0.0, nrow=length(x), ncol=length(y))
k <- matrix(0.0, nrow=length(x), ncol=length(y))
for (rep in 1:n) {
index <- which(Mod(z) < 2)
z[index] <- z[index]^p + c[index]*q
k[index] <- k[index] + 1
}
if (showplot==TRUE) { image(x,y,k,col=cols, xlab="Re(c)", ylab="Im(c)")}
if (showvals==TRUE) {return(k)}
}
Here is the plot:
mandelbrot_generator(p=2, q=1)
Does anyone know how to make an "animation" using the above code, so that it looks like this?
(https://en.wikipedia.org/wiki/Mandelbrot_set#/media/File:Mandelbrot_sequence_new.gif)
I always wondered : how are these animations made? I understand that a single picture of the Mandelbrot Set can be made, but how do these "zooming" animations work? Is this simply done by changing the "axis" (i.e. scale) of the picture?
The above picture is made from x = (-2.1, 0.8) and y = (-1.3, 1.3) - my understanding is that if we wanted to make a "zooming animation", we would "shrink" these ranges at each frame?
For example:
Frame 1: x = (-2.1, 0.8) and y = (-1.3, 1.3)
Frame 2 : x = (-1.9, 0.6) and y = (-1.1, 1.1)
Frame 3 : x = (-1.4, 0.3) and y = (0.7, 0.7)
etc.
Is this correct? Could we use this logic to make a series of "Mandelbrot frames" - and then fade/transition between these frames, giving them the illusion of being animated?
Thank you!
First, you should change a little your code like
mandelbrot_generator <- function(n){
p=2
q=1
xmin = -2.1 # minimum x value
xmax = 0.8 # maximum x value
nx = 500
ymin = -1.3 # minimum y value
ymax = 1.3 # maximum y value
ny = 500
showplot = TRUE # If TRUE then display image,
showvals = FALSE
cols = colorRampPalette(c("black","cyan","cyan3","black"))(11)
# variables
x <- seq(xmin, xmax, length.out=nx)
y <- seq(ymin, ymax, length.out=ny)
c <- outer(x,y*1i,FUN="+")
z <- matrix(0.0, nrow=length(x), ncol=length(y))
k <- matrix(0.0, nrow=length(x), ncol=length(y))
for (rep in 1:n) {
index <- which(Mod(z) < 2)
z[index] <- z[index]^p + c[index]*q
k[index] <- k[index] + 1
}
if (showplot==TRUE) { image(x,y,k,col=cols, xlab="Re(c)", ylab="Im(c)")}
if (showvals==TRUE) {return(k)}}
Then you can use the images to create a gif file, following this this thread, for instance.
To some mathematical aspects of Mandelbrot set, please see this thread, and find more on SearchOnMath.
I wonder how you can simplify these two :
plot (payroll,wins)
id = identify(payroll, wins,labels = code, n = 5)
plot (payroll,wins)
with(data, text(payroll, wins, labels = code, pos = 1, cex=0.5))
using other alternatives - pch() dan as.numeric()?
Not sure it's easier but you change pch during identification as below (taken from the R-help). Every time you click empty point change to filled-in dot.
# data simulation
data <- data.frame(payroll = rnorm(10), wins = rnorm(10), code = letters[1:10])
identifyPch <- function(x, y = NULL, n = length(x), plot = FALSE, pch = 19, ...)
{
xy <- xy.coords(x, y)
x <- xy$x
y <- xy$y
sel <- rep(FALSE, length(x))
while (sum(sel) < n) {
ans <- identify(x[!sel], y[!sel], labels = which(!sel), n = 1, plot = plot, ...)
if(!length(ans)) {
break
}
ans <- which(!sel)[ans]
points(x[ans], y[ans], pch = pch)
sel[ans] <- TRUE
}
## return indices of selected points
which(sel)
}
if(dev.interactive()) { ## use it
with(data, plot(payroll,wins))
id = with(data, identifyPch(payroll, wins))
}
I would like to add a gradient of colours following the fitted values (e.g. higher fitted values darker colours, lower fitted values lighter colours) in my three-dimensional dependence plots.
