在R中将用黄土绘制的线分成两部分

Question

我正在尝试使用命令loess 在 R 中制作最适合的一行，但问题是它应该由两部分组成，而我的数据只有一个向量...

所以我首先有这个命令：

x = 1:generaties
CI.up = as.numeric(gemiddelde)+as.numeric(ci)
CI.dn = as.numeric(gemiddelde)-as.numeric(ci)
plot(gemiddelde~x, cex=0.5,xaxt='n',ylim=c(0,1.1), xlab='Populatie(a = 100, size = 500, n = 15, tau = 10, c = 50, sigma = 0.5, phenoNumber = 5)',ylab='Fitness', main='Fitness over  generaties',col='blue',pch=16)
axis(1, at=x)
arrows(x,CI.dn,x,CI.up,code=3,length=0.05,angle=90,col='red')

这给了我一个图表

然后我尝试添加：

x1 = 1:(generaties/2)
x2 = (generaties/2):generaties
lo1 <- loess(gemiddelde[1:(generaties/2)]~x1)
lo2 <- loess(gemiddelde[(generaties/2):generaties]~x2)
lines(predict(lo1), col='red', lwd=2)
lines(predict(lo2), col='red', lwd=2)

但随后 I R 从 0 开始两条线，而不是从我的图表中途开始的第二条线。

对此我能做些什么？

Answer 1

我想你快到了 - 只需要为 lines() 命令指定一些 x 值。希望我创建的数据足以代表您的数据：

#Make up some data
set.seed(0)
generaties <- 20
gemiddelde <- 0.5+0.2*sin(1:generaties)
ci <- runif(generaties)*0.3

#your plot
x = 1:generaties
CI.up = as.numeric(gemiddelde)+as.numeric(ci)
CI.dn = as.numeric(gemiddelde)-as.numeric(ci)
plot(gemiddelde~x, cex=0.5,xaxt='n',ylim=c(0,1.1), xlab='Populatie(a = 100, size = 500, n = 15, tau = 10, c = 50, sigma = 0.5, phenoNumber = 5)',ylab='Fitness', main='Fitness over  generaties',col='blue',pch=16)
axis(1, at=x)
arrows(x,CI.dn,x,CI.up,code=3,length=0.05,angle=90,col='red')

#Adding the loess smoother
x1 = 1:(generaties/2)
x2 = (generaties/2):generaties
lo1 <- loess(gemiddelde[1:(generaties/2)]~x1)
lo2 <- loess(gemiddelde[(generaties/2):generaties]~x2)
lines(x1,predict(lo1), col='red', lwd=2)  #specify x values
lines(x2,predict(lo2), col='red', lwd=2)  #specify x values