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

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sharifah shuthairah syed abdullah 2019-1-7

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/438546-undefined-operator-for-input-arguments-of-type-cell

评论： sharifah shuthairah syed abdullah 2019-1-8

在 MATLAB Online 中打开

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

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madhan ravi 2019-1-7

r ?

sharifah shuthairah syed abdullah 2019-1-7

r = number of variable

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回答（1 个）

madhan ravi 2019-1-7

0
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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/438546-undefined-operator-for-input-arguments-of-type-cell#answer_355273

在 MATLAB Online 中打开

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

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sharifah shuthairah syed abdullah 2019-1-7

在 MATLAB Online 中打开

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 2019-1-8

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 2019-1-8

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

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