Undefined operator '==' for input arguments of type 'cell'?

i got an error

Undefined operator '==' for input arguments of type 'cell'.

Error in GeneratePopulation (line 143)

pt=find(p1==1); %cycle value utk p1

what can i do to fix it?

%% parent selection 

cp=0.4;                 % Crossover Percentage
np=2*round(cp*r/2);  % Number of Offsprings (Parents)
% np=cp*r

valp=np/2;
v=valp;


n1=length(RS(:,1));
PP=RS([1:v,n1-v+1:end],3:end);   
    
  
P=cell(1,np); % preallocate
for i=1:np
     P{i} = PP(i,:); 
end
%celldisp(P);
%vertcat(P{:}); % double array
 

%% CYCLE CROSSOVER


szcld = size(P,2); % p 
n2=length(P(:,1));
c1=zeros(1,szcld);
c2=zeros(1,szcld);
p1= P(1:v);
p2= P(1:n2-v+1:end);
pt=find(p1==1); %cycle value utk p1

for i=1:P
while (c1(pt)==0)
  c1(pt)=p1(pt);
  pt=find(p1==p2(pt));
end
left=find(c1==0);
c1(left)=p2(left)

pt=find(p4==1); %cycle value utk p1
while (c2(pt)==0)
  c2(pt)=p4(pt);
  pt=find(p2==p1(pt));
end
left=find(c2==0);
c2(left)=p1(left);

end

[madhan ravi]

r ?

[sharifah shuthairah syed abdullah]

r = number of variable

Answer 1

[madhan ravi]

P=vertcat(P{:}); % double array %% uncomment this line

[Walter Roberson]

If you do that then you might as well use PP instead of PP since PP is in matrix form and for some reason is being deliberately broken out into rows.

[madhan ravi]

True sir Walter completely missed that part , OP has to clarify further more and indeed he needs to provide the missing datas to run the code.

[Walter Roberson]

left is not a variable in the code you posted so we have no idea what it is set to.

[sharifah shuthairah syed abdullah]

this is my full coding.. could you help me

%% read data
clc
clear

%global f d;
%flow
f=[     0    90    10    23    43     0     0     0     0     0    0     0
   90     0     0     0     0    88     0     0     0     0    0     0
   10     0     0     0     0     0    26    16     0     0    0     0
   23     0     0     0     0     0     0     0     0     0    0     0
   43     0     0     0     0     0     0     0     0     0    0     0
    0    88     0     0     0     0     0     0     1     0    0     0
    0     0    26     0     0     0     0     0     0     0    0     0
    0     0    16     0     0     0     0     0     0    96    0     0
    0     0     0     0     0     1     0     0     0     0   29     0
    0     0     0     0     0     0     0    96     0     0    0    37
    0     0     0     0     0     0     0     0    29     0    0     0
    0     0     0     0     0     0     0     0     0    37    0     0];    
   
        
%distance
d=[     0    36    54    26    59    72     9    34    79    17   46    95
   36     0    73    35    90    58    30    78    35    44   79    36
   54    73     0    21    10    97    58    66    69    61   54    63
   26    35    21     0    93    12    46    40    37    48   68    85
   59    90    10    93     0    64     5    29    76    16    5    76
   72    58    97    12    64     0    96    55    38    54    0    34
    9    30    58    46     5    96     0    83    35    11   56    37
   34    78    66    40    29    55    83     0    44    12   15    80
   79    35    69    37    76    38    35    44     0    64   39    33
   17    44    61    48    16    54    11    12    64     0   70    86
   46    79    54    68     5     0    56    15    39    70    0    18
   95    36    63    85    76    34    37    80    33    86   18     0];
      
[r,c]=size(f); 

max_i=r;
max_j=r;
max_k=r;
max_q=r;

MaxIt=12;

% % % Create Empty Structure
%  empty_individual.P1=[];
%  empty_individual.P2=[];
 
 % Create Population Matrix (Array)
% pop=repmat(empty_individual,r,1);

