I am plotting a distribution of angles with the rose.diag function from the circular library. Input data are radiants. My MWE code is
library(circular);
dat<-read.csv(file.choose(),header=F);
data=unlist(dat);
rose.diag(data, bins=24)
and I get this graph:
I'm interested in showing only part of the data, from -pi/2 to pi/2, and that the length of the biggest tick equals the length of the radius of the circle as show here:
Any help would be appreciated!
EDIT
As suggested by #lawyeR here is the code with a sample of the data:
library(circular);
data<- c(-0.188,-0.742,-0.953,-0.948,-0.953,-1.187,-0.9327200,-0.855,- 0.024,1.303,-1.041,-1.068,-1.066,1.442,1.150,0.965,0.665,0.649,0.984,-1.379,-0.584,-0.573,-0.357,-0.237,-0.287,-0.486,-0.783,-0.298,0.849,1.088,-1.003,-0.952,-0.776,-0.811,-0.880);
rose.diag(data, bins=24);
Perhaps you can first draw the half circle without the distribution, using default plot functions. Then fill in the distribution without its circle:
library(circular)
data<- c(-0.188,-0.742,-0.953,-0.948,-0.953,-1.187,-0.9327200,-0.855,- 0.024,1.303,-1.041,-1.068,-1.066,1.442,1.150,0.965,0.665,0.649,0.984,-1.379,-0.584,-0.573,-0.357,-0.237,-0.287,-0.486,-0.783,-0.298,0.849,1.088,-1.003,-0.952,-0.776,-0.811,-0.880)
freq <- diff(colSums(outer( data %% (2*pi), (1:24)*pi/12,"<"))) / length(data)
r.max <- sqrt(max(freq))
#-----------------------------------------------------------------
# Plot the half circle:
lab.width <- 0.15*r.max
lab.height <- 0.15*r.max
plot( c(-r.max,r.max), c(0,0),
axes=FALSE,
ylim=c(0,r.max+lab.height),
xlim=c(-r.max-lab.width,r.max+lab.width),
xlab="", ylab="", type="l")
for ( i in 0:(5*12-1) )
{
psi <- i*pi/(5*12)
x1 <- r.max*cos(psi)
y1 <- r.max*sin(psi)
x2 <- r.max*cos(psi+pi/(5*12))
y2 <- r.max*sin(psi+pi/(5*12))
lines( c(x1,x2), c(y1,y2), type="l")
if (i %% 5 == 0) { lines( x1*c(1,0.95), y1*c(1,0.95), type="l" ) }
}
par(cex=2.0)
text( x = c(-r.max,0,r.max),
y = c(0,r.max,0),
labels = c("-pi/2","0","pi/2"),
pos = c(2,3,4))
#------------------------------------------------------------
# Plot the distribution, but without the circle:
rose.diag(data,
bins = 24,
rotation = "clock",
tcl.text = NA,
ticks = FALSE,
zero = pi/2,
control.circle = circle.control( col="white" ),
add = TRUE )
.
Related
I am analyzing difference scores with polynomial regression in R. Based on [Edwards and Parry's (1993)][1] recommendations I have been trying to combine a persp() plot with a contour() plot. I would also need to plot the first two principal axes on the contour plot. My attempts so far have only provided me with each individual plot, but I don't know how to combine them. An example for the end-result is :
Edwards & Parry (1993) example difference score visualisation
I manage to get the persp() plot just fine. I have also obtained the contour plot. I can't seem to find any way to combine the two. I have managed to make the plot in plotly using the add_surface() option in the pipeline. My problem with the output is that the surface is smooth, and the contourplot lacks the values in the plot. Basically: persp() and contour() are visualised in a way that is extremely similar to the look I'm aiming for, per the example in the source.
My current attempt (in minimalistic code) is as follows:
surface <- function(e, i){
y <- .2*e + .14*i + .08*e^2 + + .1*e*i + .2*i^2
}
e <- i <- seq(-3, 3, length= 20)
y <- outer(e, i, surface)
persp(e, i, y,
xlab = 'Explicit',
ylab = 'Implicit',
zlab = 'Depression',
theta = 45)
contour(e,i,y)
So basically my question is: how can I make a plot like Edwards and Parry (1993) make, with a similar visual style, in R. It does not have to be base-R, I'm happy with any method. I've been stuck on this problem for a week now.