I have used the example presented in dismo package:
library(dismo)
data(Anguilla_train)
angaus.tc5.lr01 <- gbm.step(data=Anguilla_train, gbm.x = 3:13, gbm.y = 2,
family = "bernoulli", tree.complexity = 5, learning.rate = 0.01,
bag.fraction = 0.5)
# Find interactions in the gbm model:
find.int <- gbm.interactions( angaus.tc5.lr01)
find.int$interactions
find.int$rank.list
I have only managed to add the same colour to the whole plot:
gbm.perspec( angaus.tc5.lr01, 7, 1,
x.label = "USRainDays",
y.label = "SegSumT",
z.label = "Fitted values",
z.range=c(0,0.435),
col="blue")
Or to add a gradient colour but not following the fitted values:
gbm.perspec( angaus.tc5.lr01, 7, 1,
x.label = "USRainDays",
y.label = "SegSumT",
z.label = "Fitted values",
col=heat.colors(50),
z.range=c(0,0.435))
I also checked the code of function gbm.perspec, and If I understood correctly the fitted values are call inside the formula as "prediction", and later on are part of the "pred.matrix" that is passed to the final plotting: persp(x = x.var, y = y.var, z = pred.matrix...), but I have no managed to access them from the gbm.perspec formula. I tried to modified the gbm.perpec function by adding "col=heat.colors(100)[round(pred.matrix*100, 0)]" into the persp() inside the function, but it does not do what I am looking for:
persp(x = x.var, y = y.var, z = pred.matrix, zlim = z.range,
xlab = x.label, ylab = y.label, zlab = z.label,
theta = theta, phi = phi, r = sqrt(10), d = 3,
ticktype = ticktype,
col=heat.colors(100)[round(pred.matrix*100, 0)],
mgp = c(4, 1, 0), ...)
I believe the solution might come from modifying the gbm.perpec function, do you know how?
Thank you for your time!
Modifying the gbm.perspec function is certainly an option, although if you use the predicted values from the gbm model and plot them onto a 3D scatterplot from another package you should be able to achieve it as well.
Here's an option using the plot3Drgl package, original code was provided by #Fabrice.
library(dismo); library(plot3Drgl); library(devEMF)
data(Anguilla_train)
angaus.tc5.lr01 <- gbm.step(data=Anguilla_train, gbm.x = 3:13, gbm.y = 2,
family = "bernoulli", tree.complexity = 5, learning.rate = 0.01,
bag.fraction = 0.5)
# Find interactions in the gbm model:
find.int <- gbm.interactions( angaus.tc5.lr01)
find.int$interactions
find.int$rank.list
d<-plot(angaus.tc5.lr01,c(1,7),return.grid=T)
x <- d$SegSumT
y <- d$USRainDays
z <- d$y
grid.lines = 30
elevation.site = loess(z ~ x*y, data=d, span=1, normalize = FALSE)
x.pred <- seq(min(x), max(x), length.out = grid.lines) # x grid
y.pred <- seq(min(y), max(y), length.out = grid.lines) # y grid
xy <- expand.grid( x = x.pred, y = y.pred) # final grid combined
z.site=matrix(predict(elevation.site, newdata = xy), nrow = grid.lines, ncol = grid.lines) # predicedt matrix
scatter3D(x, y, z, theta = 160, phi = 35, # x y z coords and angle of plot
clab = c(""), # Needs moving - label legend
colkey = list(side = 4, length = 0.65,
adj.clab = 0.15, dist = -0.15, cex.clab = 0.6, cex.axis = 0.6), # change the location and length of legend, change position of label and legend
clim = c(-4,0.1),
bty = "b", # type of box
col = ramp.col(col = c("grey", "blue"), 200),
pch = 19, cex = 0.55, # shape and size of points
xlab = "SegSumT",
xlim=c(10,20),ylim=c(0,3.5), zlim=c(-4,0.1), d= 2,
ylab = "USRaindays",
zlab= "Fitted values", #axes labels
cex.lab = 0.8, font.lab = 1, cex.axis = 0.6, font.axis= 1, # size and font of axes and ticks
ticktype = "detailed", nticks = 5, # ticks and numer of ticks
#type = "h", # vertical lines
surf = list(x = x.pred, y = y.pred, z = z.site,
facets = NA, CI=NULL))
enter image description here
By tweaking with grid.lines and reversing the x axis you should be able to produce exactly what you want.
