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Finding if a vector is a subset

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I am trying to build a function that for
a=[1 2] b=[1 3 2 9 5]
will return false
and for
a=[1 2] b=[1 2 2 9 5]
return true
What I manage to do is
function[yn] = subset1(v1,v2)
yn=0;
n=length(v1);
m=length(v2);
v=[];
if n<=m
for i=1:n
for j=1:(m-n+1)
while (v1(i)==v2(j))
v(end+1)=v1(i);
i=i+1;
j=j+1;
end
end
end
end
if length(find(v))==length(find(v1)) && find(v)==find(v1)
yn=1;
end
if n>m
for i=1:m
for j=1:(n-m+1)
while ([v2(i)]==v1(j))
v(end+1)=v2(i);
i=i+1;
j=j+1;
end
end
end
end
if length(find(v))==length(find(v2)) && find(v)==find(v2)
yn=1;
end
but it does not work in the first case
  2 Comments
Walter Roberson
Walter Roberson on 10 Apr 2018
The num2str() turns out not to be needed.

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Accepted Answer

Roger Stafford
Roger Stafford on 10 Apr 2018
The following should be faster:
m = size(a,2);
n = size(b,2);
for k = 1:n-m+1
s = all(a==b(k:k+m-1));
if s, break, end
end
Logical s will be true if any m-length section of b is equal to the a vector.
  2 Comments
Roger Stafford
Roger Stafford on 10 Apr 2018
My "faster" claim was in reference to Shattenstein's code.

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More Answers (1)

Rik
Rik on 10 Apr 2018
strfind should be an option, especially if you only have positive integer scalars, which you can just cast to char. Otherwise, the solution below might also be an option. It might not scale really well to huge vectors due to that convolution, but that is done on a binary matrix, so that should be as fast as it can be.
Another note: this uses implicit expansion, so if you don't have R2016b or newer, you'll have to use bsxfun.
a=[1 2];b1=[1 3 2 9 5];b2=[1 2 2 9 5];
%requires implicit expansion (use bsxfun on R2016a and earlier)
HasMatch=@(a,b) any(any(conv2(b'==a,logical(eye(length(a))),'same')==length(a)));
HasMatch(a,b1)
HasMatch(a,b2)
  2 Comments
Walter Roberson
Walter Roberson on 10 Apr 2018
Oh yes: the one restriction here is that strfind() will only work with row vectors, not with column vectors.

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