Indexing a vector within a bigger one in MATLAB - matlab

Question

I'm trying to find the index position of the smaller vector inside a bigger one.
I've already solved this problem using strfind and bind2dec,
but I don't want to use strfind, I don't want to convert to string or to deciamls at all.
Given the longer vector
a=[1,1,1,1,1,1,1,1,0,0,1,1,1,1,1,1,1,0,0,0,0,1,1];
I want to find the index of the smaller vector b inside a
b=[1,1,1,0,0,0];
I would expect to find as result:
result=[15,16,17,18,19,20];
Thank you

Answer 1

Here is as solution using 1D convolution. It may find multiple matches so start holds beginning indices of sub-vectors:
f = flip(b);
idx = conv(a,f,'same')==sum(b) & conv(~a,~f,'same')==sum(~b);
start = find(idx)-ceil(length(b)/2)+1;
result = start(1):start(1)+length(b)-1;

Answer 2

Solution with for loops:
a=[1,1,1,1,1,1,1,1,0,0,1,1,1,1,1,1,1,0,0,0,0,1,1];
b=[1,1,1,0,0,0];
c = [];
b_len = length(b)
maxind0 = length(a) - b_len + 1 %no need to search higher indexes
for i=1:maxind0
found = 0;
for j=1:b_len
if a(i+j-1) == b(j)
found = found + 1;
else
break;
end
end
if found == b_len % if sequence is found fill c with indexes
for j=1:b_len
c(j)= i+j-1;
end
break
end
end
c %display c

Answer 3

Does it need to be computationally efficient?
A not very efficient but short solution would be this:
a=[1,1,1,1,1,1,1,1,0,0,1,1,1,1,1,1,1,0,0,0,0,1,1];;
b=[1,1,1,0,0,0];
where = find(arrayfun(#(n) all(a(n+1:n+length(b))==b),0:length(a)-length(b)));
... gives you 15. Your result would be the vector where:where+length(b)-1.
edit: I tried it and I stand corrected. Here is a version with loops:
function where = find_sequence(a,b)
na = 0;
where = [];
while na < length(a)-length(b)
c = false;
for nb = 1:length(b)
if a(na+nb)~=b(nb)
na = na + 1; % + nb
c = true;
break
end
end
if ~c
where = [where,na+1];
na = na + 1;
end
end
Despite its loops and their bad reputation in Matlab, it's a lot faster:
a = round(rand(1e6,1));
b = round(rand(10,1));
tic;where1 = find(arrayfun(#(n) all(a(n+1:n+length(b))==b),0:length(a)-length(b)));toc;
tic;where2 = find_sequence(a,b);toc;
>> test_find_sequence
Elapsed time is 4.419223 seconds.
Elapsed time is 0.042969 seconds.

Answer 4

A neater method using for would look like this, there is no need for an inner checking loop as we can vectorize that with all...
a = [1,1,1,1,1,1,1,1,0,0,1,1,1,1,1,1,1,0,0,0,0,1,1];
b = [1,1,1,0,0,0];
idx = NaN( size(b) ); % Output NaNs if not found
nb = numel( b ); % Store this for re-use
for ii = 1:numel(a)-nb+1
if all( a(ii:ii+nb-1) == b )
% If matched, update the index and exit the loop
idx = ii:ii+nb-1;
break
end
end
Output:
idx = [15,16,17,18,19,20]
Note, I find this a bit easier to read that some of the nested solutions, but it's not necessarily faster, since the comparison is done on all elements in b each time.