By incorporating some of the code found here into the gbm.perspec() source code you can create the desired effect.
First run
# Color palette (100 colors)
col.pal<-colorRampPalette(c("blue", "red"))
colors<-col.pal(100)
Then, add z.facet.center to gbm.perspec() source code after else and change the z in the code to pred.matrixas follows,
# and finally plot the result
#
if (!perspective) {
image(x = x.var, y = y.var, z = pred.matrix, zlim = z.range)
} else {
z.facet.center <- (pred.matrix[-1, -1] + pred.matrix[-1, -ncol(pred.matrix)] +
pred.matrix[-nrow(pred.matrix), -1] + pred.matrix[-nrow(pred.matrix), -ncol(pred.matrix)])/4
# Range of the facet center on a 100-scale (number of colors)
z.facet.range<-cut(z.facet.center, 100)
persp(x=x.var, y=y.var, z=pred.matrix, zlim= z.range, # input vars
xlab = x.label, ylab = y.label, zlab = z.label, # labels
theta=theta, phi=phi, r = sqrt(10), d = 3,
col=colors[z.facet.range],# viewing pars
ticktype = ticktype, mgp = c(4,1,0), ...) #
which will give you a plot like this (please note, this is not plotted using the sample dataset which is why the interaction effect is different than the plot in the question).
Alternatively, you can create a new function. The following example modifies gbm.perspec() to give a white-to-red gradient. Simply run the code in R, then change gbm.perspec() to gbm.perspec2()
# interaction function
# Color palette (100 colors)
col.pal<-colorRampPalette(c("white", "pink", "red"))
colors<-col.pal(100)
gbm.perspec2 <- function(gbm.object,
x = 1, # the first variable to be plotted
y = 2, # the second variable to be plotted
pred.means = NULL, # allows specification of values for other variables
x.label = NULL, # allows manual specification of the x label
x.range = NULL, # manual range specification for the x variable
y.label = NULL, # and y la seminar committeebel
z.label = "fitted value", #default z label
y.range = NULL, # and the y
z.range = NULL, # allows control of the vertical axis
leg.coords = NULL, #can specify coords (x, y) for legend
ticktype = "detailed",# specifiy detailed types - otherwise "simple"
theta = 55, # rotation
phi=40, # and elevation
smooth = "none", # controls smoothing of the predicted surface
mask = FALSE, # controls masking using a sample intensity model
perspective = TRUE, # controls whether a contour or perspective plot is drawn
...) # allows the passing of additional arguments to plotting routine
# useful options include shade, ltheta, lphi for controlling illumination
# and cex for controlling text size - cex.axis and cex.lab have no effect
{
if (! requireNamespace('gbm') ) { stop('you need to install the gbm package to use this function') }
requireNamespace('splines')
#get the boosting model details
gbm.call <- gbm.object$gbm.call
gbm.x <- gbm.call$gbm.x
n.preds <- length(gbm.x)
gbm.y <- gbm.call$gbm.y
pred.names <- gbm.call$predictor.names
family = gbm.call$family
# and now set up range variables for the x and y preds
have.factor <- FALSE
x.name <- gbm.call$predictor.names[x]
if (is.null(x.label)) {
x.label <- gbm.call$predictor.names[x]
}
y.name <- gbm.call$predictor.names[y]
if (is.null(y.label)) {
y.label <- gbm.call$predictor.names[y]
}
data <- gbm.call$dataframe[ , gbm.x, drop=FALSE]
n.trees <- gbm.call$best.trees
# if marginal variable is a vector then create intervals along the range
if (is.vector(data[,x])) {
if (is.null(x.range)) {
x.var <- seq(min(data[,x],na.rm=T),max(data[,x],na.rm=T),length = 50)
} else {
x.var <- seq(x.range[1],x.range[2],length = 50)
}
} else {
x.var <- names(table(data[,x]))
have.factor <- TRUE
}
if (is.vector(data[,y])) {
if (is.null(y.range)) {
y.var <- seq(min(data[,y],na.rm=T),max(data[,y],na.rm=T),length = 50)
} else {y.var <- seq(y.range[1],y.range[2],length = 50)}
} else {
y.var <- names(table(data[,y]))
if (have.factor) { #check that we don't already have a factor
stop("at least one marginal predictor must be a vector!")
