Re: Problem: a matrix and a box...
- To: mathgroup at smc.vnet.net
- Subject: [mg27271] Re: [mg27268] Problem: a matrix and a box...
- From: Ken Levasseur <Kenneth_Levasseur at uml.edu>
- Date: Sun, 18 Feb 2001 02:52:08 -0500 (EST)
- References: <200102170830.DAA17186@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
Bedrosian: You can use the same trick as is used in cellular automata to get the "neighbors" of entries in an array: Here's a "big" array to test thing with: In[79]:= big = Array[a, {3, 3}] Out[79]= {{a[1, 1], a[1, 2], a[1, 3]}, {a[2, 1], a[2, 2], a[2, 3]}, {a[3, 1], a[3, 2], a[3, 3]}} Here is how the list of neighbors can be generated as a matrix of the same dimensions as the original with the list of neighbors that you want as the entries. In[80]:= nbs = {#, RotateLeft[#], Transpose[RotateLeft[Transpose[#]]], Transpose[RotateLeft[Transpose[RotateLeft[#]]]]} &[big] // MapThread[List, #, 2] & Out[80]= {{{a[1, 1], a[2, 1], a[1, 2], a[2, 2]}, {a[1, 2], a[2, 2], a[1, 3], a[2, 3]}, {a[1, 3], a[2, 3], a[1, 1], a[2, 1]}}, {{a[2, 1], a[3, 1], a[2, 2], a[3, 2]}, {a[2, 2], a[3, 2], a[2, 3], a[3, 3]}, {a[2, 3], a[3, 3], a[2, 1], a[3, 1]}}, {{a[3, 1], a[1, 1], a[3, 2], a[1, 2]}, {a[3, 2], a[1, 2], a[3, 3], a[1, 3]}, {a[3, 3], a[1, 3], a[3, 1], a[1, 1]}}} Sums can be computed by replacing List with Plus in MapThread. In[82]:= NeighborSums[ big_] := {#, RotateLeft[#], Transpose[RotateLeft[Transpose[#]]], Transpose[RotateLeft[Transpose[RotateLeft[#]]]]} &[big] // MapThread[Plus, #, 2] & Here's a numeric example to use as a test. In[83]:= test = Partition[Table[Random[Integer, {0, 9}], {16}], 4] Out[83]= {{2, 2, 2, 4}, {9, 8, 6, 2}, {4, 2, 6, 3}, {6, 4, 5, 0}} In[84]:= NeighborSums[test] Out[84]= {{21, 18, 14, 17}, {23, 22, 17, 18}, {16, 17, 14, 13}, {14, 13, 11, 12}} I'm not sure if this is exactly what you want, but it should be easy enough to adjust it. Ken Levasseur UMass Lowell Bedrosian Baol wrote: > Hi > this is the problem: I have a n x m matrix (with n and m ~3000 elements or > more) > > a1 a2 a3.. > b1 b2 b3.. > c1 c2 c3.. > . . . > . . . > > the algorythm is: compute the summ of the the a1, a2, b1 and b2 elements (a > ''gliding box'' 2x2), shifing the ''box'' for 1 column and repeat the sum of > the elements (now a2, a3, b2, b3), and so on... > at the and of the first line, the box return on the second line and > repeat... > > the final resuls would be another matrix like this: > > [a1+a2+b1+b2] [a2+a3+b2+b3]... > [b1+b2+c1+c2] [b2+b3+c2+c3]... > ... > > I'm not very experienced with Math4: there is a simple and easy way for > reach > the result? > > thanks > Bed
- References:
- Problem: a matrix and a box...
- From: Bedrosian Baol <Bedrosian@MailAndNews.com>
- Problem: a matrix and a box...