Finding minimum value by using min()

I want to find minimum 10 value

i wrote

for j= 1:10
for i= 1:10
for k = 1:13
 fitness (j,i,k)=(........);%10*10*13 double
end
end
minfit(j) = min(fitness(j,:));
end
disp (minfit);

in min function it is unable to find

minimum value of .

please helpme i am not able to solfitnesse

Answer 1

[Image Analyst]

To find the 10 minimum values of any array, just sort:

sortedValues = sort(fitness(:), 'descend');
tenMinValues = sortedValues(1:10);

If you don't want duplicated values, use unique

sortedValues = unique(fitness(:));
tenMinValues = sortedValues(1:10);

[Sushree Patra]

Thank you very much for the solution. It helps me to find the result.

[Steven Lord]

Alternately you could use the mink function.

[Sushree Patra]

Thank you sir Steven,

I need a source code for Free search krill herd algorithm for global optimization.

Can you help me sir?

[Image Analyst]

How is that different than the numerous other animal herd/swarm/flock/etc. algorithms?

There is no Krill Toolbox so if you can't find it in the File Exchange, and can't use one of the other PSO (Particle Swarm Optimization) algorithms, then you could ask the author (whoever invented it) for code, or write it yourself from lower level functions.

[Sushree Patra]

Thank you sir for the response.

Well, it is different in case of movement of krill and finding the next generation they specifically use free search operator and opposition based method for population intialization.

And I asked to the author but they don't reply.

So I am trying by myself.

Please help me if I am unable to solve.

[Image Analyst]

I've done all the help I can do. See the FAQ

I'm not an expert in PSO-type algorithms, and haven't used them (not needed to so far).

Answer 2

[madhan ravi]

minFit=squeeze(min(Ofitness,[],[1 2])).'
minFit=squeeze(min(min(Ofitness))).' % in older versions of matlab thank you sir Walter

[Sushree Patra]

showing error

Subscripted assignment dimensions mismatch

[Sushree Patra]

showing error

error using min

Dimension argument must be a positive integer scalar within index range.

i placed min function outside of all theloop 3 for

[madhan ravi]

attach fitness as .mat file

[Sushree Patra]

attaching the code named as rly_sen.m

[Jan]

@Sushree Patra: Please post the code. It is strange, that you get two different errors. Without seeing the code, it is impossible to fix it.

[madhan ravi]

I just ran your code but I didn't get any errors?

[Sushree Patra]

ok i post it

clc;

close all; % Close all figures (except those of imtool.)

clear; % Erase all existing variables. Or clearvars if you want.

workspace; % Make sure the workspace panel is showing.

format long g;

format compact;

fontSize = 25;

figure;

imshow('pc3.png');

hold on

hold on

%legend('Relay', 'Sink', 'Sensor');

%

x1 = [105,107,58,58,75,146,118,18,48,55,108,80,80];

y1 = [190,280,190,280,145,153,103,153,103,65,65,50,10];

hold on

A=100;

B=140;

plot(A,B, 'mo-', 'MarkerSize', 10,'LineWidth', 2);

hold on

hold on

% Make 75 points in set #2

Xmin = 60;

Xmax = 99;

x2 = Xmin+rand(1,50)*(Xmax-Xmin);

Ymin = 10;

Ymax= 260;

y2 = Ymin+rand(1,50)*(Ymax-Ymin);

plot(x2, y2, 'b*');

%numPoints2 = 50;

%x2 = 167*rand(numPoints2, 1);

%y2 = 302*rand(numPoints2, 1);

% Plot set 1 in red

plot(x1, y1, 'r.', 'MarkerSize', 13);

% Plot set 2 in blue

hold on;

plot(x2, y2, 'b*', 'MarkerSize', 5);

grid on;

for i=1:13

for j=1:50

%Find distances between every point in set 1 to every point in set #2.

distances(i,j) = pdist2([x1(i),y1(i)], [x2(j), y2(j)], 'euclidean');

end

minDistance(i) = min(distances(i,:));

end

% Find min distance

disp (distances);

dist1 = sqrt((x2-A).^2+(y2-B).^2)

%disp (dist1);

dist2 =sqrt((x1-A).^2+(y1-B).^2)

disp (minDistance);

a=1:40;

for i = 1:10

pop(i,:) = a(randperm(numel(a),10))

end

for i = 1:10

opo_pop(i,:) = 41-pop(i,:)

end

%Est(13,5)=zeros();

%for i=1:13

% b=1;

%for k=1:50

%for j=10*k-9:10*k

%if mod(k,10)==0

%b=b+1;

%end

s=250;

Et=16.7;

Eamp=1.97;

N=10;

data_agg=1;

Er = 36.1;

d0=30;

c=10;

for k = 1:13

for j = 1:10

for i = 1:10

Ets(k,j,i)=s*(Et+Eamp*(distances(k,pop(j,i)).^2));

%Es(i,b)=Est(i,b)+Et(i,k);

end

end

end

disp (Ets);

for k = 1:13

for j = 1:10

for i = 1:10

OEts(k,j,i)=s*(Et+Eamp*(distances(k,opo_pop(j,i)).^2));

%Es(i,b)=Est(i,b)+Et(i,k);

end

end

end

disp (OEts);

%disp(Es);

for j=1:10

for i=1:10

for k= 1:1

Etf(j,i,k)= N*data_agg*Et*(s*dist1(k,pop(j,i)))+N*s*data_agg*Et;

Erf = (N-1)*Er*s;

%Ef_total(i,k)=Etf(i,k)+Erf

end

end

end

Ef_total =Etf + Erf

for j=1:10

for i=1:10

for k= 1:1

OEtf(j,i,k)= N*data_agg*Et*(s*dist1(k,opo_pop(j,i)))+N*s*data_agg*Et;

