Finding if a vector is a subset

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

[David Fletcher]

There might be some mileage in investigating if existing string comparison functions will do what you need with a bit less effort.

a=[1 2] 
b=[1 3 2 9 5]
c=[1 2 2 9 5]

strfind(num2str(b),num2str(a))
strfind(num2str(c),num2str(a))

[Walter Roberson]

The num2str() turns out not to be needed.

Answer 1

[Roger Stafford]

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.

[Steven Lord]

I haven't tried this to see if it's faster, but find-ing the first element of a in b and iterating over only those starting points using the technique your code uses may help. I suspect adding that initial search would be particularly useful if a(1) is relatively rare in b.

[Roger Stafford]

My "faster" claim was in reference to Shattenstein's code.

Answer 2

[Rik]

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)

[Walter Roberson]

>> a=[1 2]; b=[1 3 2 9 5];
>> strfind(b,a)
ans =
     []
>> a=[1 2], b=[1 2 2 9 5]
a =
     1     2
b =
     1     2     2     9     5
>> strfind(b,a)
ans =
   1

This is not a documented use for strfind() but it has worked for quite some time.

You do not need to convert to char: it is happy to search on char, integer-valued doubles, logical, even floating point numbers -- but do note that it looks for bitwise exact matches, not tolerances at all.

The value returned is the indices of the matches, so you can test isempty() to see if there was a match.

[Walter Roberson]

Oh yes: the one restriction here is that strfind() will only work with row vectors, not with column vectors.