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From: alfann on 4 Jun 2010 02:43
If I have this matrix:
______________________
U1=[20 33 50 20 10 80 10; 10 1 1 1 90 10 90; 60 1 20 33 21 21 30; 50 05 50 44 50 50 10; 80 50 50 10 50 90 10; 20 0 50 60 30 10 50; 10 50 30 50 50 10 90; 80 20 10 70 80 50 30; 40 50 30 80 20 10 20]
and Now:
I want to calculate the average of the summation of the neighbors for ones values( when the value=1).

example1 (one chain):
In the above matrix, we see the ones are in :
U1(3,2)=1
and
U1(2,2)=1
and
U1(2,3)=1
and
U1(2,4)=1

now, I want calculate the summation of the neighbors of ones, and it will be:
average=(5+60+10+33+50+20+90+33+20+5)/10;
Then
U1 becomes:
U1=[20 33 50 20 10 80 10; 10 32.6 32.6 32.6 90 10 90; 60 32.6 20 33 21 21 30; 50 05 50 44 50 50 10; 80 50 50 10 50 90 10; 20 0 50 60 30 10 50; 10 50 30 50 50 10 90; 80 20 10 70 80 50 30; 40 50 30 80 20 10 20]

______________________
______________________

Example 2 (one chain):
U1=[20 33 50 20 10 80 10; 10 1 85 88 90 10 90; 60 1 20 33 21 21 30; 50 05 50 44 50 50 10; 80 50 50 10 50 90 10; 20 0 50 60 30 10 50; 10 50 30 50 50 10 90; 80 20 10 70 80 50 30; 40 50 30 80 20 10 20]

In this matrix, we see the ones are in :
U1(3,2)=1
and
U1(2,2)=1
only

now, I want calculate the summation of the neighbors of ones, and it will be:
average=(5+60+10+33+85+20)/6;
Then change the one values to the (average), which means:
U1 becomes:
U1=[20 33 50 20 10 80 10; 10 35.5 85 88 90 10 90; 60 35.5 20 33 21 21 30; 50 05 50 44 50 50 10; 80 50 50 10 50 90 10; 20 0 50 60 30 10 50; 10 50 30 50 50 10 90; 80 20 10 70 80 50 30; 40 50 30 80 20 10 20]


______________________
______________________

Example 3 (two chains) more clear:
U1=[20 33 50 20 10 80 10; 10 1 85 88 90 10 90; 60 1 20 33 21 21 30; 50 05 50 44 50 50 10; 80 50 50 10 1 1 1; 20 0 50 60 1 10 50; 10 50 30 50 50 10 90; 80 20 10 70 80 50 30; 40 50 30 80 20 10 20]

In this matrix, we see the ones are in :
First chain:
U1(3,2)=1
and
U1(2,2)=1

second chain:
U1(6,5)=1
and
U1(5,5)=1
and
U1(5,6)=1
and
U1(5,7)=1

now, I want calculate the summation of the neighbors of ones, and it will be:
For first chain:
average=(5+60+10+33+85+20)/6;
and
For second chain:
average=(50+60+10+50+50+10+50+10)/8

Then the U1 matrix becomes:
U1=[20 33 50 20 10 80 10; 10 35.5 85 88 90 10 90; 60 35.5 20 33 21 21 30; 50 05 50 44 50 50 10; 80 50 50 10 36.25 36.25 36.25; 20 0 50 60 36.25 10 50; 10 50 30 50 50 10 90; 80 20 10 70 80 50 30; 40 50 30 80 20 10 20]



How can I do that?
From: Oleg Komarov on 4 Jun 2010 07:37
alfann <alfann.net(a)hotmail.com> wrote in message <1373922205.277491.1275648270975.JavaMail.root(a)gallium.mathforum.org>...
> If I have this matrix:
> ______________________
> U1=[20 33 50 20 10 80 10; 10 1 1 1 90 10 90; 60 1 20 33 21 21 30; 50 05 50 44 50 50 10; 80 50 50 10 50 90 10; 20 0 50 60 30 10 50; 10 50 30 50 50 10 90; 80 20 10 70 80 50 30; 40 50 30 80 20 10 20]
> and Now:
> I want to calculate the average of the summation of the neighbors for ones values( when the value=1).
>
> example1 (one chain):
> In the above matrix, we see the ones are in :
> U1(3,2)=1
> and
> U1(2,2)=1
> and
> U1(2,3)=1
> and
> U1(2,4)=1
>
> now, I want calculate the summation of the neighbors of ones, and it will be:
> average=(5+60+10+33+50+20+90+33+20+5)/10;
> Then
> U1 becomes:
> U1=[20 33 50 20 10 80 10; 10 32.6 32.6 32.6 90 10 90; 60 32.6 20 33 21 21 30; 50 05 50 44 50 50 10; 80 50 50 10 50 90 10; 20 0 50 60 30 10 50; 10 50 30 50 50 10 90; 80 20 10 70 80 50 30; 40 50 30 80 20 10 20]

