<div dir="ltr"><br><div class="gmail_extra"><br><div class="gmail_quote">On Wed, Sep 24, 2014 at 2:43 PM, Moritz Lennert <span dir="ltr"><<a href="mailto:mlennert@club.worldonline.be" target="_blank">mlennert@club.worldonline.be</a>></span> wrote:<br><blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex"><span class="">On 24/09/14 14:08, Paulo van Breugel wrote:<br>
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On Wed, Sep 24, 2014 at 1:12 PM, Moritz Lennert<br></span><span class="">
It thus does not make sense to provide coordinates for count or<br>
area: which coordinates would you use if you have cells with the<br>
same category all over your map ?<br>
<br>
<br>
It makes sense if each raster cell is assigned an unique category (which<br>
I did). What I was trying was get the area per raster cell (of a latlon<br>
raster). But yes, it makes sense that in in general it wouldn't be<br>
logical to have coordinates and area at the same time. I was too much<br>
focussed on my very specific case (trying an alternative way to get<br>
raster cell size of latlon grid).<br>
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Ok, now I understand your problem, but honestly, in such a situation I believe that it is better to chose a relevant projection, reproject the grid and then just use resolution^2 as cell size...<br>
<br>
But since you already have gone down the route of calculating cell size in r.mapcalc, I guess you really wanted to do it in latlon... ;-)<span class="HOEnZb"><font color="#888888"><br>
<br></font></span></blockquote><div><br></div><div>Yes, more convenient in this case. I can compute the area directly using r.mapcalc, so no problem. I was just trying if this would be an easier / faster method. Thanks for your reply!<br></div><div> </div><blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex"><span class="HOEnZb"><font color="#888888">
Moritz<br>
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