Hi David et al,
Thanks for your help. I spent the afternoon and thought that would
work but then I realized it was giving a different answer.
I have counts ('hits') in each grid cell, and have then calculated the
proportion of the total hits represented in each cell (such that the
sum of all cells = 1). I want to take the cell with largest
proportion, add the next largest proportion to it, etc until I reach
10% of the TOTAL number of 'hits'. Quantiles unfortunately are
created using the total number of CELLS and not the total number of
HITS, if that makes sense.
Any thoughts??
Many many thanks,
Sara
_________________________________
Sara M. Maxwell, Ph.D.
On Apr 21, 2011, at 1:26 PM, David Winsemius wrote:
On Apr 21, 2011, at 3:23 PM, Sara Maxwell wrote:
I am working with a raster and want to take values assigned to each
cell and sort them from largest to smallest, then cummulatively sum
them together (in order from largest to smallest). I'll then be
coding the individual cells such that the top 10% of the largest
cell values can be visualize with one color, the next 10% with
another and so on.
I have tried a number of schemes but am having trouble figuring out
how to chose the maximum value, code it and re-search the raster
for the next highest value without replacement. I am assuming this
requires a loop, unless there is a function that will do this
automatically.
?quantile
Here is a sample dataset:
library(raster)
r <- raster(ncol=10, nrow=10)
values(r) <- runif(ncell(r))
> quantile(values(r), prob=seq(0,1,by=0.1))
0% 10% 20% 30% 40%
0.004888148 0.106378528 0.217009097 0.307201289 0.364990984
50% 60% 70% 80% 90%
0.512523817 0.593382875 0.667916094 0.722919876 0.835839832
100%
0.996683575
You will also need findInterval()
If you want to create a factor that will assign your colors. perhaps
this could be used to index a suitable color vector:
fac <- findInterval(values(r), quantile(values(r),
prob=seq(0,1,by=0.1)) )
> fac
[1] 6 10 1 10 7 7 8 10 2 9 9 1 6 2 9 1 9 4 2 2
[21] 3 4 8 9 7 1 9 2 10 5 4 9 8 1 8 10 1 11 3 5
[41] 5 6 6 5 6 7 4 7 5 3 8 6 3 4 10 4 7 7 8 9
[61] 10 4 1 8 8 8 3 7 5 1 9 5 2 7 2 10 3 8 4 9
[81] 6 6 2 6 10 5 5 4 3 6 2 2 1 3 3 3 4 7 1 5
David Winsemius, MD
West Hartford, CT
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