[R] How to compare contours
dear All
can anybody point me in to the right direction for this kind of operation??
Here an example.
Please consider an hilly matematical landscapa as
i-1
X- runif(i, min=0, max=4*pi)
Y- runif(i, min=0, max=4*pi)
Z-(cos(X)+cos(Y))/2
plot(X,Y,xlim=c(0,4*pi),ylim=c(0,4*pi), xlab=X,ylab=Y, main=c(i, points))
coscos.spl-interp.new(X,Y
,Z,xo=seq(0,4*pi,length=100),yo=seq(0,4*pi,length=100))
contour(coscos.spl,add=T,col=blue,levels=c(seq(-1,1,1/5)),labcex=0.8)
in this case contour plot derived from 1 points interpolated rappresent very
welll real trend.
now consider
par(mfrow=c(4,5) )
for (i in seq(5,195,10)) {
X- runif(i, min=0, max=4*pi)
Y- runif(i, min=0, max=4*pi)
Z-(cos(X)+cos(Y))/2
plot(X,Y,xlim=c(0,4*pi),ylim=c(0,4*pi), xlab=X,ylab=Y, main=c(i, points))
coscos.spl-interp.new(X,Y
,Z,xo=seq(0,4*pi,length=100),yo=seq(0,4*pi,length=100))
contour(coscos.spl,add=T,col=blue,levels=c(seq(-1,1,1/5)),labcex=0.8)
}
How many points are necessary to fit (at 95% c.i.) true surface???
85 points???
or more than 200???
tnx in advance!!
-
Landini Massimiliano
-
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On Tue, 25 Jan 2005 15:42:45 +0100, you wrote: |=[:o) Dear R users, |=[:o) |=[:o) Is it reasonable to transform data (measurements of plant height) to the |=[:o) power of 1/4? I´ve used boxcox(response~A*B) and lambda was close to 0.25. |=[:o) IMHO (I'm far to be a statistician) no. I think that Box Cox procedure must be a help to people that had none experience in data transforming. In fact data transforming include other methods that Box Cox procedure can't perform as rank transformation, arcsine square root percent transformation, hyperbolic inverse sine, log-log, probit, normit and logit. Transformation is not simply an application of a formula to massive data. Is preferable decide appropriate transformation knowing deepening how and from where data were collected. |=[:o) Regards, |=[:o) Christoph |=[:o) |=[:o) __ |=[:o) R-help@stat.math.ethz.ch mailing list |=[:o) https://stat.ethz.ch/mailman/listinfo/r-help |=[:o) PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html - Landini dr. Massimiliano Tel. mob. (+39) 347 140 11 94 Tel./Fax. (+39) 051 762 196 e-mail: numero (dot) primo (at) tele2 (dot) it - Legge di Hanggi: Più stupida è la tua ricerca, più verrà letta e approvata. Corollario alla Legge di Hanggi: Più importante è la tua ricerca, meno verrà capita. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
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Re: Re: [R] muliple plots with pairs (matrix of scatter plots)
On Tue, 21 Dec 2004 12:47:44 +1300, you wrote:
|=[:o) Hi
|=[:o)
|=[:o)
|=[:o) Uwe Ligges wrote:
|=[:o) Tiago R Magalhaes wrote:
|=[:o)
|=[:o) I am trying to make a graph with 4 scatter matrixes plots and
couldn't
|=[:o) do it. While trying to find a solution for this I also came across
the
|=[:o) idea of giving different values to the same argument for each of the
|=[:o) lower and upper function but couldn't do it. (Examplified below with
|=[:o) the col argument). The first problem of plotting 4 scatter matrixes
in
|=[:o) a graph is a problem of real interest for me at this point. The
second
|=[:o) problem is a matter of curiosity.
|=[:o)
|=[:o) I am using a Mac PowerBook G4 with OS 10.3.7 and R 2.0.1
|=[:o)
|=[:o)
|=[:o) Problem 1)
|=[:o) x=data.frame(a=sample(1:100, 50), b=sample(1:100,
50),c=sample(1:100,
|=[:o) 50),d=sample(1:100, 50))
|=[:o) x.list=vector('list',4)
|=[:o) for (j in 1:4) x.list[[j]]=x
|=[:o)
|=[:o) #produces a graph with four plots:
|=[:o) layout(matrix(c(1,3,2,4),2,2))
|=[:o) for (j in seq(x)){
|=[:o) plot(x.list[[j]][1:2])
|=[:o) }
|=[:o)
|=[:o) # But unfortunately the following produces a new plot everytime:
|=[:o) layout(matrix(c(1,3,2,4),2,2))
|=[:o) for (j in seq(x)){
par(new=TRUE)
|=[:o) pairs(x.list[[j]])
|=[:o) }
|=[:o) #Maybe pairs can't be used to produce a graph with multiple plots?
|=[:o)
|=[:o)
|=[:o) Yes, it uses similar constructs to put multiple plots together.
|=[:o)
|=[:o) You might want to use packages grid and gridBase to set something up
|=[:o) using viewports.
|=[:o)
|=[:o)
|=[:o)
|=[:o) I don't think that's going to work either -- pairs() makes some pretty
|=[:o) strong assumptions that it is the only plot on the page.
|=[:o)
|=[:o) One possible way to go is to use splom() instead from the lattice
|=[:o) package. For example (using your data from above) ...
|=[:o)
|=[:o) splom(~ x)
|=[:o)
|=[:o) ... and lattice plots can be embedded in grid viewports easily, for
|=[:o) example ...
|=[:o)
|=[:o) grid.newpage()
|=[:o) pushViewport(viewport(layout=grid.layout(2, 2)))
|=[:o) for (j in seq(x)) {
|=[:o) row - (j - 1) %/% 2 + 1
|=[:o) col - (j - 1) %% 2 + 1
|=[:o) pushViewport(viewport(layout.pos.col=col,
|=[:o) layout.pos.row=row))
|=[:o) print(splom(~ x.list[[j]]), newpage=FALSE)
|=[:o) popViewport()
|=[:o) }
|=[:o) popViewport()
|=[:o)
|=[:o) ... you may need to fiddle with the splom() args to get them looking
how
|=[:o) you want them.
|=[:o)
|=[:o) Paul
-
Landini dr. Massimiliano
Tel. mob. (+39) 347 140 11 94
Tel./Fax. (+39) 051 762 196
e-mail: numero (dot) primo (at) tele2 (dot) it
-
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