Try:
n = 100;
x = matrix(seq(-10, 30, length.out=500), ncol=1)
col = colorRampPalette( c("green", "white",
"red"), space="rgb")(n) breaks c(seq(-10, 0,
length.out=n/2)-0.5, 0, seq(0, 30, length.out=n/2)+0.5)
image(x, col=col, breaks=breaks)
See also today's discussion on R-Help about "red-black-green color
palette",
since most of the code above came from Earl Glyn post.
Jarek
====================================================\======
Jarek Tuszynski, PhD. o / \
Science Applications International Corporation <\__,|
(703) 676-4192 "> \
Jaroslaw.W.Tuszynski at saic.com ` \
-----Original Message-----
From: r-help-bounces at stat.math.ethz.ch
[mailto:r-help-bounces at stat.math.ethz.ch] On Behalf Of array chip
Sent: Thursday, August 04, 2005 12:31 PM
To: Thomas Lumley
Cc: r-help at stat.math.ethz.ch
Subject: Re: [R] color palette
Thanks for the suggestion. I still could not figure out how to use the
function to do my job. What's important in my job is that I have to map
white color to value 0, and then form a bue gradient for negative values,
and red gradient for positive values. The data matrix I have is not
symmetric aound 0, say 0 is at the 18 percentile, and values range from -10
to 30.
How exactly I could do my job? Bear with me, I am new to color, I don't
quite understand how the image function map the number is the data matrix to
the colors defined in the col argument.
Thanks
--- Thomas Lumley <tlumley at u.washington.edu> wrote:
> On Wed, 3 Aug 2005, array chip wrote:
>
> > Hi, I have a matrix with both positive and
> negative
> > numbers, I would like to use image() to draw a heatmap. How can I
> > can design a palette (or is
> there a
> > function already available) that treat negative numbers in a blue
> > gradient and positive numbers in
> a
> > red gradient and treat 0 as white?
> >
>
> ?colorRampPalette
>
> -thomas
>
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