R help - Nov 2013 - divergent colors around zero in levelplot()

I would like to produce a levelplot with divergent colors such that increasingly
negative values of Z get darker in the first color and increasingly
positive values get darker in the second color.  this is common in cartography.
I have tried tinkering with the col.regions argument but the best I can do
is to get the split in the middle of my range of Z, but in my particular case
range(Z) is (-1,12).

I am using R 3.0.2 on OSX 10.9

Here is an example

x <- y <- c(1:25) 
grid <- expand.grid(x=x,y=y)
grid$z <- sort(runif(625,min=-1,max=12))
levelplot(z ~ x*y,grid)   # produces the default pink and blue but the split is
at ~5.5

# do something clever here
# e.g., my.colors <- <create a palette that splits at zero>

levelplot(z ~ x*y,grid,col.regions=my.colors)  # so there should be some light
pink at the bottom and the rest increasingly intense blue

Ideas appreciated.  Thanks in advance.



Don McKenzie
Research Ecologist
Pacific Wildland Fire Sciences Lab
US Forest Service

Affiliate Professor 
School of Environmental and Forest Sciences 
University of Washington 
 
dmck at uw.edu

Use the Rcolorbrewer package.

-- Bert

On Fri, Nov 22, 2013 at 8:43 PM, Don McKenzie <dmck at u.washington.edu>
wrote:
> I would like to produce a levelplot with divergent colors such that
increasingly negative values of Z get darker in the first color and increasingly
> positive values get darker in the second color.  this is common in
cartography. I have tried tinkering with the col.regions argument but the best I
can do
> is to get the split in the middle of my range of Z, but in my particular
case range(Z) is (-1,12).
>
> I am using R 3.0.2 on OSX 10.9
>
> Here is an example
>
> x <- y <- c(1:25)
> grid <- expand.grid(x=x,y=y)
> grid$z <- sort(runif(625,min=-1,max=12))
> levelplot(z ~ x*y,grid)   # produces the default pink and blue but the
split is at ~5.5
>
> # do something clever here
> # e.g., my.colors <- <create a palette that splits at zero>
>
> levelplot(z ~ x*y,grid,col.regions=my.colors)  # so there should be some
light pink at the bottom and the rest increasingly intense blue
>
> Ideas appreciated.  Thanks in advance.
>
>
>
> Don McKenzie
> Research Ecologist
> Pacific Wildland Fire Sciences Lab
> US Forest Service
>
> Affiliate Professor
> School of Environmental and Forest Sciences
> University of Washington
>
> dmck at uw.edu
>
> ______________________________________________
> R-help at r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.


-- 

Bert Gunter
Genentech Nonclinical Biostatistics

(650) 467-7374

On Fri, 22 Nov 2013, Don McKenzie wrote:
> I would like to produce a levelplot with divergent colors such that
increasingly negative values of Z get darker in the first color and increasingly
> positive values get darker in the second color.  this is common in
cartography. I have tried tinkering with the col.regions argument but the best I
can do
> is to get the split in the middle of my range of Z, but in my particular
case range(Z) is (-1,12).
>
> I am using R 3.0.2 on OSX 10.9
>
> Here is an example
>
> x <- y <- c(1:25)
> grid <- expand.grid(x=x,y=y)
> grid$z <- sort(runif(625,min=-1,max=12))
> levelplot(z ~ x*y,grid)   # produces the default pink and blue but the
split is at ~5.5
>
> # do something clever here
> # e.g., my.colors <- <create a palette that splits at zero>
>
> levelplot(z ~ x*y,grid,col.regions=my.colors)  # so there should be some
light pink at the bottom and the rest increasingly intense blue
>
> Ideas appreciated.  Thanks in advance.
One approach is to limit the range of colors (to match the range of the 
data) as you suggest above. The other approach is to extend the range of 
the legend (beyond the range of the data). For example:

levelplot(z ~ x*y, grid, at = seq(-12, 12, length = 100))

This produces a legend that is symmetric around zero.

For other/better diverging color palettes, you can use the RColorBrewer 
package (as suggested by Bert) or the colorspace package (see e.g., its 
graphical choose_color() tool).
>
>
> Don McKenzie
> Research Ecologist
> Pacific Wildland Fire Sciences Lab
> US Forest Service
>
> Affiliate Professor
> School of Environmental and Forest Sciences
> University of Washington
>
> dmck at uw.edu
>
> ______________________________________________
> R-help at r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>

On Tue, Nov 26, 2013 at 1:02 AM, C. Alina Cansler <acansler at uw.edu>
wrote:
> Don,
>
> This looks helpful:
> https://stat.ethz.ch/pipermail/r-help/2011-March/272361.html
Yes, he's a helpful chap.

 The fundamental problem here is the colour palette. When I was a boy
all we had was a pen plotter with four coloured pens, and of course
you could stick different coloured pens in the different pen slots and
draw four different coloured lines. So your graphics package just said
which number pen it was going to use and the colour that came out was
up to you. Most R graphics functions still have this concept, although
there may be over 100 pens and you don't end up swearing when one runs
out of ink. A numeric value is converted to an integer and the integer
does a lookup in a palette to get the colour.

 What you really want, and what my colourscheme package did, was to
create functions that let users map values directly to colours. You
could then plot points and lines using that function which meant
totally controlled value-to-colour mappings that could be used across
different plots if desired. That worked because points and lines lets
you specify a colour value directly using something like col="#FF23EC"
(as well as allowing col=23 and doing a palette lookup).

 But the image function (and probably levelplot) doesn't allow that so
there's various tricks to make functional colour lookups work. I would
convert the image matrix values to a matrix of colours, then create a
matrix of the values 1:(n*m), and then image() that 1:(n*m) matrix
using the colour matrix as a palette. That way each cell had its own
palette entry, and you controlled that colour using the value-colour
function.

 Or you could just use ggplot which I'm pretty sure has the same
concept of mapping values to colours.

Barry

On Tue, Nov 26, 2013 at 9:58 AM, Prof Brian Ripley
<ripley at stats.ox.ac.uk> wrote:
>>   But the image function (and probably levelplot) doesn't allow
that so
>
> Mis-information alert!   The help says
>
>       col: a list of colors such as that generated by ?rainbow?,
>            ?heat.colors?, ?topo.colors?, ?terrain.colors? or similar
>            functions.
>
> and look at what they generate.  Or see e.g. ?col2rgb .
>
> Although base graphics has the concept of a palette of colours, AFAIK it
> has always been bolted on top of a general colour specification,
> originally RGB and for many years already RGBA.

 Yes image allows you to specify col=, but it always specifies a
palette. The matrix values are scaled from 1:length(col) and looked up
in that palette. You can't call image with z as matrix of colours and
get those colours, nor set col to a matrix of colours and a see those
colours laid out. This is unlike points() where specifying col= as a
vector of the same length as the number of points gives you a 1:1
mapping of points to colours.

 To do image() with a 1:1 mapping of cell values to colours requires a
tiny bit of hoop-jumping.

Barry