How to call index of vector from matrix?

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ha ha
ha ha el 31 de Mzo. de 2018
Comentada: ha ha el 3 de Abr. de 2018
Hello, I have: n-by-3 matrix A, where xi,yi,zi are non-integer(e.g., x=-1.23;y=2.45;z=123.11):
A contain 10million element
A=[ x1 y1 z1 % element 1
x2 y2 z2 % element 2
x3 y3 z3 % element 3
x4 y4 z4 % element 4
.......
xn yn zn] % element n , where n=10,000,000
B=[ x3 y3 z3 % B contain 2 elements
x70 y70 z70]
Question: How to call index of matrix B from matrix A?(i don't wanna use "find(ismember(A,B,'rows'));" , since time consuming with large matrix A & B)
result=[3;70]
  1 comentario
Image Analyst
Image Analyst el 2 de Abr. de 2018
I don't know what "call index of matrix" or "call index of vector" means. Perhaps you can get a native English speaker to look it over. And what would you want? Would you want a 2-by-1 vector of rows where the numbers are found? Like output = [3; 70]? Or what??? Or do you want null/empty because row 3 followed by row 70 never ever appear in A?

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Rik
Rik el 31 de Mzo. de 2018
The code below executes in 0.35 seconds. This will scale about linearly with the number of rows in B.
%Generate data
A=rand(10000000,3);
B=A([3 70],:);
tic
%loop through B to find row indices
result=NaN(size(B,1),1);
for n=1:length(result)
true_row=...
A(:,1)==B(n,1) & ...
A(:,2)==B(n,2) & ...
A(:,3)==B(n,3);
row_idx=find(true_row);
if numel(row_idx)==0
%row not found, keep NaN
elseif numel(row_idx)>1
%more than 1 row found, throw warning
warning('no unique match for row')
%you can use the line below as well, if you just want to use the
%first row that matches.
%result(n)=row_idx(1);
else
result(n)=row_idx;
end
end
toc
  3 comentarios
Rik
Rik el 2 de Abr. de 2018
Comparing many elements just is very time consuming. Use the profiler or tic and toc to figure out which method is fastest for you application. I don't think there are many ways to improve beyond ismember or the method I outline here. You might check to see if arrayfun or parfor yield any speedup.
ha ha
ha ha el 3 de Abr. de 2018
@Rik Wisselink. Thank so much

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