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Finding if a vector is a subset

30 ビュー (過去 30 日間)
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 件のコメント
Walter Roberson
Walter Roberson 2018 年 4 月 10 日
The num2str() turns out not to be needed.

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採用された回答

Roger Stafford
Roger Stafford 2018 年 4 月 10 日
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 件のコメント
Roger Stafford
Roger Stafford 2018 年 4 月 10 日
My "faster" claim was in reference to Shattenstein's code.

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その他の回答 (1 件)

Rik
Rik 2018 年 4 月 10 日
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 件のコメント
Walter Roberson
Walter Roberson 2018 年 4 月 10 日
Oh yes: the one restriction here is that strfind() will only work with row vectors, not with column vectors.

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