Rapid Vector Matching/Search Problem

2014 8 20
2 个回答
回答已采纳
更新时间：2014 8 23
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1 个投票

I have a matrix (we'll call A) of size m x n, and a vector (we'll call B) that is a 1 x n, and I am interested in finding the one unique index (and I know there is just one) where A(index,:) equals B, is there a way for me to quickly make the determination in MATLAB besides using the following code:
for i = 1 : m
if ((isequal(A(i,:), B)))
indexIntrest = i;
break;
end
end
Thanks

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 采纳的回答

Joseph Cheng
Joseph Cheng 2014-8-20
编辑：Joseph Cheng 2014-8-20

1 个投票

you'll need to use the function ismember()
such as:
found_row = ismember(A,B,'rows')
which will return a logical array where a 1 will represent a match. then by doing something like this
A= randi(2,100,4);
indexIntrest=find(ismember(A,[ 2 1 1 1],'rows'))
will return the indexes that match.

3 个评论
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Kyle Johnston
Kyle Johnston 2014-8-20
I am familiar with "ismember" and I was hoping for something faster then that. I have also tried implementing fitctree, assigning each row a "class" and then using the resulting tree to quickly find the location of the matching row. But that is also too slow.
if it helps, A is resulting from the function perms(1:n) in MATLAB, and B is a member of the set that A represents.
I've settled on the following for right now:
space = factorial(n)/n;
for i = (space*(n- B(1)) + 1) : space*(n- B(1) + 1)
if (isequal(A(i,:),B))
indexIntrest = i
break;
end
end
And by the way, the example with ismember will yield:
[found_row, indexInterest] = ismember(A,B,'rows')
making the find given in the answer above not necessary
Joseph Cheng
Joseph Cheng 2014-8-21
I did it that way just in case you had more than 1. When i do your ismember(A,B,'rows') it does not come back with the index of which row.. that is why i used find to come up with the index.
Joseph Cheng
Joseph Cheng 2014-8-21
Alright finally grasped how the perms() function lays out the permutations. Needed to eat a cookie and it came to me.
clc
clear all
n=10;
A= perms(1:n);
B=A(randi(length(A),1),:);
space = factorial(n-1);
tic
for i = (space*(n- B(1)) + 1) : space*(n- B(1) + 1)
if (isequal(A(i,:),B))
index3 = i
break;
end
end
looptime = toc;
index4 = 1;
tic
seq = [n:-1:1];
for i =1:n-1
space=factorial(n-i);
Subind = find(seq==B(i)); %determine which section
index4= index4+space*(Subind-1);
seq = A(index4,i+1:end); %contains order of next column
end
calctime = toc;
disp([A(index4,:);B])
disp(['loop time: ' num2str(looptime)])
disp(['calc time: ' num2str(calctime)])
disp(['loop - calc: ' num2str(looptime-calctime)])

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更多回答（1 个）

Oleg Komarov
Oleg Komarov 2014-8-21

2 个投票

You can try to use bsxfun():
find(all(bsxfun(@eq, A,B),2))
What happens is every row of A is tested with @eq for element-wise equality against B, producing a m x n logical matrix. Then, all() tests each row if all elements are true, and find() returns the position.

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