> On 11 Apr 2017, at 20:55, dncdd <dn...@aliyun.com <mailto:dn...@aliyun.com>>
> wrote:
>
> Thank you Ismail SEZEN.
> The link you give is filled.contour code which only works with my first mini
> data .
filled.contour + lattice::levelplot solution can handle either matrix or a 3
column (x,y,z) data.frame by using the formula z ~ x + y.
>
> The code in the link for R 3.3.2 is :
> ****** code ***** in **** link **** for R 3.3.2 ****
> panel.filledcontour <- function(x, y, z, subscripts, at, col.regions =
> cm.colors,
> col = col.regions(length(at) - 1), ...)
> {
> stopifnot(require("gridBase"))
> z <- matrix(z[subscripts],
> nrow = length(unique(x[subscripts])),
> ncol = length(unique(y[subscripts])))
> if (!is.double(z)) storage.mode(z) <- "double"
> opar <- par(no.readonly = TRUE)
> on.exit(par(opar))
> if (panel.number() > 1) par(new = TRUE)
> par(fig = gridFIG(), omi = c(0, 0, 0, 0), mai = c(0, 0, 0, 0))
> cpl <- current.panel.limits()
> plot.window(xlim = cpl$xlim, ylim = cpl$ylim,
> log = "", xaxs = "i", yaxs = "i")
> # paint the color contour regions
> .filled.contour(as.double(do.breaks(cpl$xlim, nrow(z) - 1)),
> as.double(do.breaks(cpl$ylim, ncol(z) - 1)),
> z, levels = as.double(at), col = col)
> # add contour lines
> contour(as.double(do.breaks(cpl$xlim, nrow(z) - 1)),
> as.double(do.breaks(cpl$ylim, ncol(z) - 1)),
> z, levels = as.double(at), add=T,
> col = "gray", # color of the lines
> drawlabels=F # add labels or not
> )
> }
> plot.new()
>
> print(levelplot(volcano, panel = panel.filledcontour,
> col.regions = terrain.colors,
> cuts = 10,
> plot.args = list(newpage = FALSE)))
> *** END *** code *** in *** link *** for R 3.3.2 ***
>
> first mini data
> which should be a three dimensinal data either and the data in matrix is not
> a function of rdn and tdn
> which means z matrix is not function of x,y.
Here we are not interested in z is function of x and y but at the and, if you
want to plot a 3D data on a flat surface, you need x and y for each z in the 3D
space.
>
> rdn<-c(0.8,1.8,2.8)
> tdn<-c(1,2,3,4,5,6,7,8,9)
>
> idn<-matrix(c(0.3, 0.3, 0.3, 0.2, 0.2, 0.4, 0.1, 0.1, 0.5, 0, 0.2, 0.5,
> 0, 0.3, 0.6, 0, 0.4, 0.6, 0, 0.4, 0.6, 0, 0.5, 0.7, 0, 0.5, 0.7), nrow=9,
> ncol=3, byrow=T)
>
> And the matrix looks like(3*9 = 27 data elements):
>
> 0.3, 0.3, 0.3,
> 0.2, 0.2, 0.4,
> 0.1, 0.1, 0.5,
> 0, 0.2, 0.5,
> 0, 0.3, 0.6,
> 0, 0.4, 0.6,
> 0, 0.4, 0.6,
> 0, 0.5, 0.7,
> 0, 0.5, 0.7
As I mentioned above, you have z values for each x and y values in the space.
one of elements of z might be NA/NaN but at last you have.
>
>
> Well, now I realized that the second data might (my current problem) be afour
> dimensional data:
>
> r1dn<-c(0.8,1.8,2.8)
> r2dn<-c(0.8,1.8,2.8)
> tdn<-c(0.1,0.2,0.3,0.4,0.5,0.6,0.7,0.8,0.9)
Thing are changed now. We are living in a 4D space (at least that we can sense)
and there are only 2 ways to visualize 4D data on a flat surface. These are
your only chance.
1- 3D perspective plotting (one of rgl, misc3d, scatterplot3d). Plot 'points'
in the 3D space and set 4th dimension as color for each point. Does it make
sense? I don’t know, it’s up to you.
2- for the (x,y,z,v) dimensions, plot contours of (x,y,v) for each z. I mean,
assume z is height, slice it and plot each slice by filled.contour/levelplot or
what ever you want.
>
> And (3*3*9 = 81 data elements):
>
> 0.8 1.8 2.8
> 0.8 1.8 2.8 0.8 1.8 2.8 0.8 1.8 2.8
>
> --------------- 81 ---- elements ------three matrix----------------
>
> 0.3, 0.3, 0.3, 0.3, 0.3, 0.5, 0.3, 0.3, 0.3,
> 0.2, 0.2, 0.4, 0.2, 0.4, 0.4, 0.4, 0.2, 0.5,
> 0.1, 0.1, 0.5, 0.2, 0.3, 0.5, 0.4, 0.4, 0.5,
> 0, 0.2, 0.5, 0.2, 0.2, 0.6, 0.4, 0.5, 0.6,
> 0, 0.3, 0.6, 0.3, 0.3, 0.6, 0.5, 0.5, 0.7,
> 0, 0.4, 0.6, 0.2, 0.5, 0.7, 0.5, 0.6, 0.7,
> 0, 0.4, 0.6, 0, 0.5, 0.6, 0.5, 0.6, 0.9,
> 0, 0.5, 0.7, 0, 0.6, 0.8, 0.5, 0.7, 0.8,
> 0, 0.5, 0.7 0, 0.6, 0.8 0.5, 0.8, 0.9
>
> The three matrix is not the function of r1dn, r2dn, tdn. r1dn, r2dn, tdn can
> be labels. So there are four dimensional data. x is r1dn, y is r2dn, z is tdn
> and the three matrix is, let's say, vdn.
> Four dimension: r1dn r2dn tdn fdn as x,y,z,v. And v is not the function of
> x,y,z. So there are might need a 3d filled.contour. But I did not find it.
> All the code I found is that x,y,z and z is a function of x,y. Another
> situation I found is that x,y,z,v and v is function of x,y,z. But in my data,
> v is not a function of x,y,z.
Actually, you need to detail (in your mind and to us) what do you mean by being
function of. A point can be represented by 4 points in 4D space (x,y,z,t).
According to this; v is short for value (measurement); v(x,y,z,t) can represent
a value in space and time. So this is a 5D data I assume. We mostly use one or
2 or 3 of this dimensions. So, v is function of (x, y, z, t) and change by (x,
y, z, t). For instance, if v is not function of time (t), it is always constant
and does not change in time. Hence, I dont need 4th dimension (t here) and I
don’t need to plot v in a 4D space. This is what I know about being function of.
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