My attempt in plotly (to compare it to my desired end-result) is:
if(!"plotly" %in% installed.packages){install.packages('plotly')}
library(plotly)
plot_ly(z = ~y) %>% add_surface(x = ~e, y= ~i, z= ~y,
contours = list(
z = list(
show=TRUE,
usecolormap=FALSE,
highlightcolor="#ff0000",
project=list(z=TRUE)
)
)
) %>%
layout(
scene=list(
xaxis = list(title = "Explicit"),
yaxis = list(title = "Implicit"),
zaxis = list(title = "Depression")
)
)
[1]: Edwards, J. R., & Parry, M. E. (1993). On the use of polynomial regression as an alternative to difference scores. Academy of Management Journal, 36(6), 1577–1613. https://doi.org/10.2307/256822
I have found an answer and I will share it here. It seems it cannot be done in base-R. But the RSM-package allows for the addition of contour lines to the base of the plot.
In this answer I will give a minimal example of:
the persp() plot
the contour lines in the base
addition of x=y and x=-y axis
calculation and addition of the first and second principal axis
The only thing I could not solve is that the lines now are drawn over the surface. I don't know how to solve it.
library(rsm)
x <- seq(-3,3,by=0.25)
y <- seq(-3,3,by=0.25)
d <- expand.grid(x=x,y=y)
z <- c(data=NA,1089)
b0 = .140; b1 = -.441; b2 = -.154; b3 = .161 ; b4 =-.106; b5 = .168
k=1
for (i in 1:25) {
for (j in 1:25) {
z[k]=b0+b1*x[i]+b2*y[j]+b3*x[i]*x[i]+b4*x[i]*y[j]+ b5*y[j]*y[j]
k=k+1
} }
data.lm <- lm(z~poly(x,y,degree=2),data=d)
res1 <- persp(data.lm,x~y,
zlim=c(-2,max(z)),
xlabs = c('X','Y'),
zlab = 'Z',
contour=list(z="bottom"),
theta=55,
phi=25)
# draw x=y line (lightly dotted)
xy_pos <- matrix(c(-3,-3,3,3),ncol=2,byrow = T)
lines(trans3d(xy_pos[,2], xy_pos[,1], z=-2, pmat = res1$`y ~ x`$transf),
lty = 3,
col = 'darkgrey')
# draw x=-y line (lightly dotted)
xy_neg <- matrix(c(-3,3,3,-3),ncol=2,byrow = T)
lines(trans3d(xy_neg[,2], xy_neg[,1], z=-2, pmat = res1$`y ~ x`$transf),
lty = 3,
col = 'darkgrey')
# Find stationary points:
X0 <- (b2*b4 - 2*b1*b5) / (4*b3*b5 - b4^2)
Y0 <- (b1*b4 - 2*b2*b3) / (4*b3*b5 - b4^2)
# First Principal Axis
p11 = (b5-b3+sqrt((b3-b5)^2+b4^2))/b4
p10 = Y0 - p11*X0
Ypaf1 = p10 + p11*x
# plot first principal axis (full line)
xypaf1 <- matrix(c(Ypaf1[1], -3, Ypaf1[25], 3),ncol=2, byrow=T)
lines(trans3d(xypaf1[,2], xypaf1[,1], z=-2, pmat = res1$`y ~ x`$transf),
lty = 1,
col = 'black')
# Second Principal Axis
p21 = (b5-b3-sqrt((b3-b5)^2+b4^2))/b4
p20 = Y0 - p21*X0
Ypaf2 = p20 + p21*x
# plot second principal axis (dashed line)
xypaf2 <- matrix(c(Ypaf2[1], -3, Ypaf2[25], 3),ncol=2, byrow=T)
lines(trans3d(xypaf2[,2], xypaf2[,1], z=-2, pmat = res1$`y ~ x`$transf),
lty = 2,
col = 'black')
I have written my own code to simulate the Conway maxwell distribution sample.
