Re: ListCorrelate[] ??

*To*: mathgroup at smc.vnet.net*Subject*: [mg30182] Re: ListCorrelate[] ??*From*: "Mariusz Jankowski" <mjkcc at usm.maine.edu>*Date*: Wed, 1 Aug 2001 02:19:49 -0400 (EDT)*Organization*: University of Southern Maine*References*: <9jtkeq$894$1@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

Brent, "Brent Pedersen" <bpederse at nature.berkeley.edu> wrote in message news:9jtkeq$894$1 at smc.vnet.net... > with a real square matrix H, I calculate the sum of each cell and its 8 > neighbors as: > Htot=ListCorrelate[kern, H ,{2,-2}]; (thanks to the help of some of > you) > where: > kern = Table[1, {3}, {3}]; > You probably should use Htot=ListCorrelate[kern, H ,{2, 2}]. For a 3x3 kernel ListCorrelate[kern,H,{2,-2}] == ListCorrelate[kern,H,{2,2}], but this is not true for larger kernels. > now, for each 9 cell neighborhood around (and including cell H[[i,j]], I > wish to calculate: > Sum[(H[[m,n]]/Htot[[i,j]])^c,{m,i-1,i+1},{n,j-1,j+1}]; > (when c=1, this will be 1.) > of course, this code won't work as is but the idea is there.I can write > loops containing a Sum[] similar to that above,but i am sure that > ListCorrelate[] operates much faster (which is important since H is > 256*256).in short, i need to output a 256*256 matrix where each cell > contains the result of the Sum[] function above. > > ListCorrelate[kern, H/Htot,{2,-2}]; does not work because it contains > H[[i,j]]/Htot[[i,j]] not H[[i,j]]/Htot[[m,n]]; > Right, correlation will not return the same result, but here is a reasonable solution: Table[(Tr[Flatten[(#1/Tr[Flatten[#1]] )^c]] & )[ H[[{i - 1, i, i + 1},{j - 1, j, j + 1}]] ], {i, 2, 255}, {j, 2, 255}] or Htot=ListCorrelate[kern,H,{2,2}]; Table[(Tr[Flatten[(#1/Htot[[i,j]])^c]] & )[ H[[{i - 1, i, i + 1},{j - 1, j, j + 1}]] ], {i, 2, 255}, {j, 2, 255}] where Tr[Flatten[...]] is just a fast way to calculate Sum. The second solution may be faster since you are precomputing the averages. If you REALLY need to have a 256x256 result, then you should pad the original data prior to the calculations shown above and change the iterators in Table to 1,256 A question. The calculation you desire is a form of edge enhancement/detection. The calculation using two subsequent ListCorrelates which you rejected also enhances edges. Why is one better than the other? Mariusz -- ====================================== Mariusz Jankowski University of Southern Maine email: mjkcc at usm.maine.edu

**Follow-Ups**:**Re: Re: ListCorrelate[] ??***From:*Brent Pedersen <bpederse@nature.berkeley.edu>