Why do you count the 5 twice? BTW here's a solution (if I'm correct), up to you to shorten the code:

% rows and cols for each 1
[r, c] = find(U1 == 1);
% four connected neighbours
neighB = [r-1 c; r+1 c; r c-1; r c+1];
% Exclude those out of matrix boundaries
IDX = any(neighB < 1,2) | neighB(:,1) > size(U1,1) | neighB(:,2) > size(U1,2);
% Take all the neigbours
neighB = U1(sub2ind(size(U1),neighB(~IDX,1),neighB(~IDX,2)));
% Average excluding the ones
IDXones = neighB == 1;
average = mean(neighB(~IDXones));
% Substitute into U1
U1(IDXones) = average;

Oleg
From: alfann on 4 Jun 2010 04:10
Thanks for that,
but the result is not right.
apply it for this example:
U1=[20 33 50 20 10 80 10; 10 1 1 1 90 10 90; 60 1 20 33 21 21 30; 50 05 50 44 50 50 10; 80 50 50 10 50 90 10; 20 0 50 60 30 10 50; 10 50 30 50 50 10 90; 80 20 10 70 80 50 30; 40 50 30 80 20 10 20]

then apply the code which you wrote it:

% rows and cols for each 1
[r, c] = find(U1 == 1);
% four connected neighbours
neighB = [r-1 c; r+1 c; r c-1; r c+1];
% Exclude those out of matrix boundaries
IDX = any(neighB < 1,2) | neighB(:,1) > size(U1,1) | neighB(:,2) > size(U1,2);
% Take all the neigbours
neighB = U1(sub2ind(size(U1),neighB(~IDX,1),neighB(~IDX,2)));
% Average excluding the ones
IDXones = neighB == 1;
average = mean(neighB(~IDXones));
% Substitute into U1
U1(IDXones) = average;


then
the result is not correct.


Note:
If the neigbour is equal to 1, we must neglect it.
From: Oleg Komarov on 4 Jun 2010 09:38
> % Take all the neigbours
> neighB = U1(sub2ind(size(U1),neighB(~IDX,1),neighB(~IDX,2)));

> then
> the result is not correct.

Substitute the row with:
neighB = U1(unique(sub2ind(size(U1),neighB(~IDX,1),neighB(~IDX,2))));

Oleg
From: Oleg Komarov on 4 Jun 2010 09:40
"Oleg Komarov" <oleg.komarovRemove.this(a)hotmail.it> wrote in message <huavjt$n6i$1(a)fred.mathworks.com>...
> > % Take all the neigbours
> > neighB = U1(sub2ind(size(U1),neighB(~IDX,1),neighB(~IDX,2)));
>
> > then
> > the result is not correct.
>
> Substitute the row with:
> neighB = U1(unique(sub2ind(size(U1),neighB(~IDX,1),neighB(~IDX,2))));
>
> Oleg

There's another error, I'll repost the whole code:
U1=[20 33 50 20 10 80 10; 10 1 1 1 90 10 90; 60 1 20 33 21 21 30; 50 05 50 44 50 50 10; 80 50 50 10 50 90 10; 20 0 50 60 30 10 50; 10 50 30 50 50 10 90; 80 20 10 70 80 50 30; 40 50 30 80 20 10 20]
% rows and cols for each 1
[r, c] = find(U1 == 1);
% four connected neighbours
neighB = [r-1 c; r+1 c; r c-1; r c+1];
% Exclude those out of matrix boundaries
IDX = any(neighB < 1,2) | neighB(:,1) > size(U1,1) | neighB(:,2) > size(U1,2);
% Take all the neigbours
neighB = U1(unique(sub2ind(size(U1),neighB(~IDX,1),neighB(~IDX,2))));
% Average excluding the ones
IDXones = neighB == 1;
average = mean(neighB(~IDXones));
% Substitute into U1
U1(U1 == 1) = average;

Oleg
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