This is the pmf (Guikema & Goffelt, 2008):
However, I have met some problem to plot the density plot.
rcomp <- function(n,lamb,v)
{
u <- runif(n)
w <- integer(n)
for(i in 1:n) {
z=sum(sapply( 0:100, function(j) (( ((lamb)^j) / (factorial(j)) )^v) ))
x <- seq(1, 50, 1) #seq of 1 to 50, increase by 1
px <- (((lamb^x)/factorial(x))^v)/z
# px is pmf of re-parameter conway maxwell
w[i] <- if (u[i] < px[1]) 0 else (max (which (cumsum(px) <= u[i])))
}
return (w)
}
dcomp <- function(x,lamb,v) {
z=sum(sapply( 0:100, function(j) (( ((lamb)^j) / (factorial(j)) )^v) ))
px <- (((lamb^x)/factorial(x))^v)/z
return(px)
}
As I wanna plot the density plot to check whether lamb or v is location parameter, the plot I get is weird.
x = rcomp(100,6,0.2); pdf = dcomp(x,6,0.2)
x1 = rcomp(100,6,0.5); pdf1 = dcomp(x1,6,0.5)
x2 = rcomp(100,6,0.7); pdf2 = dcomp(x2,6,0.7)
plot(x2, pdf2, type="l", lwd=1,lty=1,col="blue")
How could I solve this problem?
Source: Guikema & Goffelt (2008), A Flexible Count Data Regression Model for Risk Analysis. Risk Analysis 28(1): 215.
You have to sort the values of the x coordinate if you want a graph to connect the points in their axis order.
Note, however, that there might be better ways to graph the density you want. See the red curve. I first create a vector x of values within a certain range and then compute the PDF for those values. These pairs (x, y) are what function lines plots.
set.seed(2673) # Make the results reproducible
x2 <- rcomp(100, 6, 0.7)
x2 <- sort(x2)
pdf2 <- dcomp(x2, 6, 0.7)
plot(x2, pdf2, type = "l", lwd = 1, lty = 1, col = "blue")
x <- seq(0, 50, length.out = 100)
y <- dcomp(x, 6, 0.2)
lines(x, y, type = "l", col = "red")
I have data cdecn:
set.seed(0)
cdecn <- sample(1:10,570,replace=TRUE)
a <- rnorm(cdecn,mean(cdecn),sd(cdecn))
I have created a plot which displays the cumulative probabilities.
aprob <- ecdf(a)
plot(aprob)
I am wondering how I can switch the x-axis and y-axis to get a new plot, i.e., the inverse of ECDF.
Also, for the new plot, is there a way to add a vertical line through where the my curve intersects 0?
We can do the following. My comments along the code is very explanatory.
## reproducible example
set.seed(0)
cdecn <- sample(1:10,570,replace=TRUE)
a <- rnorm(cdecn,mean(cdecn),sd(cdecn)) ## random samples
a <- sort(a) ## sort samples in ascending order
e_cdf <- ecdf(a) ## ecdf function
e_cdf_val <- 1:length(a) / length(a) ## the same as: e_cdf_val <- e_cdf(a)
par(mfrow = c(1,2))
## ordinary ecdf plot
plot(a, e_cdf_val, type = "s", xlab = "ordered samples", ylab = "ECDF",
main = "ECDF")
## switch axises to get 'inverse' ECDF
plot(e_cdf_val, a, type = "s", xlab = "ECDF", ylab = "ordered sample",
main = "'inverse' ECDF")
## where the curve intersects 0
p <- e_cdf(0)
## [1] 0.01578947
## highlight the intersection point
points(p, 0, pch = 20, col = "red")
## add a dotted red vertical line through intersection
abline(v = p, lty = 3, col = "red")
## display value p to the right of the intersection point
## round up to 4 digits
text(p, 0, pos = 4, labels = round(p, 4), col = "red")
cdecn <- sample(1:10,570,replace=TRUE)
a <- rnorm(cdecn,mean(cdecn),sd(cdecn))
aprob <- ecdf(a)
plot(aprob)
# Switch the x and y axes
x <- seq(0,1,0.001754386)
plot(y=knots(aprob), x=x, ylab = "Fn(y)")
# Add a 45 degree straight line at 0, 0
my_line <- function(x,y,...){
points(x,y,...)
segments(min(x), y==0, max(x), max(y),...)
