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Finding if a vector is a subset

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Harel Harel Shattenstein
Harel Harel Shattenstein el 10 de Abr. de 2018
Comentada: Roger Stafford el 10 de Abr. de 2018
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 comentarios
David Fletcher
David Fletcher el 10 de Abr. de 2018
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
Walter Roberson el 10 de Abr. de 2018
The num2str() turns out not to be needed.

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Respuesta aceptada

Roger Stafford
Roger Stafford el 10 de Abr. de 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 comentarios
Steven Lord
Steven Lord el 10 de Abr. de 2018
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
Roger Stafford el 10 de Abr. de 2018
My "faster" claim was in reference to Shattenstein's code.

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Más respuestas (1)

Rik
Rik el 10 de Abr. de 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 comentarios
Walter Roberson
Walter Roberson el 10 de Abr. de 2018
>> 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
Walter Roberson el 10 de Abr. de 2018
Oh yes: the one restriction here is that strfind() will only work with row vectors, not with column vectors.

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