我想创建以下类型的情节。但是不知道是否有任何包已经存在,因为我找不到一个。 数据:
myd <- data.frame (X = 1:5, Y = c(0.8, 0.6, 0.7, 0.75, 0.1),
clockd = c(12.05, 12.25, 12.45, 1.30, 2.1))
clockd是时间12.05是12点钟过去几分钟。我不知道在R中输入时间的适当单位。
plot (myd$X, myd$Y)
修改 红色和绿色箭头表示时钟中钟表的小时和小时位置。
答案 0 :(得分:6)
clockplot<-function(x, y, h, m, r, border="black", col="lightblue",
col.hour="darkblue", col.min="red"){
#x and y are the coordinates of the clock
#h and m the hour (base 12) and minutes
# r the radius of the clock
t<-seq(0,2*pi,by=0.01)
x.c<-r*cos(t)+x
y.c<-r*sin(t)+y
t.h<-pi/2 - 2*pi*(h-m/60)/12
x.h<-0.5*r*cos(t.h)+x
y.h<-0.5*r*sin(t.h)+y
t.m<-pi/2 - 2*pi*m/60
x.m<-r*cos(t.m)+x
y.m<-r*sin(t.m)+y
polygon(x.c,y.c,col=col, border=border)
segments(x,y,x.h,y.h,col=col.hour)
segments(x,y,x.m,y.m,col=col.min)
}
myd <- data.frame (X = 1:5, Y = c(0.8, 0.6, 0.7, 0.75, 0.1),
clockd = c(12.05, 12.25, 12.45, 1.30, 2.1))
myd$hour<-myd$clockd%/%1
myd$min<-myd$clockd%%1 *100
plot(myd$X, myd$Y, type="l", asp=1)
apply(myd,1,function(x)clockplot(x[1],x[2],x[4],x[5], r=0.25))
答案 1 :(得分:5)
我在@GregSnow使用来自my.symbols
包的TeachingDemo
函数的建议。这样你可以使用一小块(ish)代码来完成它,但你可以对图形参数进行很大的控制,比如时钟圆的大小和填充,箭头的外观,你自定义绘图的方式等。用于计算位置我修改了@agstudy代码,以便正确提取分钟数。
myd <- data.frame (X = 1:5, Y = c(0.8, 0.6, 0.7, 0.75, 0.1),
clockd = c(12.05, 12.25, 12.45, 1.30, 2.1))
hour <- round(myd$clockd)#takes hours by ignoring decimals
minute <- 100*(myd$clockd - trunc(myd$clockd,2))#takes decimals
#for getting the angle I'm subtracting from pi/2
#thats because pi/2 orients the arrow into 0 degree position, pointing up
hourAngle <- pi/2 - (hour/12*2*pi)
minuteAngle <- pi/2 - (minute/60*2*pi)
#now all the plotting
plot(myd$X, myd$Y, type="l", xaxt="n", xlab="", ylab="",
xlim=c(0.5,5.5), ylim=c(0,1), col="gray")#standard plot, no x axis
axis(1, at=myd$X, labels=myd$X)#custom x-axis
require(TeachingDemo)
my.symbols(myd$X, myd$Y, ms.arrows, angle=hourAngle, add=T,
col="blue", symb.plots=TRUE, adj=0)
my.symbols(myd$X, myd$Y, ms.arrows, angle=minuteAngle, add=T,
col="red", symb.plots=TRUE, adj=0)
my.symbols(myd$X, myd$Y, ms.polygon, n=250, add=T,
r=1.1, col="gray")
答案 2 :(得分:4)
提供您的数据:
read.table(text = 'col1 col2
0012 0001245',head=T,colClasses=c('character','numeric'))
myd <- data.frame (X = 1:5,
Y = c(0.8, 0.6, 0.7, 0.75, 0.1),
clockd = c(12.05, 12.25, 12.45, 1.30, 2.1))
使用格子+网格我们可以这样做:
library(lattice)
library(grid)
xyplot(Y~X,data=myd,
panel=function(x,y,...)
{
panel.fill(col='yellow')
panel.lines(x,y)
h_min <- do.call(rbind,strsplit(as.character(myd$clockd),'[.]'))
hours <- as.numeric(h_min[,1])
minutes <- as.numeric(h_min[,2])
lapply(seq_along(x),
function(i)drawClock(x[i],y[i],hour = hours[i], minute = minutes[i]))
})
从this采用drawClock并根据需要提出这个问题:
drawClock <- function(x0,y0,hour, minute) {
t <- seq(0, 2*pi, length=13)[-13]
x <- cos(t)
y <- sin(t)
# Circle with ticks
grid.circle(x=x0, y=y0, default="native",
r=unit(0.4,'in'),gp=gpar(fill='blue',alpha=0.5))
# grid.segments(x, y, x*.9, y*.9, default="native")
# Hour hand
hourAngle <- pi/2 - (hour + minute/60)/12*2*pi
grid.segments(x0, y0,
x0+.06*cos(hourAngle), y0+.06*sin(hourAngle),
default="native", gp=gpar(lex=4))
# Minute hand
minuteAngle <- pi/2 - (minute)/60*2*pi
grid.segments(x0, y0,
x0+.08*cos(minuteAngle), y0+.08*sin(minuteAngle),
default="native", gp=gpar(lex=2))
}
答案 3 :(得分:2)
查看TeachingDemos包中的my.symbols
函数,了解有助于创建此类情节的工具。