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

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O Mueller
O Mueller 2021-4-29
评论： O Mueller 2021-4-29
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!
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O Mueller
O Mueller 2021-4-29
yeah there is one more indexing column i omitted, let me fix to make the code consistent
O Mueller
O Mueller 2021-4-29
编辑：O Mueller 2021-4-29
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)

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采纳的回答

Jan
Jan 2021-4-29
编辑：Jan 2021-4-29
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;
  4 个评论
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Jan
Jan 2021-4-29
编辑：Jan 2021-4-29
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
O Mueller 2021-4-29
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

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