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

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 : )

Answer 1

[Image Analyst]

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.

[Abdulatif Alabdulatif]

true!

also,I have this array to create and it takes 47.967158 seconds:

n = 12500/2;

t=1; v=1;

ext= n+1;

count = -n+1;

tic;

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

A3 = flipud(A3);

toc

Is it make sense or it is too long?! I have to create two arrays and it takes the same amount of time which is mean 47.967158*2 = ..... every time I run the function!!

Is it possible to avoid that?

[Image Analyst]

It didn't run. What is m? What is A3 - can I use the same sample values I used? Also please read this so I don't have to spend time fixing your formatting: http://www.mathworks.com/matlabcentral/answers/13205-tutorial-how-to-format-your-question-with-markup

[Abdulatif Alabdulatif]

Sorry regards that : )

m is a variable and A3 is an empty array.

A3 = [];
m = 12500;
n = m/2;

t=1; v=1;

ext= n+1;

count = -n+1;

tic;

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

A3 = flipud(A3);

toc

[John D'Errico]

Sigh. It is NOT find that is so slow. It IS the fact that you are growing the array A3 dynamically.

I'm not going to try to figure out what it is you are doing. HOWEVER I am sure that if you explain what this code does, then we can help you do it MUCH faster.

[Abdulatif Alabdulatif]

I build A3 array to perform mapping for some outputs. This is a sample of A3 array that I have:

12499,-1
12498,-2
12497,-3
12496,-4
12495,-5
12494,-6
12493,-7
12492,-8
12491,-9
12490,-10
12489,-11
12488,-12
12487,-13
12486,-14

For example: I have to find the corresponding value of 12487 which is -13. BUT you have to consider that I have to create this array every time running the function "which I do not like". PLUS searching operation inside that array which is consists of 6250 values.

[Image Analyst]

John hit the nail on the head. To fix, you know in advance what n and m are so you know what the max size of A3 will be. You should be able to preallocate A3 before you get to the loop with zeros(). Even if you don't know what the actual values are yet, just simply making a big array of all zeros will dramatically speed this up.

[Image Analyst]

Abdulatif's "Answer" moved here since it's not an answer to his original question.

preallocate A3 array does not solve the issue.

I think set up the array for permanent rather than make it every time will solve the issue. BUT I am bot sure if there is any way to do that!!

[Image Analyst]

Then give us some way to run the same code you're running, because apparently the little snippets we're using don't illustrate the problem.

[Abdulatif Alabdulatif]

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

[Image Analyst]

I don't have the sym() function. I think that's in the Symbolic Toolbox, which I don't have. You didn't list it in the Products section, so I did it for you. What I did was to just delete the word sym everywhere so that the variables were regular numbers and then ran it. It ran fine and was virtually instant. This is my output:

X =
        1127       21853       18724         995       12191       14286           1
Total =
    26
Total1 =
        4464
Elapsed time is 0.000069 seconds.
Elapsed time is 0.000069 seconds.
Elapsed time is 0.000003 seconds.
          26        4464
ans =
          26        4464

So it looks like my time is like 300 nanoseconds, not 47 seconds. Not sure what's going wrong with yours. Is it any faster if you have them be regular numbers instead of sym numbers?

[Abdulatif Alabdulatif]

perfect!!

when I removed sym(), it works like a tick! : )

I am wondering if there is any interpret for why this is happen?!

Thanks for all : )

[Image Analyst]

I have no ideas about SYM since I don't have that toolbox. Can you mark my Answer as "Accepted" now? Thanks.