On Oct 25, 2009, at 1:51 AM, Kang Min wrote:

Hi Milton,

The matrix can be generated using

p = matrix(1:50, nrow=5)

If I just use levelplot(p), it gives me a graph that is vertical. How
can I rotate it so it becomes horizontal?
I cannot do

q = t(p); levelplot(q)

because this is representing a map from a piece of land, transposing
the data will flip the map upside down laterally.

You can reverse the order of the indices. I cannot quite figure out which reversal will accomplish your goals because your description seems ambiguous but take a look at the output of htese and you should be able to apply hte strategy to your situation:

 library(lattice)
#as a courtesy to the helpeRs it's nice to reference the package in which you function resides

 p = matrix(1:50, nrow=5)
 q2 = p[5:1, ] ; levelplot(q2)

 levelplot(p)

 q = t(p)
 levelplot(q[10:1, ])

-- David


Thanks,
Kang Min

On Oct 25, 12:40 pm, milton ruser <milton.ru...@gmail.com> wrote:
Hi Kang,

Could you send a reproducible sample-code?

Bests

miltinho



On Sat, Oct 24, 2009 at 11:32 PM, Kang Min <ngokang...@gmail.com> wrote:
Hi all,

I have a matrix with 5 rows and 10 columns, which represent the grids
on a rectangular map.
I used the code below to plot, but it gives me the map with the 10
columns as y-axis, and the 5 rows as the x-axis, and the (0,0) point
is at the usual bottom left hand corner. My map starts with the (0,0)
at the top left hand corner.

How can I rotate the map 90 degrees clockwise so the (0,0) starts at
the top left? That means I need the 10 columns to be x-axis. I cannot transpose the matrix because the map would be laterally flipped over.

levelplot(quadsec.mat, colorkey=list(space="bottom"), scales=list
(tick.number=10), aspect=c("iso"))

Thanks, I'm ready to give more information if needed.

Kang Min

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and provide commented, minimal, self-contained, reproducible code.

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David Winsemius, MD
Heritage Laboratories
West Hartford, CT

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