R-beta: image(x,y,z)?

Bill Simpson (wsimpson@uwinnipeg.ca)
Tue, 28 Apr 1998 10:28:21 -0500 (CDT)


Date: Tue, 28 Apr 1998 10:28:21 -0500 (CDT)
From: Bill Simpson <wsimpson@uwinnipeg.ca>
To: r-help <r-help@stat.math.ethz.ch>
Subject: R-beta: image(x,y,z)?

Ok now I've read in my dataframe.  The file is set up like this:
x y z
1 1 1 
1 2 2
1 3 5
2 1 3
2 2 9
2 3 2
3 1 8
3 2 4
3 3 7

I can get at data$x, data$y, data$z. I want to do an image plot.  Ideally
image (or a relative of image) would accept the vectors data$x, data$y,
data$z as arguments. (After all, if you can do plot(x,y) on vectors x and
y, why can't you do image(x,y,z) on vectors x, y, and z?) However it
doesn't work like that.

I have been fooling around with image. The examples show something like
x<-seq(0,1)
y<-x
func<-function(x,y) {x/(x+y)}
z<-outer(x,y,"func")
image(z)
or 
image(x,y,z)

The bit with outer is required to make z a matrix.

I tried this idea:
 x<-seq(1,4)
 y<-x
 z<-seq(1,16)
 func<-function(x,y) {z}
 z<-outer(x,y,"func")
# z
#     [,1] [,2] [,3] [,4]
#[1,]    1    5    9   13
#[2,]    2    6   10   14
#[3,]    3    7   11   15
#[4,]    4    8   12   16

image(x,y,z,col=gray(16:1/16))

I still have the problem that in reality my x and y vectors are not in
ascending order as required by image.
Anyway, can someone please explain how to get my data into a form
acceptable to image?  Thanks very much!

Bill Simpson

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