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From: Andreea on 14 Jul 2010 06:02
Hello

I have an incredibly large binary matrix.In this matrix let's call it A, I need to find the sequences of 1.But I need specific sequences.(like 3 ones one after another for example)
For this I created a matrix of ones with one column.---->this matrix must have a specific nr of rows that I introduce from the kb.

example
A=[1 0 1 1 1 1 1 0 0 0 0........,1] infinite
pattern=[1 1 1,1]
matformed=[ _ _ _,1] of A

How I did it:

I break matrix A using a for loop in a smaller matrixes, based on the no of rows I need.Than I compare using isequal(mat_formed,mat ones).

The problem:
It takes ages. Is there a faster way to compare matrixes?

The purpose:
Find the no. of intervals that fit.

Thank you
From: Oleg Komarov on 14 Jul 2010 06:33
"Andreea " <andreea.costache(a)ymail.com> wrote in message <i1k1uu$64f$1(a)fred.mathworks.com>...
> Hello
>
> I have an incredibly large binary matrix.In this matrix let's call it A, I need to find the sequences of 1.But I need specific sequences.(like 3 ones one after another for example)
> For this I created a matrix of ones with one column.---->this matrix must have a specific nr of rows that I introduce from the kb.
>
> example
> A=[1 0 1 1 1 1 1 0 0 0 0........,1] infinite
> pattern=[1 1 1,1]
> matformed=[ _ _ _,1] of A
>
> How I did it:
>
> I break matrix A using a for loop in a smaller matrixes, based on the no of rows I need.Than I compare using isequal(mat_formed,mat ones).
>
> The problem:
> It takes ages. Is there a faster way to compare matrixes?
>
> The purpose:
> Find the no. of intervals that fit.
>
> Thank you

You can give a look at findseq: http://www.mathworks.com/matlabcentral/fileexchange/28113-findseq

This function retrieves the repeated value, the star index and the end index of the sequence and the length of the sequence along the specified dimension.

Example:

A = [1 0 0 0 1 0 1 0 1 1 0 0; 1 1 1 1 0 0 0 0 1 0 1 0];

% Find sequences along the second dimension (horizontally)
Out = findseq(A,2);

% Select only sequences of ones longer than 2
IDX = Out(:,1) == 1 & Out(:,4) > 2;
S = Out(IDX,:)
S =
1 2 8 4

So, you have one sequence of ones which starts at position 2 and ends at position 8
and is long 4 numbers (horizontally).

Oleg
From: us on 14 Jul 2010 06:39
"Andreea " <andreea.costache(a)ymail.com> wrote in message <i1k1uu$64f$1(a)fred.mathworks.com>...
> Hello
>
> I have an incredibly large binary matrix.In this matrix let's call it A, I need to find the sequences of 1.But I need specific sequences.(like 3 ones one after another for example)
> For this I created a matrix of ones with one column.---->this matrix must have a specific nr of rows that I introduce from the kb.
>
> example
> A=[1 0 1 1 1 1 1 0 0 0 0........,1] infinite
> pattern=[1 1 1,1]
> matformed=[ _ _ _,1] of A
>
> How I did it:
>
> I break matrix A using a for loop in a smaller matrixes, based on the no of rows I need.Than I compare using isequal(mat_formed,mat ones).
>
> The problem:
> It takes ages. Is there a faster way to compare matrixes?
>
> The purpose:
> Find the no. of intervals that fit.
>
> Thank you

one of the many solutions

m=[1,1,0,1,1,1,0,1,1,0,1,0];
p=[1,1]; % <- pattern...
ix=strfind(m,p)
% ix = 1 4 5 8 % <- note: overlap(!)...
% -or-
s=char(m+'0');
ix=regexp(s,char(p+'0'))
% ix = 1 4 8 % <- note: NO overlap(!)...
% -or- donwload this FEX submission

http://www.mathworks.com/matlabcentral/fileexchange/1518

us
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