%% calculate fitness function

for no=1:MaxIt

ID=randperm(r); %PERMUTATION r NOMBOR ..r = 3, 1 2 3, 1 3 2...
x=zeros(r,r); 
n=length(ID);
x1=ID;

for i=1:n
    for j=1:r
    x(j,x1(j))=1;
    B(no).mat=x; % utk simpan memori bila buat 0,1 permuation tu dlm (i),BINARY UNTUK RELATION     
    end
  x=zeros(r,r);  
end
 

    z=0;xa=B(no).mat;
for i=1:max_i
    for j=1:max_j
        for k=1:max_k           
            for q=1:max_q               
                 z= z+ f(i,k).*d(j,q).*xa(i,j).*xa(k,q);
            end           
        end
    end
end
  no;
 % G(no,:)=[no z]; %STOR SEMUA VALUE Z 1 HINGGA MaxI DLM MATRIX G  
  F(no,:)=[no z ID]; %STOR SEMUA VALUE Z dan ID 1 HINGGA MaxIt DLM MATRIX F
  

end
%end

 format compact % Suppresses the display of blank lines.
 
% 
       zmin=min(F(:,2));
       zmax=max(F(:,2));
%      ii=find(F(:,2)==zmin) ;
%      x_initial = B(ii).mat;   
     %xbin= B(ii).mat % xbin = initial value for facility location for minimun z
     %Iter_zvalue = F  %matrix F ada no iter..col 1 dan z
     Iter_zvalue = F;  %matrix F ada no iter, z dan ID
 %    K = G %matrix F ada no iter, z dan ID
     
%      %permutation untuk min value.. 
%      b=find(F(:,2)==zmin); 
%      ID_for_minz=F(b,:)


% 
%% Rank Selection.

  
RS = sortrows(F,2);
RS(:,1) = 1:size(F,1);


%% parent selection 

cp=0.4;                 % Crossover Percentage
np=2*round(cp*r/2);  % Number of Offsprings (Parents)
% np=cp*r

valp=np/2;
v=valp;


n1=length(RS(:,1));
PP=RS([1:v,n1-v+1:end],3:end);   
    
  
P=cell(1,np); % preallocate
for i=1:np
     P{i} = PP(i,:); 
end
celldisp(P);
P=vertcat(P{:});% double array
 

%% CYCLE CROSSOVER (long)


szcld = size(P,2); % p 
n2=length(P(:,1));
c1=zeros(1,szcld);
c2=zeros(1,szcld);
p1= P(1:v);
p2= P(1:n2-v+1:end);
pt=find(p1==1); %cycle value utk p1

for i=1:P
while (c1(pt)==0)
  c1(pt)=p1(pt);
  pt=find(p1==p2(pt));
end
left=find(c1==0);
c1(left)=p2(left)

pt=find(p2==1); %cycle value utk p1
while (c2(pt)==0)
  c2(pt)=p2(pt);
  pt=find(p2==p1(pt));
end
left=find(c2==0);
c2(left)=p1(left);

end

[Walter Roberson]

Your PP matrix is 2d. Exactly what part of the 2d array should be tested for value equal 1?

[sharifah shuthairah syed abdullah]

each line in PP represent the result that i get from iteration and i want to used the result to make a crossover..

for example

result 1 = 4 8 3 1 2 11 12 7 6 10 9 5

result 2= 11 4 7 5 12 3 9 1 2 8 6 10

result 3= 7 6 12 3 10 8 11 5 4 1 2 9

result 4= 5 2 6 4 9 8 12 11 7 1 10 3

so i eant to used result 1 as p1 and result 4 as p2 in the crssover and repeated untill it used all the result in PP.. however i didnt want to do it manually.. thats why i want to make it as array so that (i) can automatically change in the loop variable

[Walter Roberson]

Why result1 and result4 ? You look like you are doing a traveling salesman problem in which every row is a permutation of 1:12, so all of the rows will contain a 1 . I do not understand what you expect

pt=find(p1==1);

to return?

[sharifah shuthairah syed abdullah]

i didnt do TSP..im doing quadrattic assignment problem..its a little bit different but still to find a minimum result..

im trying to do cycle crossover..

pt=find(p1==1);

im tying to fine the cycle value on p1..

[Walter Roberson]

So you want to find all the linear indices in p1 at which the value is 1? Not the column number in each row?

[sharifah shuthairah syed abdullah]

when u said it like that.. its seem that my coding is wrong.. i want to find a linear indices in p1 but i dont want all the value is equal to 1.

and i want to read PP not as matrix but as an answer .. each row is an answer / parent