Creating half a polar plot (rose diagram) with circular package

Question

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);

Thanks for your answer.I've edited the post to add a sample of the data. — michltm, Oct 16 '15 at 12:36
Good. I edited your question and added another tag, hopefully to attract an answer or two. — lawyeR, Oct 16 '15 at 14:53

mra68 · Accepted Answer · 2015-10-19T21:36:05.937

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 )

.

That's an excellent idea! Thanks for your help. I'd have a last question: what would be the best way to add -pi/2, 0, pi/2 on the outer circle? — michltm, Oct 19 '15 at 09:14
I've edited my answer. You can use `text` to add the labels. — mra68, Oct 19 '15 at 21:38

Creating half a polar plot (rose diagram) with circular package

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