How to replace all for-loops ?

Question

Sadiq Akbar 2023-1-10

0
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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1891062-how-to-replace-all-for-loops

评论： Jan 2023-1-11

I had posted a function here for vectorization. Now this is the same function and even I tried myself to replace all for-loops like previous but I didn't succeed because in this the outer loop has enclosed all the for-loops and instead of pop being a vector, now it is a matrix due to which I failed to do so. The code is:

clear all; clc
u=[1  2 10 20]; 
dim=length(u);
pop=rand(10,dim);
C = size(pop,2);
P=C/2;
M=2*C;
e = zeros(size(pop,1),1);
for ii = 1:size(pop,1)
    % calculate xo
    b = pop(ii,:);
    xo=zeros(1,M);
    for k=1:M
        for i=1:P
            xo(1,k)=xo(1,k)+u(i)*exp(-1i*(k-1)*pi*cos(u(P+i)));
        end
    end
    
    % calculate xe
    xe=zeros(1,M);
    for k=1:M
        for i=1:P
            xe(1,k)=xe(1,k)+b(i)*exp(-1i*(k-1)*pi*cos(b(P+i)));
        end
    end
    
    abc=0.0;
    for m1=1:M
        abc=abc+(abs(xo(1,m1)-xe(1,m1))).^2; 
    end
    abc=abc/M; 
    
    e(ii)=abc
    
end

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Dyuman Joshi 2023-1-10

在 MATLAB Online 中打开

To add on @Rik's comment, I vectorised a part of the code, and here is the comparison.

You can see that double for loop is faster than the vectorized code.

u   = [1  2 10 20]; 
pop = rand(10, numel(u));
P = width(pop)/2;
M = width(pop)*2;
H = height(pop);
e1=vector(P,M,H,pop,u);
e2=twofor(P,M,H,pop,u);
%comparing results from both methods
result_comparison=isequal(e1,e2)
result_comparison = logical
   1
tic
for i=1:1e3
    y=vector(P,M,H,pop,u);
end
t1=toc;
fprintf('time taken by vector method = %0.4f seconds', t1)
time taken by vector method = 0.1316 seconds
tic
for i=1:1e3
    z=twofor(P,M,H,pop,u);
end
t2=toc;
fprintf('time taken by double for loop method = %0.4f seconds', t2)
time taken by double for loop method = 0.0789 seconds
function e = vector(P,M,H,pop,u)
e=zeros(H,1);
xo=zeros(1,M);
xe=zeros(1,M);
for ii = 1:H
    b=pop(ii,:);
    xo=(exp(-1i*(0:M-1)'*pi*cos(u(P+1:2*P)))*u(1:P)')';
    xe=(exp(-1i*(0:M-1)'*pi*cos(b(P+1:2*P)))*b(1:P)')';
    e(ii)=sum(abs(xo-xe).^2)/M;    
end
end
function e = twofor(P,M,H,pop,u)
e = zeros(H,1);
for ii = 1:H
    b = pop(ii,:);
    xo = zeros(1,M);
    xe = zeros(1,M);
    abc = 0;
    for jj=1:M
        for kk=1:P
            xo(1,jj) = xo(1,jj) + u(kk) * exp(-1i*(jj-1) * pi * cos(u(P+kk)));
            xe(1,jj) = xe(1,jj) + b(kk) * exp(-1i*(jj-1) * pi * cos(b(P+kk)));
        end
        abc = abc+(abs(xo(1,jj)-xe(1,jj))).^2; 
    end
    
    e(ii) = abc/M;    
end
end

Dyuman Joshi 2023-1-11

编辑：Dyuman Joshi 2023-1-11

在 MATLAB Online 中打开

As @Rik mentioned, tic toc will only give an approximate idea of execution time. Results from running tic-toc a single time will always vary. That is why I ran in individual for loops to get a better idea of their execution time.

