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我正在使用circular
库中的rose.diag
函数绘制角度分布。输入数据是辐射值。我的 MWE 代码是
library(circular);
dat<-read.csv(file.choose(),header=F);
data=unlist(dat);
rose.diag(data, bins=24)
我感兴趣的是仅显示从 -pi/2 到 pi/2 的部分数据,并且最大刻度的长度等于圆半径的长度,如下所示:
如有任何帮助,我们将不胜感激!
编辑
根据#lawyeR的建议,这里是带有数据示例的代码:
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);
最佳答案
也许您可以先使用默认绘图函数绘制没有分布的半圆。然后填写不带圆圈的分布:
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 )
关于r - 使用圆形包创建半个极坐标图(玫瑰图),我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/33168771/
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