[R] Re : Half of a heatmap

[Rosenfeld, Jeffrey]

I have correlation values between -1 and 1 and I need to keep the sign.

Jeff


-----Original Message-----
From: Brad McNeney [mailto:mcneney at gmail.com]
Sent: Mon 6/25/2007 1:01 PM
To: Rosenfeld, Jeffrey
Cc: Neil Shephard; r-help at stat.math.ethz.ch
Subject: Re: [R] Re : Half of a heatmap
 
Just for the record, LDheatmap can display any upper-triangular matrix of
measures between 0 and 1, not just LD measures. Users with their own matrix
should pass it as the first argument (gdat) to the function.

-b

--
Brad McNeney
Statistics and Actuarial Science
Simon Fraser University
Burnaby, BC, Canada


On 6/25/07, Rosenfeld, Jeffrey <rosenfel at cshl.edu> wrote:
>
> Thank you for your reply.  I have looked at LDheatmap, but it does not
> seem to do what I want and seems to only work well for LD data.  I was
> looking for something that would produce a figure identical to what
> heatmap.2 gives me, including the proper X and Y-axis labels and a
> dendogram, except that it would only have half of the map.  Preferably, it
> would have the Color Key in the place where the other triangle of the
> heatmap would be, to save space.
>
>
> Jeff
>
>
> -----Original Message-----
> From: Neil Shephard [mailto:nshephard at gmail.com]
> Sent: Mon 6/25/2007 6:14 AM
> To: r-help at stat.math.ethz.ch
> Cc: Rosenfeld, Jeffrey
> Subject: Re : [R] Half of a heatmap
>
> > I am trying to produce a heatmap of pairwise correlations, but since the
> matrix is
> > symmetric, I only need either the upper or the lower triangle.  I have
> scoured the
> > web and R documentation, but I have not been able to find a way to
> produce such a
> > figure.  Is there a simple way to produce a heat map with only the part
> above or
> > below the diagonal?
>
> You might want to check out the LDheatmap() package which can generate
> the plots you describe.  The help indicates that it accepts a matrix
> of pair-wise linkage disequilibrium measures, one of which is R^2 (the
> correlation coefficient between loci), but I suspect you could simply
> pass it a matrix of correlation coefficents.
>
> Hope that helps,
>
> Neil
> --
> "In mathematics you don't understand things. You just get used to
> them."  - Johann von Neumann
>
> Email - nshephard at gmail.com / n.shephard at sheffield.ac.uk
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