# Find pattern in vector while ignoring/skipping certain indices

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Hello,

Is there an efficient way to search for a specific pattern in a mat vector while ignoring some indices in the pattern?

For example, I need to search for a 9-element pattern [0 4 X 0 6 Y 0 8 Z] in a mat vector, where X, Y, Z can be any values.

I currently have a loop based approach but is there a faster vectorized approach?

Thank you.

##### 3 Comments

### Answers (4)

Image Analyst
on 11 Jun 2022

I think this should work but for your given pattern, and a vector of 100 million elements of random values, I never did see a match. And I ran it several times. Never found a match so hopefully you believe there should be a match somehow and you're not just using random integers like I did.

% Create sample data.

vec = randi(8, 1, 100000000);

% Define the pattern. Nan = "don't care".

pattern = [0 4 nan 0 6 nan 0 8 nan]

% Define a mask for what values we want to check.

mask = ~isnan(pattern)

lastIndex = length(vec) - length(pattern);

% Scan along the vector looking for matches.

for k = 1 : lastIndex

% Print out progress every 100 thousand window locations.

if mod(k, 100000) == 0

fprintf('k = %d of %d (%.1f%%)\n', k, lastIndex, 100*k/lastIndex);

end

% Extract the window.

thisWindow = vec(k : k+length(pattern)-1);

% Compare this window to our pattern but only at the mask = true locations.

if isequal(pattern(mask), thisWindow(mask))

% Found a match. Report where it was.

fprintf('Match at k = %d where vec = [%d, %d, %d, %d, %d, %d, %d, %d, %d]\n', k, thisWindow)

end

end

fprintf('Done!\n');

##### 1 Comment

Image Analyst
on 11 Jun 2022

Matt J
on 11 Jun 2022

Edited: Matt J
on 11 Jun 2022

vec=[0 4 1 0 6 5 0 8 7, 3 3 3 , 0 4 2 0 6 4 0 8 6]; %patterns start at i=1 and i=13

pat = [0 4 nan 0 6 nan 0 8 nan];

pat=pat(:); vec=vec(:)';

m=numel(vec); n=numel(pat);

include=find(~isnan(pat));

idx=0:m-n;

sequences = cell2mat(arrayfun(@(i)vec(i+idx),include,'uni',0));

matchlocations=find(all(sequences==pat(include),1) )

##### 0 Comments

per isakson
on 11 Jun 2022

I assume it's a vector of integers.

Steve Amphlett showed this trick at comp.soft-sys.matlab twenty years ago.

%% Create sample data

pat = [0,4,nan,0,6,nan,0,8,nan];

msk = true(1,numel(pat));

msk(isnan(pat)) = false;

pat(not(msk)) = 0;

vec = randi([-8,8],1,1e6);

vec(101:109) = [0,4,11,0,6,12,0,8,13];

vec(701:709) = [0,4,14,0,6,15,0,8,16];

%

%% Search matches

tic

z = conv(vec,pat(end:-1:1));

hit = find(abs(z==sum(pat.^2)))-numel(pat)+1;

%%

% hit may contain false hits.

for ix = hit

v9 = vec(ix:ix+8);

if all( v9(msk) == pat(msk) )

disp(ix)

end

end

toc

##### 0 Comments

Voss
on 11 Jun 2022

Edited: Voss
on 11 Jun 2022

% the pattern:

pat = [0 4 NaN 0 6 NaN 0 8 NaN];

% create some data containing the pattern:

data = randn(1,10000);

idx = find(~isnan(pat));

for ii = 100:100:9900

data(ii+idx-1) = pat(idx);

end

% find the pattern in the data:

idx = find(~isnan(pat));

result = find(all(data((0:numel(data)-numel(pat)).'+idx) == pat(idx),2));

% display the result:

disp(result);

##### 0 Comments

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