What is the faster mean to find a value of an array?

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Abdulatif Alabdulatif 2014-5-5

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评论： Image Analyst 2014-5-7

采纳的回答： Image Analyst

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Hi all,

I am looking for the fastest search inside an array. For example A3 array has 6250 elements:

[i,j] = find(A3(:,1) == Total);
Total = A3(i,j+1);

This method takes about a minutes to decide if the value "Total" is in the array A3 or not! which is long time :(

Is there any solution for the problem?

Thanks : )

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Image Analyst 2014-5-5

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Get a faster computer. Look at my results:

% Create sample array of 6250 elements
A3 = [randi(9, [1250,4]), ones(1250,1)];
numberOfElements = numel(A3)
Total = 8;
tic;
[i,j] = find(A3(:,1) == Total);
Total = A3(i,j+1);
toc
Elapsed time is 0.000033 seconds.

Is that 33 microseconds I'm getting? And you're getting minute(s)? Something doesn't add up.

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Abdulatif Alabdulatif 2014-5-6

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This is the function: It is not a professional coding!

I am concern about the time in both sections 3 & 4 they take about

Elapsed time is 47.096965 seconds. && Elapsed time is 47.003354 seconds.

The two while loops run faster in another function which is strange!!

You can use this input as X = [1127 21853 18724 995 12191 14286 1]

or [15171 4712 18725 12951 6588 13701 1]

function [result] = test(X)
% *************  define a set of variables section1 ***************  
A3 = zeros(6251,2);
A4 = zeros(6251,2);
result=zeros(1,2);
R = sym(16667); M = sym(25000); m =sym(12500);
d1 =sym(0); d2 =sym(0); d3 =sym(0); d4 =sym(0);
Total =sym(0); 
Total1 =sym(0);
t = 1;
v=1;
% ************** do some calculations section 2***************
d1 = X(t,v)*R;
d2 = X(t,v+1)*R^2;
d3 = X(t,v+3)*R;
d4 = X(t,v+4)*R^2;
d1 = d1+d2;
d3= d3+d4;
Total = mod(d1,m)
Total1 = mod(d3,m)
%************** define a negative range section3************
tic;
n = m/2;
t=1;
v=1;
ext=n+1;
count = -n+1;
while n<m-1
     A3(t,v)=ext;
     A3(t,v+1) = count;
     t = t+1;
     ext = ext+1;
     count = count+1;
     n = n+1;
end
toc
%********* define the range of decimal points section 4 ***********
tic;
n = m/2;
t=1;
v=1;
ext= n+1;
count = n-1;
while n<m-1
     A4(t,v)=ext;  
     A4(t,v+1) = count;
     t = t+1;
     ext = ext+1;
     count = count-1;
     n = n+1;
end
toc
%disp(A4);
%************** detect negative numbers************
dia1=0;
dia2=0;
i=0;
j=0;
tic;
if (Total<12500 && Total >6250)  
    %fprintf('I am in if 1.1\n');
 [i,j] = find(A3(:,1) == Total);
  dia1 = A3(i,j+1);
else
    %fprintf('I am in if 1.2\n');
    dia1=Total;
end
toc
t=1;
i=0;
j=0;
k=1;
if isequal(X(t,v+6),k)
   % fprintf('I am in if 2.1\n');
    dia2=Total1;
else   
    %fprintf('I am in if 2.2\n');
 [i,j] = find(A4(:,1) == Total1);
  dai2 = A4(i,j+1); 
    %A5(t,v+1)=Total1;
end
%******** assign the results ****************
result(1,1)=dia1;
result(1,2)=dia2;
disp(result);
end