}
lines(my_line(x=x, y = knots(aprob)))
The "straight line at x==0" bit makes me suspect that you want a QQplot:
qqnorm(a)
qqline(a)
I would like to produce this kind of graph:
However, I don't know how to do it using R. I was wondering if someone knew a solution to do it in R?
I would use the package rgl.
library(rgl)
# load your data
X= c(1:6)
Y=seq(10,70, 10)
Z=c(-70, -50, -30, -20, -10, 10)
# create an empty plot with the good dimensions
plot3d(1,1,1, type='n', xlim=c(min(X),max(X)),
ylim=c(min(Y),max(Y)),
zlim=c(min(Z),max(Z)),
xlab="", ylab="", zlab="", axe=F )
# draw your Y bars
for(i in X){ segments3d(x = rep(X[i],2), y = c(0,Y[i]), z=0, lwd=6, col="purple")}
# do the same for the Z bars
plot3d(X,0,Z, add=T, axe=F, typ="n")
for(i in X){segments3d(x = rep(X[i],2), y = 0, z= c(0,Z[i]), lwd=6, col="blue" )}
# draw your axis
axes3d()
mtext3d(text = "Time (days)", edge = "y+", line =3, col=1 )
mtext3d(text = "Change %", edge = "z++", line = 5, col=1 )
However I have found the width of the bars restricted to 6. That could be a limit. Better looking when you have more data.
Hope it could help.
I am using package vioplot. I would like to ask, how can I create violinplot in different colours.
This is my reproducible example:
# Violin Plots library(vioplot)
x1 <- mtcars$mpg[mtcars$cyl==4]
x2 <- mtcars$mpg[mtcars$cyl==6]
x3 <- mtcars$mpg[mtcars$cyl==8]
vioplot(x1, x2, x3,
names=c("4 cyl", "6 cyl", "8 cyl"), col="gold")
title("Violin Plots of Miles Per Gallon")
Thank you.
It is not possible to have many colors. But it is not difficult to hack the function vioplot and edit the source code. Here steps you should follow to accomplish this:
copy the initial function:
my.vioplot <- vioplot()
edit this function:
edit(my.vioplot)
Search the word "polygon" and and replace col by col[i]
Do a test in the beginning of function for the case you give a single color. and add this line :
if(length(col)==1) col <- rep(col,n)
For example using your data :
vioplot(x1, x2, x3, names=c("4 cyl", "6 cyl", "8 cyl"), col="gold")
title("Violin Plots of Miles Per Gallon")
my.vioplot(x1, x2, x3, names=c("4 cyl", "6 cyl", "8 cyl"), col=c("gold","red","blue"))
title("Violin Plots of Miles Per Gallon multi colors")
To expand on agstudy's answer and correct one thing, here is the complete and new vioplot script.
Use source("vioplot.R") instead of library(vioplot) in your script to use this multicolor version instead. This one will repeat any colors until it reaches the same number of datasets.
library(sm)
vioplot <- function(x,...,range=1.5,h=NULL,ylim=NULL,names=NULL, horizontal=FALSE,
col="magenta", border="black", lty=1, lwd=1, rectCol="black", colMed="white", pchMed=19, at, add=FALSE, wex=1,
drawRect=TRUE)
{
# process multiple datas
datas <- list(x,...)