} else {have.factor <- TRUE}
}
pred.frame <- expand.grid(list(x.var,y.var))
names(pred.frame) <- c(x.name,y.name)
pred.rows <- nrow(pred.frame)
#make sure that the factor variable comes first
if (have.factor) {
if (is.factor(pred.frame[,2])) { # swap them about
pred.frame <- pred.frame[,c(2,1)]
x.var <- y.var
}
}
j <- 3
# cycle through the predictors
# if a non-target variable find the mean
for (i in 1:n.preds) {
if (i != x & i != y) {
if (is.vector(data[,i])) {
m <- match(pred.names[i],names(pred.means))
if (is.na(m)) {
pred.frame[,j] <- mean(data[,i],na.rm=T)
} else pred.frame[,j] <- pred.means[m]
}
if (is.factor(data[,i])) {
m <- match(pred.names[i],names(pred.means))
temp.table <- table(data[,i])
if (is.na(m)) {
pred.frame[,j] <- rep(names(temp.table)[2],pred.rows)
} else {
pred.frame[,j] <- pred.means[m]
}
pred.frame[,j] <- factor(pred.frame[,j],levels=names(temp.table))
}
names(pred.frame)[j] <- pred.names[i]
j <- j + 1
}
}
#
# form the prediction
#
#assign("pred.frame", pred.frame, pos=1)
prediction <- gbm::predict.gbm(gbm.object,pred.frame,n.trees = n.trees, type="response")
#assign("prediction", prediction, pos=1, immediate =T)
# model smooth if required
if (smooth == "model") {
pred.glm <- glm(prediction ~ ns(pred.frame[,1], df = 8) * ns(pred.frame[,2], df = 8), data=pred.frame,family=poisson)
prediction <- fitted(pred.glm)
}
# report the maximum value and set up realistic ranges for z
max.pred <- max(prediction)
message("maximum value = ",round(max.pred,2),"\n")
if (is.null(z.range)) {
if (family == "bernoulli") {
z.range <- c(0,1)
} else if (family == "poisson") {
z.range <- c(0,max.pred * 1.1)
} else {
z.min <- min(data[,y],na.rm=T)
z.max <- max(data[,y],na.rm=T)
z.delta <- z.max - z.min
z.range <- c(z.min - (1.1 * z.delta), z.max + (1.1 * z.delta))
}
}
# now process assuming both x and y are vectors
if (have.factor == FALSE) {
# form the matrix
pred.matrix <- matrix(prediction,ncol=50,nrow=50)
# kernel smooth if required
if (smooth == "average") { #apply a 3 x 3 smoothing average
pred.matrix.smooth <- pred.matrix
for (i in 2:49) {
for (j in 2:49) {
pred.matrix.smooth[i,j] <- mean(pred.matrix[c((i-1):(i+1)),c((j-1):(j+1))])
}
}
pred.matrix <- pred.matrix.smooth
}
# mask out values inside hyper-rectangle but outside of sample space
if (mask) {
mask.trees <- gbm.object$gbm.call$best.trees
point.prob <- gbm::predict.gbm(gbm.object[[1]],pred.frame, n.trees = mask.trees, type="response")
point.prob <- matrix(point.prob,ncol=50,nrow=50)
pred.matrix[point.prob < 0.5] <- 0.0
}
#
# and finally plot the result
#
if (!perspective) {
image(x = x.var, y = y.var, z = pred.matrix, zlim = z.range)
} else {
z.facet.center <- (pred.matrix[-1, -1] + pred.matrix[-1, -ncol(pred.matrix)] +
pred.matrix[-nrow(pred.matrix), -1] + pred.matrix[-nrow(pred.matrix), -ncol(pred.matrix)])/4
# Range of the facet center on a 100-scale (number of colors)
z.facet.range<-cut(z.facet.center, 100)
persp(x=x.var, y=y.var, z=pred.matrix, zlim= z.range, # input vars
xlab = x.label, ylab = y.label, zlab = z.label, # labels
