将3D散点图转换为3D线性图并根据颜色进行分离

Question

一段时间以来，我一直在努力在R中绘制3D图表。我想我非常接近我想要的。我之前曾问过question。我现在只需要知道如何将带点的散点图转换为线性。我的意思是，只要我能把这些观点联系起来，对我来说是很棒的。我现在的样子如下： 我需要连接3D视角更好的点。我需要为每种颜色单独添加一条线，并想在图表中添加图例。

我将数据定义为：

df <- data.frame(a1 = c(489.4,  505.8,  525.8,  550.2,  576.6),
a2 = c(197.8,  301,    389.8,  502,    571.2),
b1 = c(546.8,  552.6,  558.4,  566.4,  575),
b2 = c(287.2,  305.8,  305.2,  334.4,  348.6), c1 = c(599.6,  611.4,  
623.6,  658,    657.4), c2 = c(318.8,  423.2,  510.8,  662.4,  656),
d1 = c(616,    606.8,  600.2,  595.6,  595),
  d2 = c(242.4,  292.8,  329.2,  378,    397.2),
e1 = c(582.4,  580,    579,    579,    579),
e2 = c(214,    255.4,  281.8,  303.8,  353.8))

colnames(df) <- rep(c("V1", "V2"), 5)
df.new <- rbind(df[, c(1, 2)],df[, c(3, 4)],df[, c(5, 6)],               
df[, c(7, 8)],df[, c(9, 10)])
df.new$Group <- factor(rep(c("a","b","c","d","e"), each = 5))
df.new$Class <- rep(c(1:5), 5)
x=df.new$Class
y=V1
z=V2

下面是我的代码：

 library(scatterplot3d)  #colors
 colors <- c("#999999", "#E69F00", "#56B4E9","#1B9E77", "#D95F02")
 colors <- colors[as.numeric(df.new$Group)]#Others
 xlabs <- c("[7,9]", "[10,12]", "[16,18]", "[19,21]", "[22,24]")
 scatterplot3d(x,y,z, pch = 16, color=colors,main="Title",xlab 
 ="Intervals",ylab = "", zlab = "Total time",     x.ticklabs=xlabs)
 text(8, 2.4, "c",cex = 1)
 text(9, 2, "c",cex = 1)

如果有人可以帮助我解决我一直在努力的这个问题，我将非常感谢。我知道这里有一个type = 1，但它只能构成一个统一的图。

Answer 1

尝试一下：

sd<-scatterplot3d(x,y,z, pch = rep(16:12, each=5), color=colors,main="Title",xlab 
                  ="Intervals",ylab = "", zlab = "Total time", x.ticklabs=xlabs)
sd$points3d(x[1:5],y[1:5],z[1:5], col="purple", type="l")
sd$points3d(x[6:10],y[6:10],z[6:10], col="orange", type="l")
sd$points3d(x[11:15],y[11:15],z[11:15], col="blue", type="l")
sd$points3d(x[16:20],y[16:20],z[16:20], col="green", type="l")
sd$points3d(x[21:25],y[21:25],z[21:25], col="red", type="l")
legend("right", legend = levels(df.new$Group), col= levels(as.factor(colors)),pch = rep(16:12, each=1))