n <- length(datas)
if(missing(at)) at <- 1:n
# pass 1
#
# - calculate base range
# - estimate density
#
# setup parameters for density estimation
upper <- vector(mode="numeric",length=n)
lower <- vector(mode="numeric",length=n)
q1 <- vector(mode="numeric",length=n)
q3 <- vector(mode="numeric",length=n)
med <- vector(mode="numeric",length=n)
base <- vector(mode="list",length=n)
height <- vector(mode="list",length=n)
baserange <- c(Inf,-Inf)
# global args for sm.density function-call
args <- list(display="none")
if (!(is.null(h)))
args <- c(args, h=h)
for(i in 1:n) {
data<-datas[[i]]
# calculate plot parameters
# 1- and 3-quantile, median, IQR, upper- and lower-adjacent
data.min <- min(data)
data.max <- max(data)
q1[i]<-quantile(data,0.25)
q3[i]<-quantile(data,0.75)
med[i]<-median(data)
iqd <- q3[i]-q1[i]
upper[i] <- min( q3[i] + range*iqd, data.max )
lower[i] <- max( q1[i] - range*iqd, data.min )
# strategy:
# xmin = min(lower, data.min))
# ymax = max(upper, data.max))
#
est.xlim <- c( min(lower[i], data.min), max(upper[i], data.max) )
# estimate density curve
smout <- do.call("sm.density", c( list(data, xlim=est.xlim), args ) )
# calculate stretch factor
#
# the plots density heights is defined in range 0.0 ... 0.5
# we scale maximum estimated point to 0.4 per data
#
hscale <- 0.4/max(smout$estimate) * wex
# add density curve x,y pair to lists
base[[i]] <- smout$eval.points
height[[i]] <- smout$estimate * hscale
# calculate min,max base ranges
t <- range(base[[i]])
baserange[1] <- min(baserange[1],t[1])
baserange[2] <- max(baserange[2],t[2])
}
# pass 2
#
# - plot graphics
# setup parameters for plot
if(!add){
xlim <- if(n==1)
at + c(-.5, .5)
else
range(at) + min(diff(at))/2 * c(-1,1)
if (is.null(ylim)) {
ylim <- baserange
}
}
if (is.null(names)) {
label <- 1:n
} else {
label <- names
}
boxwidth <- 0.05 * wex
# setup plot
if(!add)
plot.new()
if(!horizontal) {
if(!add){
plot.window(xlim = xlim, ylim = ylim)
axis(2)
axis(1,at = at, label=label )
}
box()
for(i in 1:n) {
# plot left/right density curve
polygon( c(at[i]-height[[i]], rev(at[i]+height[[i]])),
c(base[[i]], rev(base[[i]])),
col = col[i %% length(col) + 1], border=border, lty=lty, lwd=lwd)
if(drawRect){
# plot IQR
lines( at[c( i, i)], c(lower[i], upper[i]) ,lwd=lwd, lty=lty)
# plot 50% KI box
rect( at[i]-boxwidth/2, q1[i], at[i]+boxwidth/2, q3[i], col=rectCol)
# plot median point
points( at[i], med[i], pch=pchMed, col=colMed )
}
}
}
else {
if(!add){
plot.window(xlim = ylim, ylim = xlim)
axis(1)
axis(2,at = at, label=label )
}
box()
for(i in 1:n) {
# plot left/right density curve
polygon( c(base[[i]], rev(base[[i]])),
c(at[i]-height[[i]], rev(at[i]+height[[i]])),
col = col[i %% length(col) + 1], border=border, lty=lty, lwd=lwd)
if(drawRect){
# plot IQR
lines( c(lower[i], upper[i]), at[c(i,i)] ,lwd=lwd, lty=lty)
# plot 50% KI box
rect( q1[i], at[i]-boxwidth/2, q3[i], at[i]+boxwidth/2, col=rectCol)
# plot median point
points( med[i], at[i], pch=pchMed, col=colMed )
}
}
}
invisible (list( upper=upper, lower=lower, median=med, q1=q1, q3=q3))
}
Don't forget geom_violin in the ggplot2 package. There are examples of how to change the fill colour in the docs: http://docs.ggplot2.org/0.9.3/geom_violin.html
Plotting the vectors 1-by-1 seem easier than modifying the function:
require(vioplot)
yalist = list( rnorm(100), rnorm(100, sd = 1),rnorm(100, sd = 2) )
plot(0,0,type="n",xlim=c(0.5,3.5), ylim=c(-10,10), xaxt = 'n', xlab ="", ylab = "Pc [%]", main ="Skanderbeg")
for (i in 1:3) { vioplot(na.omit(yalist[[i]]), at = i, add = T, col = c(1:3)[i]) }
axis(side=1,at=1:3,labels=3:1)