theta=theta, phi=phi, r = sqrt(10), d = 3,
col=colors[z.facet.range],# viewing pars
ticktype = ticktype, mgp = c(4,1,0), ...) #
}
}
if (have.factor) {
# we need to plot values of y for each x
factor.list <- names(table(pred.frame[,1]))
n <- 1
#add this bit so z.range still works as expected:
if (is.null(z.range)) {
vert.limits <- c(0, max.pred * 1.1)
} else {
vert.limits <- z.range
}
plot(pred.frame[pred.frame[,1]==factor.list[1],2],
prediction[pred.frame[,1]==factor.list[1]],
type = 'l',
#ylim = c(0, max.pred * 1.1),
ylim = vert.limits,
xlab = y.label,
ylab = z.label, ...)
for (i in 2:length(factor.list)) {
#factor.level in factor.list) {
factor.level <- factor.list[i]
lines(pred.frame[pred.frame[,1]==factor.level,2],
prediction[pred.frame[,1]==factor.level], lty = i)
}
# now draw a legend
if(is.null(leg.coords)){
x.max <- max(pred.frame[,2])
x.min <- min(pred.frame[,2])
x.range <- x.max - x.min
x.pos <- c(x.min + (0.02 * x.range),x.min + (0.3 * x.range))
y.max <- max(prediction)
y.min <- min(prediction)
y.range <- y.max - y.min
y.pos <- c(y.min + (0.8 * y.range),y.min + (0.95 * y.range))
legend(x = x.pos, y = y.pos, factor.list, lty = c(1:length(factor.list)), bty = "n")
} else {
legend(x = leg.coords[1], y = leg.coords[2], factor.list, lty = c(1:length(factor.list)), bty = "n", ncol = 2)
}
}
}
I am attempting to reproduce a corrgram (below; Fig 1) using Zuur et al (2010) reproducible R code (below) showing the frequency with which pairs of water- bird species both have zero abundance. The colour and the amount that a circle has been filled correspond to the proportion of observa- tions with double zeros. The diagonal running from bottom left to the top right represents the percentage of observations of a variable equal to zero..
I have adapted this code for my data but I am experiencing the same problem after running the code for both datasets. When I run the code, the circles inside the corrgram are not filling in, and remain empty (below; Figure 2).
I am however confused as to why I am hitting this problem. If anyone has a solution as to why this occurs, then I would be deeply appreciative for your help.
Data: By Zuur et al (2010)
The data is too large to include with this post but it can be found in the supporting materials section called ElphickBirdData.txt
R Code: Zuur et al (2010)
RiceField <- read.table(file="ElphickBirdData.txt", header = TRUE)
AllS <- c(
"TUSW", "GWFG", "WHGO", "CAGO", "MALL",
"GADW", "GWTE", "CITE", "UNTE", "AMWI", "NOPI",
"NOSH", "RIDU", "CANV", "BUFF", "WODU", "RUDU",
"EUWI", "UNDU", "PBGB", "SORA", "COOT", "COMO",
"AMBI", "BCNH", "GBHE", "SNEG", "GREG", "WFIB",
"SACR", "AMAV", "BNST", "BBPL", "KILL", "LBCU",
"GRYE", "LEYE", "LBDO", "SNIP", "DUNL", "WESA",
"LESA", "PEEP", "RUFF", "UNSH", "RBGU", "HEGU",
"CAGU", "GUSP")
#Determine species richness
Richness <- colSums(RiceField[,AllS] > 0, na.rm = TRUE)
#Remove all covariates
Birds <- RiceField[,AllS]
#To reduce the of variables in the figure, we only used the
#20 species that occured at more than 40 sites.
