[R] Need a hint
Gregory Snow
Greg.Snow at intermountainmail.org
Thu Feb 23 23:22:20 CET 2006
Here are a couple of quick thoughts on your problem.
1. Use alpha channels (may require you to produce all your graphs as pdf
files).
Fill each of your criteria categories with a mostly transparent color,
e.g. the full contour of z[1] between 20 and 30 is 20% opaque and the
full contour(s) of z[2] < 40 is 20% opaque. Then where they overlap
will be 40% opaque and stand out (and if you have 5 critera then where
they all overlap will be 100% opaque.
2. create a dataframe with all your z's predicted over a regular grid of
x and y values (possibly the same set as used for the contours), then
create a logical variable that ands together all your critera, e.g.:
New <- transform(old, z.combined = 20 < z1 & z1 < 30 & z2 < 40)
Then do a levelplot with the new logical variable as the response (maybe
do as.numeric on it first), then overlay your contours on top of the
levelplot.
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.snow at intermountainmail.org
(801) 408-8111
> -----Original Message-----
> From: r-help-bounces at stat.math.ethz.ch
> [mailto:r-help-bounces at stat.math.ethz.ch] On Behalf Of Mike Saunders
> Sent: Thursday, February 23, 2006 3:02 PM
> To: R Help
> Subject: [R] Need a hint
>
> R community:
>
> I have been creating code for plotting nomographs, or
> multiple, overlain contour plots of z-variables on a common
> x- and y- variable. My input has been a matrix with observed
> x, y, and multiple z variables; I then create a trend surface
> using trmat for each z-variable. So far so good.
>
> One application I have for these, requires shading a portion
> of the nomogram that meets criteria for some of the
> z-variables (i.e., z[1] must be between 20 and 30, z[2] must
> be less than 40, etc.). My solution was to use a logical
> comparison on each contour surface provided by trmat, sum the
> "logical surfaces" up and see if they were less than the
> total number of criteria. It works, but it is quite
> inefficient even if I vectorize the code somewhat; for
> example if I specify a gridsize of 200 in trmat, have 5 z
> variables, and 1 criteria for each, I will have well over
> 200,000 comparisons to make! So I am looking for hints or
> maybe an entirely different approach to speed this up.
>
> I attached the crit.region function below along with my write
> up on how it works. Can somebody give me some ideas on how
> to proceed?
>
> Thanks,
> Mike
>
> Mike R. Saunders
> Forest Biometrician
> Cooperative Forest Research Unit
> University of Maine
> 5755 Nutting Hall
> Orono, ME 04469-5755
>
> 207-581-2763 (O)
> 207-581-2833 (F)
>
>
> # The following function selects a region that meets a set of
> # criteria defined in terms of z-variables in a list from
> nomogram # or a similarly formatted list. This function
> basically is a set # of logical comparisons on z-values at
> each xy-coordinate. As such, # the function is rasterized
> and can take considerable time when # each z-variable matrix
> is quite large. Parameters for the # function are:
> #
> # 1) x (Required) Either a list consisting of a vector
> # of gridded x-coordinates, a vector of
> # gridded y-coordinates and matrices of
> # each z-variable, or a vector of just
> # the gridded x-coordinates.
> # 2) y (Optional) A vector of gridded y-coordinates.
> # 3) z (Optional) A matrix or data.frame of z-variates
> # that correspond to the gridded
> # xy-coordinates.
> # 4) critmat (Required) A matrix or data.frame with rows equal
> # to the number of z-variables and 2
> # columns. The first column corresponds
> # to the minimum value allowed for each
> # z-variable, the second to the maximum
> # value. If there is no minimum or
> # maximum for a variable, NA should be
> # used in the appropriate row and column.
> #
> # This function returns the critical area as a matrix of NA
> and 1 # with dimension equal to a z-variable matrix. The
> function also # returns a message if there is no critical
> area solution.
> #
> # [Future versions of this function will try to improve its #
> computational speed.] #
> crit.region<-function(x,y=NULL,z=NULL,critmat) {
> if(all(missing(y),missing(z))) {
>
> stopifnot(class(x)=="list",sum(lapply(x,class)[1:2]!="numeric"
)==0,sum(sapply(x,class)[3:length(x)]!="matrix")==>
0,length(x[[1]])==dim(x[[3]])[1],length(x[[2]])==dim(x[[3]])[2
> ],length(x)>4)
> y<-x[[2]]
> z<-x[c(3:length(x))]
> x<-x[[1]]
> } else if(any(missing(y),missing(z))) {
> stop("y and z are both required unless x is properly
> formatted list")
> } else
> stopifnot(class(y)=="numeric",class(z)=="list",length(x)==dim(
> z[[1]])[1],length(y)==dim(z[[1]])[2],sum(sapply(z,class)!="mat
rix")==0)
> w<-length(z)
> zrange<-sapply(z,range,na.rm=T)
>
> stopifnot(class(critmat)%in%c("matrix","data.frame"),dim(critm
at)==c(w,2))
> critarea<-matrix(data=0,nrow=dim(z[[1]])[1],ncol=dim(z[[1]])[2])
> for(i in 1:w) {
> minz<-ifelse(is.na(critmat[i,1]),zrange[1,i],critmat[i,1])
> maxz<-ifelse(is.na(critmat[i,2]),zrange[2,i],critmat[i,2])
> critarea<-critarea+apply(z[[i]],c(1,2), function(x)
> ifelse(x>minz & x<maxz,1,0))
> }
> critarea<-apply(critarea,c(1,2), function(x) ifelse(x==w,1,NA))
> if(sum(critarea,na.rm=T)==0) message("Critical region is
> empty set!")
> return(critarea)
> }
>
>
>
> [[alternative HTML version deleted]]
>
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