OErf = (N-1)*Er*s;

%Ef_total(i,k)=Etf(i,k)+Erf

end

end

end

OEf_total =OEtf + OErf

for j=1:10

for i =1:10

for k = 1:13

fitness(j,i,k)= (c*pop(j,i)*1)+(c*(Ets(k,j,i)*1+Ef_total(j,i,1)*1))+(c*sqrt((distances(k,pop(j,i))-d0).^2))+(c*1);

end

end

%minFit(j) = min(fitness(j,:,:));

end

minFit = min(fitness,[],[1,2]);

disp(fitness);

for j=1:10

for i =1:10

for k = 1:13

Ofitness(j,i,k)= (c*opo_pop(j,i)*1)+(c*(OEts(k,j,i)*1+OEf_total(j,i,1)*1))+(c*sqrt((distances(k,opo_pop(j,i))-d0).^2))+(c*1);

end

end

OminFit(j) = min(Ofitness(j,:));

end

disp(Ofitness);

disp(minFit);

minfit=minFit(:);

disp(OminFit);

ominfit=OminFit(:);

Int_Fit = [minfit;ominfit]

intfitness = sort(Int_Fit)

[madhan ravi]

REmove this section of the code!:

for j=1:10
    for i =1:10
        for k = 1:13
            Ofitness(j,i,k)= (c*opo_pop(j,i)*1)+(c*(OEts(k,j,i)*1+OEf_total(j,i,1)*1))+(c*sqrt((distances(k,opo_pop(j,i))-d0).^2))+(c*1);
        end
    end
     OminFit(j) = min(Ofitness(j,:));
 end

[Sushree Patra]

Error using min

Dimension argument must be a positive integer scalar within indexing range.

Error in rly_sen (line 128)

minFit = min(fitness,[],[1,2]);

[madhan ravi]

wierd this is what i got:

>> minFit
minFit(:,:,1) =
          14831476.9829354
minFit(:,:,2) =
          52886274.7233204
minFit(:,:,3) =
          20688650.8562262
minFit(:,:,4) =
          53651138.6228232
minFit(:,:,5) =
          7829219.62925941
minFit(:,:,6) =
          18287522.4496381
minFit(:,:,7) =
          10960447.7077643
minFit(:,:,8) =
          27724536.6404127
minFit(:,:,9) =
          19579242.1922717
minFit(:,:,10) =
          29888312.5238552
minFit(:,:,11) =
           24860479.583199
minFit(:,:,12) =
          31616660.2405204
minFit(:,:,13) =
          47582942.7897235
>> 

[Sushree Patra]

if i am usin min function like this

minFit(j) = min(Ofitness(j,:));

then no error but it is not calculate the minimum value correctly

[Sushree Patra]

is it calculated correctly the minimum fitness value

[madhan ravi]

Ofitness is a 3D matrix , which minimum do you want to return? minimum of each row/column or the minimum of each page?

[Sushree Patra]

minimum 10 fitness value of each column of each page

[Sushree Patra]

first loop i.e j is for 10 set

second loop is for with in each set 10 value

thired loop is for 13 fixed node

i calculated the distance of each value with 13 fixed node of 10 set that is 10*10*13

now i have the find 10 minimum of all of the values and stored in minfit

[madhan ravi]

min(Ofitness) % that's it outside loop 

[Sushree Patra]

but it shows minimum value with respect to all the 13 node

is it possible to find only 10 minimum fitness value whhich is stortest among all

[Sushree Patra]

Among all best 10 minimum might be calculate by sort them and choose first 10

[Sushree Patra]

now minFit = 1*10*13double

my objective =1*10double

[madhan ravi]

minFit=squeeze(min(Ofitness,[],[1 2])).' % after this I don't have any clue what you want

[Walter Roberson]

madhan, the syntax min(Array,[],[vector]) is new as of r2018b .

[madhan ravi]

Thank you so much sir Walter , is there any alternatives for it in older versions?

[Walter Roberson]

multiple min calls or reshape followed by min

[madhan ravi]

Thank you sir Walter ?

[Sushree Patra]

Thank you sir madhan, I apply the solution you gave but now minFit value equal to the fitness value

not find minimum of 10 set.

Let me clarify the objective

10 set

each set have 10 number

total number = 100

then that 100 connect to 13 then we find fitness and Ofitness

fitness & Ofitness = 10*10*13

now calculate only 10 minimum values of that 10 set means set 1 having 10*13 fitness we have to find the minimum among all. like wise for 10 set i.e. my 10 minimum

[Sushree Patra]

Please help me sir, I am also trying but recently i started matlab so kindly help me to solve the problem.

Thank you very much for your solutions and time

[madhan ravi]

forget about everything demonstrate a neat and clean example with 2 by 2 matrix with 2 pages with your desired result it'll save time and avoid confusions

[Sushree Patra]

(,1)=
A=[1,5;6,2]
A'=[9,5;3,6]
(,2)=
B=[9,2;6,3]
B'=[7,3;8,4]

Now i have to find out 2 minimum value one from (,1)another from (,2)

min=[1,2]

[madhan ravi]

what does A' and B' represent? the pages?

[Sushree Patra]

yes that are different pages

[madhan ravi]

A(:,:,1)=[1,5;6,2];
A(:,:,2)=[9,5;3,6];
B(:,:,1)=[9,2;6,3];
B(:,:,2)=[7,3;8,4];

result=[min(A(:)) min(B(:))]

[Sushree Patra]

Thank you sir but in my code how it will fix