#As a result, N = 20. Else it becomes a mess.
Birds2 <- Birds[, Richness > 40]
N <- ncol(Birds2)
AllNames <- names(Birds2)
A <- matrix(nrow = N, ncol = N)
for (i in 1:N){
for (j in 1:N){
A[i,j] <- sum(RiceField[,AllS[i]]==0 & RiceField[,AllS[j]]==0, na.rm=TRUE)
}}
A1 <- A/2035
print(A1, digits = 2)
rownames(A1) <- AllNames
colnames(A1) <- AllNames
library(lattice)
library(RColorBrewer)
panel.corrgram.2 <- function(x, y, z, subscripts, at = pretty(z), scale = 0.8, ...)
{
require("grid", quietly = TRUE)
x <- as.numeric(x)[subscripts]
y <- as.numeric(y)[subscripts]
z <- as.numeric(z)[subscripts]
zcol <- level.colors(z, at = at, ...)
for (i in seq(along = z))
{
lims <- range(0, z[i])
tval <- 2 * base::pi *
seq(from = lims[1], to = lims[2], by = 0.01)
grid.polygon(x = x[i] + .5 * scale * c(0, sin(tval)),
y = y[i] + .5 * scale * c(0, cos(tval)),
default.units = "native",
gp = gpar(fill = zcol[i]))
grid.circle(x = x[i], y = y[i], r = .5 * scale,
default.units = "native")
}
}
levelplot(A1,xlab=NULL,ylab=NULL,
at=do.breaks(c(0.5,1.01),101),
panel=panel.corrgram.2,
scales=list(x=list(rot=90)),
colorkey=list(space="top"),
col.regions=colorRampPalette(c("red","white","blue")))
#Grey colours
levelplot(A1.bats,xlab=NULL,ylab=NULL,
at=do.breaks(c(0.5,1.01),101),
panel=panel.corrgram.2,
scales=list(x=list(rot=90)),
colorkey=list(space="top"),
col.regions=colorRampPalette(c(grey(0.8),grey(0.5),grey(0.2))))
Figure 1.
Figure 2
The cause of your problem is that grid.circles daubs grid.polygon with white. You can solved it by changing order of grid.circle and grid.polygon (or add gp = gpar(fill=NA) to grid.circle() ).
panel.corrgram.2.2 <- function(x, y, z, subscripts, at = pretty(z), scale = 0.8, ...)
{
require("grid", quietly = TRUE)
x <- as.numeric(x)[subscripts]
y <- as.numeric(y)[subscripts]
z <- as.numeric(z)[subscripts]
zcol <- level.colors(z, at = at, ...)
for (i in seq(along = z))
{
lims <- range(0, z[i])
tval <- 2 * base::pi *
seq(from = lims[1], to = lims[2], by = 0.01)
grid.circle(x = x[i], y = y[i], r = .5 * scale, # change the order
default.units = "native")
grid.polygon(x = x[i] + .5 * scale * c(0, sin(tval)),
y = y[i] + .5 * scale * c(0, cos(tval)),
default.units = "native",
gp = gpar(fill = zcol[i]))
}
}
levelplot(A1,xlab=NULL,ylab=NULL,
at=do.breaks(c(0.5,1.01),101),
panel=panel.corrgram.2.2,
scales=list(x=list(rot=90)),
colorkey=list(space="top"),
col.regions=colorRampPalette(c("red","white","blue")))