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Re: Add terms surrounded by zero together in matrix

  • To: mathgroup at smc.vnet.net
  • Subject: [mg59174] Re: Add terms surrounded by zero together in matrix
  • From: "Carl K. Woll" <carlw at u.washington.edu>
  • Date: Sun, 31 Jul 2005 01:30:45 -0400 (EDT)
  • Organization: University of Washington
  • References: <dcccur$3j7$1@smc.vnet.net> <dcf34j$lf9$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

"Paul Abbott" <paul at physics.uwa.edu.au> wrote:
> In article <dcccur$3j7$1 at smc.vnet.net>,
> "mchangun at gmail.com" <mchangun at gmail.com> wrote:
>
>> I think this is a rather tough problem to solve.  I'm stumped and would
>> really appreciated it if someone can come up with a solution.
>>
>> What i want to do is this.  Suppose i have the following matrix:
>>
>> 0       0       0       1       0
>> 0       0       1       2       0
>> 0       0       0       2       1
>> 1       3       0       0       0
>> 0       0       0       0       0
>> 0       0       0       0       0
>> 0       0       1       1       0
>> 5       0       3       0       0
>> 0       0       0       0       0
>> 0       0       0       3       1
>>
>> I'd like to go through it and sum the elements which are surrounded by
>> zeros.  So for the above case, an output:
>>
>> [7 4 5 5 4]
>>
>> is required.  The order in which the groups surrounded by zero is
>> summed does not matter.
>>
>> The elements are always integers greater than 0.
>
> Have a look at the literature on percolation clusters and in the
> MathGroup archive.
>
> I would have thought that you could use
>
>  << Statistics`ClusterAnalysis`
>
> to solve this, but I could not get it to work right. The basic idea I
> had was to locate the non-zero entries in
>
>  dat = {{0, 0, 0, 1, 0}, {0, 0, 1, 2, 0}, {0, 0, 0, 2, 1},
>         {1, 3, 0, 0, 0}, {0, 0, 0, 0, 0}, {0, 0, 0, 0, 0},
>         {0, 0, 1, 1, 0}, {5, 0, 3, 0, 0}, {0, 0, 0, 0, 0},
>         {0, 0, 0, 3, 1}};
>
> using Position:
>
>  pos = Position[dat, n_Integer /; n > 0]
>
> and then apply FindClusters to these positions. For example,
>
>  cl = FindClusters[pos, DistanceFunction -> BrayCurtisDistance]
>

Paul,

Using Statistics`ClusterAnalysis` is a great idea. Here are some 
modifications to your approach to get it to work. First, use SparseArray to 
get the positions:

In[13]:=
datarules=Most[ArrayRules[SparseArray[dat]]]

Out[13]=
{{1, 4} -> 1, {2, 3} -> 1, {2, 4} -> 2, {3, 4} -> 2,

{3, 5} -> 1, {4, 1} -> 1, {4, 2} -> 3, {7, 3} -> 1,

{7, 4} -> 1, {8, 1} -> 5, {8, 3} -> 3, {10, 4} -> 3,

{10, 5} -> 1}

Next, define a distance function which yields 0 for identical elements, 1 
for adjacent elements, and a big number (I used 10) for nonadjacent 
elements.

myDistance[d1_, d2_] := Clip[Total[Abs[d1 - d2]], {0, 1}, {10, 10}]

Then, use FindClusters with method Agglomerate:

In[17]:=
FindClusters[datarules,Method->Agglomerate,DistanceFunction->myDistance]

Out[17]=
{{1, 1, 2, 2, 1}, {1, 3}, {1, 1, 3}, {5}, {3, 1}}

Finally, total the cluster values:

In[18]:=
Total/@%

Out[18]=
{7, 4, 5, 5, 4}

Here is a function to do the above:

In[25]:=
ClusterSums[array_]:=Module[{datarules,clusteredvalues},
datarules=Most[ArrayRules[SparseArray[array]]];
clusteredvalues=
FindClusters[datarules,Method->Agglomerate,
DistanceFunction->(Clip[Total[Abs[Subtract[##]]],{0,1},{10,10}]&)];
Total/@clusteredvalues]

And a check on our data:

In[26]:=
ClusterSums[dat]

Out[26]=
{7, 4, 5, 5, 4}

As a bonus, the approach used by ClusterSums will be able to handle arrays 
with any dimension.

Now the question is how does this perform compared to the other approaches 
offered on mathgroup?

[snip]

 -- 
> Paul Abbott                                      Phone: +61 8 6488 2734
> School of Physics, M013                            Fax: +61 8 6488 1014
> The University of Western Australia         (CRICOS Provider No 00126G)
> AUSTRALIA                               http://physics.uwa.edu.au/~paul
>        http://InternationalMathematicaSymposium.org/IMS2005/
>

Carl Woll
Wolfram Research 



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