Best way to look for vector within a matrix

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Lorenzo
Lorenzo 2013-4-30
Dear all, I have a matrix M (lot of lines, 7 columns) and a vector v (again lot of lines but not the same ad M and 3 columns) and I need to find the lines of M for which the first 3 columns are equal to the column of a certain line of v.
What I'm currently doing is:
for m=1:length(v)
n=find(M(:,1) == v(m,1) & M(:,2) == v(m,2) & M(:,3) == v(m,3));
if isempty(n)
results(m,1:4)=nan;
else
results(m,1:4)=M(n,4:7);
end
end
Now, since M is quite big and so is v (in terms of line numbers), is there any way to achieve this in a quicker way than what I'm doing?
  1 个评论
Jan
Jan 2013-4-30
编辑：Jan 2013-4-30
Please specify "a lot": It could be 100 or 10 billion, but this obviously matters for a solution.
The values of M and v matter also: If they are integers in the range of [0, 255], combining them to a UNIT32 would allow a simple ISMEMBER call.

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回答（3 个）

Jan
Jan 2013-4-30
[Lia, Locb] = ismember(M(:, 3), v, 'rows');
fullb = locb(Locb ~= 0);
result(fullb) = M(Lia, 4:7);
I cannot test this currently. Perhaps the indexing must be changed. But the general method should be clear.
Lorenzo
Lorenzo 2013-4-30
Thank you Jan; sorry for the information I missed. My matrices/vectors have around 550k rows and are made of rationals.
In your answer, what is locb (without capital L)?
Thanks again!
  1 个评论
Cedric
Cedric 2013-4-30
编辑：Cedric 2013-4-30
There are actually two small typos in the first two lines:
[Lia, Locb] = ismember(M(:,1:3), v, 'rows');
fullb = Locb(Locb ~= 0);
If you want to understand, build a simpler test case:
>> M = randi(4, 20, 7)
M =
1 2 4 1 3 3 1
2 4 2 4 1 1 1
2 4 2 3 3 3 1
2 3 1 4 4 4 4
1 3 2 4 1 1 4
2 1 2 2 1 3 2
1 3 1 4 2 4 1
1 3 4 4 4 2 3
1 4 2 2 4 3 4
4 2 4 4 1 2 4
1 4 4 1 1 4 4
4 1 1 1 1 1 4
1 1 3 1 3 3 1
3 1 2 2 2 3 1
3 4 1 2 2 2 2
3 4 2 1 1 4 1
2 1 1 3 1 2 4
4 1 1 4 4 2 2
3 3 3 2 3 3 2
4 2 1 4 3 2 1
>> v = randi(4, 10, 3)
v =
3 2 2
1 2 4
4 1 4
1 3 4
3 4 3
2 2 3
3 4 2
1 4 2
1 4 3
1 1 3
>> [Lia, Locb] = ismember(M(:, 1:3), v, 'rows');
>> [Lia, Locb]
ans =
1 2
0 0
0 0
0 0
0 0
0 0
0 0
1 4
1 8
0 0
0 0
0 0
1 10
0 0
0 0
1 7
0 0
0 0
0 0
0 0
>> class(Lia)
ans =
logical
>> class(Locb)
ans =
double
As you can see, Lia is a vector of logical indices which indicate the location of matches in M (at places where its elements are 1/true). In parallel, Locb is a vector of corresponding locations (given as row numbers) in v. The first element of Lia tells you that row 1 of M(:,1:3) has a match in v, and the first element of Locb tells you that it is row 2 of v.
Now if you want to extract matching rows of M, you can use the vector of logical indices Lia:
>> M(Lia,1:3)
ans =
1 2 4
1 3 4
1 4 2
1 1 3
3 4 2
or
>> M(Lia,:)
ans =
1 2 4 1 3 3 1
1 3 4 4 4 2 3
1 4 2 2 4 3 4
1 1 3 1 3 3 1
3 4 2 1 1 4 1
If you want to extract relevant rows of v, however, you have first to extract non-zero row numbers from Locb, and you can index v with these elements:
>> fullb = Locb(Locb ~= 0)
fullb =
2
4
8
10
7
>> v(fullb,:)
ans =
1 2 4
1 3 4
1 4 2
1 1 3
3 4 2

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Lorenzo
Lorenzo 2013-4-30
Thank you!

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