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circular filter

4 messages · Christof Bigler, Peter Wolf, Uwe Ligges

Christof Bigler
# 
I try to find a circular filter that I can export to be used in a 
spatial software.
Assuming, we have a matrix, representing 9x9 regularly spaced points  
with the center point 'filter[5, 5]'. In this example, I want to find a 
function that weighs all neighbor points within a distance of d=4 units 
with 1:

 > filter <- matrix(0, 9, 9)
 > filter <- function() ...
 > filter
       [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9]
  [1,]    0    0    0    0    1    0    0    0    0
  [2,]    0    0    1    1    1    1    1    0    0
  [3,]    0    1    1    1    1    1    1    1    0
  [4,]    0    1    1    1    1    1    1    1    0
  [5,]    1    1    1    1    1    1    1    1    1
  [6,]    0    1    1    1    1    1    1    1    0
  [7,]    0    1    1    1    1    1    1    1    0
  [8,]    0    0    1    1    1    1    1    0    0
  [9,]    0    0    0    0    1    0    0    0    0

Finally, I want to use a larger matrix, e.g. with 61x61 points. Is 
there a simple function around that I could use to this end?

Thanks!
Christof
Peter Wolf
# 
try:

filter.matrix.center<-function(n=9,size=5){
  x<-matrix(1,n,n)
  center<-(n+1)/2
  (abs(row(x)-center)+abs(col(x)-center)) < size
}
filter.matrix.center()

some tests:

 > 0+filter.matrix.center(5,2)
     [,1] [,2] [,3] [,4] [,5]
[1,]    0    0    0    0    0
[2,]    0    0    1    0    0
[3,]    0    1    1    1    0
[4,]    0    0    1    0    0
[5,]    0    0    0    0    0

 > 0+filter.matrix.center()
      [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9]
 [1,]    0    0    0    0    1    0    0    0    0
 [2,]    0    0    0    1    1    1    0    0    0
 [3,]    0    0    1    1    1    1    1    0    0
 [4,]    0    1    1    1    1    1    1    1    0
 [5,]    1    1    1    1    1    1    1    1    1
 [6,]    0    1    1    1    1    1    1    1    0
 [7,]    0    0    1    1    1    1    1    0    0
 [8,]    0    0    0    1    1    1    0    0    0
 [9,]    0    0    0    0    1    0    0    0    0

Peter Wolf

--------------------------------------------------------------------------------------------------------
Christof Bigler wrote:
I try to find a circular filter that I can export to be used in a 
spatial software.
Assuming, we have a matrix, representing 9x9 regularly spaced points  
with the center point 'filter[5, 5]'. In this example, I want to find a 
function that weighs all neighbor points within a distance of d=4 units 
with 1:

 > filter <- matrix(0, 9, 9)
 > filter <- function() ...
 > filter
      [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9]
 [1,]    0    0    0    0    1    0    0    0    0
 [2,]    0    0    1    1    1    1    1    0    0
 [3,]    0    1    1    1    1    1    1    1    0
 [4,]    0    1    1    1    1    1    1    1    0
 [5,]    1    1    1    1    1    1    1    1    1
 [6,]    0    1    1    1    1    1    1    1    0
 [7,]    0    1    1    1    1    1    1    1    0
 [8,]    0    0    1    1    1    1    1    0    0
 [9,]    0    0    0    0    1    0    0    0    0

Finally, I want to use a larger matrix, e.g. with 61x61 points. Is there 
a simple function around that I could use to this end?

Thanks!
Christof

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Uwe Ligges
#
Christof Bigler wrote:
I don't know whether there is a function makeYourFilter(), but I'm 
pretty sure it's easy to write it. Unfortunately, I don't know which 
distance you are talking about.

Uwe Ligges
Peter Wolf
# 
filter.matrix.center  implements Manhattan or L 1 distance.
If you want to define neighbor points
by Euklidean distances (L 2) use filter.matrix.center(p=2):

filter.matrix.center.p <- function(n=9,size=5,p=2){
   x<-matrix(1,n,n)
   center<-(n+1)/2
   (abs(row(x)-center)^p+abs(col(x)-center)^p)^(1/p) < size
}

example:

 > 0+filter.matrix.center.p(9,4.1)
    [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9]
[1,]    0    0    0    0    1    0    0    0    0
[2,]    0    0    1    1    1    1    1    0    0
[3,]    0    1    1    1    1    1    1    1    0
[4,]    0    1    1    1    1    1    1    1    0
[5,]    1    1    1    1    1    1    1    1    1
[6,]    0    1    1    1    1    1    1    1    0
[7,]    0    1    1    1    1    1    1    1    0
[8,]    0    0    1    1    1    1    1    0    0
[9,]    0    0    0    0    1    0    0    0    0

Peter Wolf



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Peter Wolf wrote: