how can I reduce the execution time of the given code?

Question

Neetha Mary 2016-1-27

0
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https://ww2.mathworks.cn/matlabcentral/answers/265383-how-can-i-reduce-the-execution-time-of-the-given-code

评论： Stephen23 2016-2-4

i have the following code which takes a 311 kb JPEG image as input. It takes more than half hour to execute this code.

clc;
clear all;
close all;
workspace;
image= imread('1.jpg');
% extracting green channel
t2 = image(:,:,2);
[row, col]=size(t2);
k=(row-16+1)*(col-16+1);
i1 = 1
% feature extraction
    for i=1:row-15
        for j=1:col-15
            X(1,i1) = i
            X(2,i1) = j
            AC = t2(i:i+15,j:j+15)
            f(1) = mean2(AC)
            B = mat2cell(AC, [8 8],[8 8])
            s1 = mean2(B{1,1})
            s2 = mean2(B{1,2})
            s3 = mean2(B{2,1})
            s4 = mean2(B{2,2})
            f(2) = (s1/(4 * f(1) + 0.01))
            f(3) = (s2/(4 * f(1) + 0.01))
            f(4) = (s3/(4 * f(1) + 0.01))
            f(5) = (s4/(4 * f(1) + 0.01))
            f(6) = s1 - f(1)
            f(7) = s2 - f(1)
            f(8) = s3 - f(1)
            f(9) = s4 - f(1)
            m1 = max([f(6),f(7),f(8),f(9)])
            m2 = min([f(6),f(7),f(8),f(9)])
            X(3,i1) = floor(f(1))
            X(4,i1) = floor(255 * f(2)) 
            X(5,i1) = floor(255 * f(3))
            X(6,i1) = floor(255 * f(3))
            X(7,i1) = floor(255 * f(5))
            X(8,i1) = floor(255 * ((f(6) - m2)/(m1 - m2 + 0.01))) 
            X(9,i1) = floor(255 * ((f(7) - m2)/(m1 - m2 + 0.01)))
            X(10,i1) = floor(255 * ((f(8) - m2)/(m1 - m2 + 0.01)))
            X(11,i1) = floor(255 * ((f(9) - m2)/(m1 - m2 + 0.01)))      
            i1 = i1 + 1
           end
    end
    Y = X;
    for i1 = 11:-1:3
        m1 = Y;
        f = Y(i1,:);
        m2 = zeros(1,256);
        for j = 1:k
            m2(f(j)+1) = m2(f(j)+1) + 1;
        end
        % Convert to cumulative values
        for i = 2:256
            m2(i) = m2(i) + m2(i - 1);
        end
        % Sort the array
        for j = k:-1:1
            Y(:,m2(f(j)+1))= m1(:,j);
            m2(f(j)+1) = m2(f(j)+1) - 1;
        end     
    end
P(1,:) = Y(1,:);
P(2,:) = Y(2,:);
for j=1:k-1
    P(3,j) = sqrt(((P(1,j+1) - P(1,j))*(P(1,j+1) - P(1,j)))+((P(2,j+1) - P(2,j))*(P(2,j+1)-P(2,j))));
    P(3,j) = floor(P(3,j)); 
end
AC =sort(P(3,:));

how can i reduce execution time.

5 个评论
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Stephen23 2016-1-29

编辑：Stephen23 2016-1-30

I know what mat2cell is doing, I just pointed out that is not necessary. See my answer to know how you can remove the slowest operation from your code.

Stephen23 2016-2-4

@Neetha Mary: it is considered polite on this forum to accept the answer that best resolves your question.

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Answer 1

Stephen23 2016-1-29

2
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/265383-how-can-i-reduce-the-execution-time-of-the-given-code#answer_207837

编辑：Stephen23 2016-1-29

在 MATLAB Online 中打开

test.m

After adding semicolons to all of the lines I ran the code with a small image and using the profiler, and found that this one line takes 40% of the total processing time:

B = mat2cell(AC, [8 8],[8 8]);

The cell array B is used on the following four lines, like this:

s1 = mean2(B{1,1})

Removing mat2cell and using basic matrix indexing would remove this bottle-neck. You can access those blocks directly using indexing, without using mat2cell, which will be twice as fast (the test script is attached below):

Elapsed time is 12.305447 seconds.
Elapsed time is 6.675523 seconds.

By removing the mat2cell I halved the time for that operation. You should replace all of these lines:

            AC = t2(i:i+15,j:j+15)
            f(1) = mean2(AC)
            B = mat2cell(AC, [8 8],[8 8])
            s1 = mean2(B{1,1})
            s2 = mean2(B{1,2})
            s3 = mean2(B{2,1})
            s4 = mean2(B{2,2})

with four lines like this:

s1 = mean2(t2(i+0:i+7,j+0:j+7))
s2 = mean2(t2(i+0:i+7,j+8:j+15))
etc

and one like this:

f(1) = mean2(t2(i:i+15,j:j+15))

Doing this will speed up the slowest operation in the whole code. Then you can use the profiler to check if there are other lines that can be sped up.