How to search a matrix for all elements of a vector?

I am trying to search an element list for the occurance of nodes and output the elements where the nodes are occuring.

I want to find a fast integrated method, as for loops are just horribly slow.

    x_mask=N(:,2) > X(i,1)-X(i,4)/2 & N(:,2)< X(i,1)+X(i,4)/2;
    NX=N(x_mask,:);
    y_mask=NX(:,3)> X(i,2)-X(i,5)/2 & NX(:,3)< X(i,2)+X(i,5)/2;
    NY=NX(y_mask,:);
    z_mask=NY(:,4)> X(i,3)-X(i,6)/2 & NY(:,4)< X(i,3)+X(i,6)/2;
    NZ=NY(z_mask,:); %NZ is the Vector that contains the nodes 22500x1 of them
    %(it also still contains the x,y,z coordinates but those can be ignored) 
    % i.e.:
%       1	0.849999300000000	0.273165000000000	-0.517777800000000
%       2	0.859999200000000	0.273165700000000	-0.517777800000000
%       3	0.849999500000000	0.223106800000000	-0.395308600000000
%       4	0.859999400000000	0.223107700000000	-0.395308600000000
%       5	0.849999700000000	0.180807400000000	-0.291975300000000
%       8	0.859999600000000	0.180808400000000	-0.291975300000000
%       9	0.849999800000000	0.146298800000000	-0.207777800000000
%       11	0.859999800000000	0.146299700000000	-0.207777800000000
%       13	0.849999900000000	0.119606900000000	-0.142716100000000
%       14	0.859999900000000	0.119607700000000	-0.142716100000000
%       16	0.850000000000000	0.100751900000000	-0.0967901000000000
%       17	0.860000000000000	0.100752600000000	-0.0967901000000000
%...
    pow=zeros(length(E),2); 
    
    for n=1:length(NZ)
        for e=1:length(E) 
            %E contains the elements-Node table 71300x8 (+ 1 indexing column that can be ignored)
            % i.e. 1	57122	57123	57125	57124	71892	71889	71890	71891
            %      2	56110	56062	56060	56059	71888	71885	71886	71887
            %      3	56176	56172	56165	56171	71884	71881	71882	71883
            %      4	55532	55533	55534	55535	71880	71877	71878	71879
            %      5	55172	55173	55211	55213	71876	71873	71874	71875
            %      6	52416	52411	52412	52414	71872	71869	71870	71871
            %      7	56714	56715	56683	56705	71868	71865	71866	71867
            %      8	54553	54554	54559	54556	71864	71861	71862	71863
            %      9	54906	53303	53304	54917	71860	71857	71858	71859
            %....
            if(~isempty(find(E(e,2:9)==NZ(n,1))))
                pow(e,2)=pow(e,2)+1; %i need to know how many nodes of NZ are part of an element
            end
        end
    end
    

I am trying to use some sort of find to search the matrix, but as the dimensions do not agree.

Are there functions that can do this? Maybe i should go for tree searching?

Thank you so much for your help!

[O Mueller]

if you know another post that already answers this question i am happy for a link!

[Jan]

If E is [71300 x 8], this must fail: E(e,2:9).

Do I understand correctly, that pow(k, 2) contains the number of occurences of the elements of NZ in E(:, 2:9)? The current code increases the element in pow by 1 if there is any match in a row of E, but does not care about the number of matchs.

Using length() is dangerous in the general case, because it does not measure a specific dimension, but take the longest one.

Such comments are horrible, if you ask others for help:

 %NZ is the Vector that contains the nodes 22500x1 of them
 %(it also still contains the x,y,z coordinates but those can be ignored) 

For the efficiency of the code it does matter if the arrays are matrices or vectors. With such ambiguous explanations, the readers have to guess how the real input data look like.

[O Mueller]

yeah there is one more indexing column i omitted, let me fix to make the code consistent

[O Mueller]

pow(e,2) contains occurances of Nodes in NZ in element E(e,1)

Element E(e,1) is made up from nodes E(e,2:9)

Answer 1

[Jan]

Start with replacing:

if(~isempty(find(E(e,2:9)==NZ(n,1))))

by

if any(E(e, 2:9) == NZ(n,1))

Or even better: %i need to know how many nodes of NZ are part of an element

for n = 1:size(NZ, 1)
    for e = 1:size(E, 1)
        pow(e, 2) = pow(e, 2) + any(E(e, 2:9) == NZ(n,1));
    end
end

Then test, if omitting the inner loop is faster:

for n = 1:size(NZ, 1)
    pow(:, 2) = pow(:, 2) + any(E(:, 2:9) == NZ(n), 2);
end

Avoid the repeated indexing in pow(:, 2) by using a temporary variable:

tmp = 0;
EE  = E(:, 2:9);
for n = 1:size(NZ, 1)
    tmp = tmp + any(EE == NZ(n), 2);
end
pow(:, 2) = tmp;

[O Mueller]

Yes the functionality of

B=any(A=y,2)

was exactly what i was looking for!

Thank you!

[O Mueller]

If i could get rid of the outer loop it would be even nicer, but the performance gain is already visisble!

[Jan]

Why do you think that omitting the outer loop is "nicer"? Creating large intermediate arrays is usually a drawback for the run time. In your case it would be a 22500 x 71300 x 8 array, which need 12 GB of RAM.

Before I try it, please explain, if this code really does, what you need: Should pow(i) be increased by 1 if a matching value contains anywhere in the row? Or do the number of occurences count? Are the values of the rows unique? For such cases intersect and histcounts or a fast accumarray approach might be efficient.

Can you provide some example data for NZ and E?

[O Mueller]

so The code ran and did what i wanted.

I say it would be great if there is no loop, because it still needs to loop through all 22500 nodes, and i need to do this process for every experiment.

I have 8GB RAM so i guess its out of the option.

Currently the code is acceptable, you dont need to try :)

In case you want the challenge:

• rows of E are unique
• I need to know how many times any Node from NZ occurs per row of E
• it is not important what location in row of E it occurs
• it is important in what row there are how many occurances (0 to 8)

I attached 2 mat files, for ease of use NZ is a bit smaller(only 8600 long) in this sample