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我想绘制： production.ts(31, .002, 10,12,125313.93,211,95,x,"2014-02-01","2014-05-14",z,y)作为 x,y,z 的函数

就像 Mathematica 的这个情节一样，（如果可能在 R 中）：http: //i.stack.imgur.com/3PRaf.png

我有一个功能：

library("lubridate"); library("rgl")

production.ts <- function(a, b, z, c, d, e, 
                    f, g, h, j, r, k) {
  elapsed <- (4-z)*10 + (4-c)
  un.days <- 100 - elapsed
  gone.days <- day(as.Date(h))
  rem.days <- day(as.Date(j))
  r.days <- as.numeric(as.Date(j) - as.Date(h))

  m.r <- f/100*d
  inputs <- d * a * (gone.days - 1)/365 + r
  prin <- m.r + inputs 
  costs <- (r.days/365 * r + 1) * prin 
  added.p <- a/100*d + r 
  due <- d * 1-un.days
  tomr.f <- 1- due + k^2
  acct.paid <- (d - due)*tomr.f
  net <- added.p + due + acct.paid
  pv.net <- net/(1+r*(e-30-day(as.Date(j)))/365)
  end <- d - due - acct.paid
  more.add.p <- end*a*(rem.days-1)/365
  rem <- (f-g)/100 * end
  total.fv <- pv.net + rem + more.add.p 
  out <- costs - total.fv

  out 
}


x<-seq(-10,10,by=.1)
y<-seq(0,1000,by=.1)
z<-seq(0,90,by=.1)

我努力了： func.3d<-Vectorize(production.ts(31, .002, 10,12,125313.93,211,95,x,"2014-02-01","2014-05-14",z,y))

c <- func.3d; c <- cut(c,breaks=64); cols <- rainbow(64)[as.numeric(c)]

open3d()
plot3d(x, y, z, col=cols,type="s",size=1)

但这会绘制线条并且颜色与函数应输出的值不一致。

有谁知道我该怎么做？谢谢，我真的很感谢你的时间！

4

1 回答 1

3

Like this?

x<-seq(-10,10,length=100)
y<-seq(0,1000,length=100)
z<-seq(0,90,length=100)
df <- expand.grid(x=x,y=y,z=z)

f <- function(x,y,z) {production.ts(31, .002, 10,12,125313.93,211,95,x,"2014-02-01","2014-05-14",z,y)}
df$c <- f(df$x,df$y,df$z)
c <- cut(df$c,breaks=64)
cols <- rainbow(64)[as.numeric(c)]
open3d()
plot3d(df$x, df$y, df$z, col=cols,type="p",size=1)

Your code was not plotting lines. When you pass x, y, and z like that to plot3d(...) it cycles through all the elements together, so x[1],y[1],z[1] is a point, x[2],y[2],z[2] is another point, and so on. Since the vectors are different lengths, the shorter ones are recycled to fill out to the length of the longest. The visual effect of this is that the points lie on a line.

You want yo plot every combination of x, y, and z, and give each point a color based on that combination. The code above does that. The plot does not quite look like yours, but I can't tell if that is because of the way you have defined your function.

Also, the way you defined x, y, and z there would be 201 X 10001 X 901 = 1,811,191,101 points, which is too many to handle. The code above plots 1,000,000 points.

Finally, plotting spheres (type="s") is very expensive and unnecessary in this case.

于 2014-03-21T19:48:06.300 回答