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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

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.

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)

Walter Roberson
on 10 Apr 2018

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