Hi

I'm afraid not!  

 
heatmap.2(as.matrix(d),Rowv=as.dendrogram(hc.gene),Colv=1:4,scale="row",
trace="none",col=greenred.colors(79))

Produces exactly the same results as if "Colv=FALSE" - my columns get
re-ordered (by a dendrogram no less).  Clearly this isn't meant to be
happening.... (by the way, d is NOT a square matrix either)

Cheers

Mick 

-----Original Message-----
From: Sean Davis [mailto:[EMAIL PROTECTED] 
Sent: 02 September 2004 14:39
To: michael watson (IAH-C)
Cc: [EMAIL PROTECTED]
Subject: Re: [R] Problems with heatmap.2


Didn't test this, but I think you probably want to do  
dendrogram="row",Colv=1:n
where n is the number of samples.

Sean

On Sep 2, 2004, at 9:21 AM, michael watson (IAH-C) wrote:

> Hi
>
> When I give the command:
>
>>
> heatmap.2(as.matrix(d),Rowv=as.dendrogram(hc.gene),Colv=FALSE,scale="r
> o
> w
> ",trace="none",col=greenred.colors(79))
>
> The resulting heatmap has re-ordered my columns!  This is time-course 
> data, and I don't want my columns re-ordered!  Note from the help:
>
>     Rowv: determines if and how the _row_ dendrogram should be
>           reordered.  Either a 'dendrogram' or a vector of values used
>           to reorder the row dendrogram or 'FALSE' to suppress
>           reordering or by default, 'NULL', see _Details_ below.
>
>     Colv: determines if and how the _column_ dendrogram should be
>           reordered.  Has the options as the 'Rowv' argument above and
>           _additionally_ when 'x' is a square matrix, 'Colv = "Rowv"'
>           means that columns should be treated identically to the 
> rows.
>
> I have specifically set "Colv=FALSE" in my command.
>
> Help?  What am I doing wrong?
>
> Cheers
> Mick
>
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