# How to find an exact sequence of values?

4 vues (au cours des 30 derniers jours)
Pete le 29 Nov 2019
Hi everyone,
as the title says I'm looking to find an exact sequence of values in a matrix. I have a matrix that I'm turning into a logical array as below. Now I want to be able to say if the consecutive ones in the column are less than X (as in e - see below) they should also be turned into zeroes as well.
I tried to work with find(contain()) but that did nothing. Also ismember() did not turn out the results I was looking for.
Turning matrix into binary representation
for k = 1:1:size(p,1)
for l = 1:1:(size(p,2))
if p(k,l) > z
p(k,l) = 1;
else
p(k,l) = 0;
end
end
end
Logical array
0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0
0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 1 1 1 0 0 0 0 0 0 0 0 0
0 0 0 1 1 1 1 0 0 0 0 0 0 0 0 0
0 0 0 1 1 1 1 0 0 0 0 0 0 0 0 0
0 0 1 0 1 1 1 1 0 0 0 0 0 0 0 0
0 0 1 0 1 1 1 0 0 0 0 1 0 0 0 0
0 0 1 0 1 0 0 0 0 0 0 1 0 0 0 0
0 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0
0 0 1 0 1 0 0 0 0 0 0 1 0 0 0 0
0 0 1 0 1 0 0 0 0 0 0 1 1 0 0 0
0 0 1 0 1 0 0 0 0 0 0 1 1 0 0 0
0 0 0 0 1 0 0 0 0 0 0 0 1 1 0 0
0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0
0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 1
Size of ones I would like to be able to filter
e = ones(x,1)
Thank you for your help everyone! It's greatly appreciated.
Cheers,
Peter
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dpb le 2 Déc 2019
Show us what you tried with runlength -- can't imagine it would not work to do the job...and actually, what regexp pattern you used. I'm no whizard on regular expressions but there are those here who are.
Pete le 3 Déc 2019
Luna's approach worked very well. Thank you for your input everyone!

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### Réponse acceptée

Luna le 2 Déc 2019
Modifié(e) : Luna le 2 Déc 2019
Try Loren's findpattern2. Here is link:
create vector x numbers of ones and run findpattern2.
e = ones(1,x);
for i = 1:size(LogicalArray,2)
indices = findpattern2(LogicalArray(:,i),e);
if ~isempty(indices) % if it finds x elements of this pattern make them zero.
for k=1:numel(indices)
LogicalArray(indices(k):indices(k)+x,i) = false; % from indice to pattern length will be zero
end
end
end
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Pete le 3 Déc 2019
That worked perfectly! Thank you
Luna le 3 Déc 2019

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### Plus de réponses (1)

Image Analyst le 3 Déc 2019
If you want to use functions already built in to the toolbox, you can use bwareafilt:
binaryImage = logical([...
0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0
0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 1 1 1 0 0 0 0 0 0 0 0 0
0 0 0 1 1 1 1 0 0 0 0 0 0 0 0 0
0 0 0 1 1 1 1 0 0 0 0 0 0 0 0 0
0 0 1 0 1 1 1 1 0 0 0 0 0 0 0 0
0 0 1 0 1 1 1 0 0 0 0 1 0 0 0 0
0 0 1 0 1 0 0 0 0 0 0 1 0 0 0 0
0 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0
0 0 1 0 1 0 0 0 0 0 0 1 0 0 0 0
0 0 1 0 1 0 0 0 0 0 0 1 1 0 0 0
0 0 1 0 1 0 0 0 0 0 0 1 1 0 0 0
0 0 0 0 1 0 0 0 0 0 0 0 1 1 0 0
0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0
0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 1])
x = 4; % Whatever....
% Get rid of stretches of 1's in each row that are less than x long.
for row = 1 : size(binaryImage, 1)
binaryImage(row, :) = bwareafilt(binaryImage(row, :), [x, inf]);
end
The result is:
binaryImage =
17×16 logical array
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 1 1 1 1 0 0 0 0 0 0 0 0 0
0 0 0 1 1 1 1 0 0 0 0 0 0 0 0 0
0 0 0 0 1 1 1 1 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
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