How can I find 4 or more character pattern in number array
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I have an array like this A=[1 2 3 2 5 12 3 9 12 3 5 6 3 2 5 11 10 9] (array size (1,275)
I am trying to find 4 or more character pattern in this array.
how can I do this?
6 件のコメント
回答 (6 件)
Wayne King
2013 年 1 月 23 日
編集済み: Wayne King
2013 年 1 月 23 日
See Loren's blog here:
You can use strfind with an array of numbers
A = [1 2 3 2 5 12 3 9 12 3 5 6 3 2 5 11 10 9];
% find
B = [9 12 3];
K = strfind(A,B);
K is the starting index of the pattern in the array, A.
A(K:K+1+length(K))
7 件のコメント
Laura Proctor
2013 年 1 月 23 日
If you define the pattern that you're looking for as x, for example
x = [3 9 12 3]
then you can find the starting index value for this pattern by using the following code:
n = length(x);
ind = 1:length(A);
for k = 1:n
i1 = find(A==x(k));
ind = intersect(ind,i1-k+1);
end
4 件のコメント
Image Analyst
2013 年 1 月 23 日
That doesn't make sense. You have to be searching A for some pattern. Otherwise, you might as well just pick any 4 adjacent indexes from A at random.
Cedric
2013 年 1 月 23 日
編集済み: Cedric
2013 年 1 月 23 日
And here is a funny solution:
A = [1 2 3 2 5 12 3 9 12 3 5 6 3 2 5 12 3 9] ; % Notes
p = [5 12 3 9] ; % Pattern
nA = numel(A) ; np = numel(p) ;
buffer = ~any(spdiags(repmat(A(:), 1, np), 0:np-1, nA, nA) - ...
spdiags(repmat(p, nA, 1), 0:np-1, nA, nA), 2) ;
loc = find(full(buffer(1:nA-np+1)))
This code gives loc = 5, 15.
Cheers,
Cedric
0 件のコメント
Image Analyst
2013 年 1 月 23 日
If you have the Image Processing Toolbox you can use normxcorr though it looks like Loren's method is simpler:
% Define sample data.
A=[1 2 3 2 5 12 3 9 12 3 5 6 3 2 5 11 10 9]
% Define the sequence of numbers we want to find.
patternToFind = [5 12 3 9]
% Compute the normalized cross correlation.
normCrossCorr = normxcorr2(patternToFind, A)
% Find index where the sequence starts.
% This is where the normalized cross correlation = 1.
startingIndexOfSequence = find(normCrossCorr >= 0.999999) - length(patternToFind) + 1
0 件のコメント
Walter Roberson
2013 年 1 月 23 日
At each step, K,
conv(A(K+4:end), -1./A(K:K+3), 'valid')
should, I think, become within round-off of 0 at each point at which there is a match.
Or,
B = A(K:K+3);
T = A(K+4:end);
find(T(1:end-3) == B(1) & T(2:end-2) == B(2) & T(3:end-1) == B(3) & T(4:end) == B(4), 1, 'first')
There is a vectorized solution for the entire similarity search all at once, that involves constructing a comparison array (it might have to be multidimensional); it might become impractical for larger input vectors.
0 件のコメント
Jan
2013 年 1 月 24 日
According to Wayne King's answer:
data = randi([1,12], 1, 275);
for k = 1:length(data) - 3
search = data(k:k+3);
match = k - 1 + strfind(data(k:end), search);
if length(match) > 1
fprintf('Match: [ ');
fprintf('%d ', search);
fprintf(']: \n ');
fprintf(' %d', match);
fprintf('\n');
end
end
0 件のコメント
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