If you want to see the stark difference between run times, increase the number of iterations. I have increased the count to 1e5. You can see the (obvious) difference in execution times now.

u   = [1  2 10 20];
pop = rand(10, numel(u));
P = width(pop)/2;
M = width(pop)*2;
H = height(pop);
e1=vector(P,M,H,pop,u);
e2=twofor(P,M,H,pop,u);
%comparing results from both methods
result_comparison=isequal(e1,e2)
result_comparison = logical
   1
tic
for i=1:1e5
    y=vector(P,M,H,pop,u);
end
t1=toc;
fprintf('time taken by vector method = %0.4f seconds', t1)
time taken by vector method = 9.4559 seconds
tic
for i=1:1e5
    z=twofor(P,M,H,pop,u);
end
t2=toc;
fprintf('time taken by double for loop method = %0.4f seconds', t2)
time taken by double for loop method = 5.7637 seconds
function e = vector(P,M,H,pop,u)
e=zeros(H,1);
xo=zeros(1,M);
xe=zeros(1,M);
for ii = 1:H
    b=pop(ii,:);
    xo=(exp(-1i*(0:M-1)'*pi*cos(u(P+1:2*P)))*u(1:P)')';
    xe=(exp(-1i*(0:M-1)'*pi*cos(b(P+1:2*P)))*b(1:P)')';
    e(ii)=sum(abs(xo-xe).^2)/M;    
end
end
function e = twofor(P,M,H,pop,u)
e = zeros(H,1);
for ii = 1:H
    b = pop(ii,:);
    xo = zeros(1,M);
    xe = zeros(1,M);
    abc = 0;
    for jj=1:M
        for kk=1:P
            xo(1,jj) = xo(1,jj) + u(kk) * exp(-1i*(jj-1) * pi * cos(u(P+kk)));
            xe(1,jj) = xe(1,jj) + b(kk) * exp(-1i*(jj-1) * pi * cos(b(P+kk)));
        end
        abc = abc+(abs(xo(1,jj)-xe(1,jj))).^2; 
    end
    
    e(ii) = abc/M;    
end
end

Sadiq Akbar 2023-1-11

Thanks a lot dear Dyuman Joshi for your kind response. Yes, i want to accept your answer and give vote to others also but how because vote icon is not visible with others?

Jan 2023-1-11

I've moved Dyuman Joshi's comment to the answers section, such that it can be accepted now.

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Answer 1

Dyuman Joshi 2023-1-11

1
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1891062-how-to-replace-all-for-loops#answer_1146390

移动：Jan 2023-1-11

在 MATLAB Online 中打开

That would be something like this -

e1=vector;
e2=twofor;
%comparing results from both methods
result_comparison=isequal(e1,e2)
result_comparison = logical
   1
fprintf('time taken by vector method = %e seconds', timeit(@vector))
time taken by vector method = 3.112200e-04 seconds
fprintf('time taken by double for loop method = %e seconds', timeit(@twofor))
time taken by double for loop method = 2.392200e-04 seconds
function e = vector
rng(1)
u   = [1  2 10 20];
pop = rand(10, numel(u));
P = width(pop)/2;
M = width(pop)*2;
H = height(pop);
e=zeros(H,1);
xo=zeros(1,M);
xe=zeros(1,M);
for ii = 1:H
    b=pop(ii,:);
    xo=(exp(-1i*(0:M-1)'*pi*cos(u(P+1:2*P)))*u(1:P)')';
    xe=(exp(-1i*(0:M-1)'*pi*cos(b(P+1:2*P)))*b(1:P)')';
    e(ii)=sum(abs(xo-xe).^2)/M;    
end
end
function e = twofor
rng(1)
u   = [1  2 10 20];
pop = rand(10, numel(u));
P = width(pop)/2;
M = width(pop)*2;
H = height(pop);
e = zeros(H,1);
for ii = 1:H
    b = pop(ii,:);
    xo = zeros(1,M);
    xe = zeros(1,M);
    abc = 0;
    for jj=1:M
        for kk=1:P
            xo(1,jj) = xo(1,jj) + u(kk) * exp(-1i*(jj-1) * pi * cos(u(P+kk)));
            xe(1,jj) = xe(1,jj) + b(kk) * exp(-1i*(jj-1) * pi * cos(b(P+kk)));
        end
        abc = abc+(abs(xo(1,jj)-xe(1,jj))).^2; 
    end
    
    e(ii) = abc/M;    
end
end

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How to replace all for-loops ?

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