How to pre-allocate a struct in Matlab

I want to pre-allocate stuct e1 and struct e3 in the following loop. I would be thankfull if someone can help.

clc;
clear;
close all; 
x=[-30;-25;-20;20;25;30;30;25;20;-20;-25;-30];
y=[-25;-25;-25;-25;-25;-25;25;25;25;25;25;25];


 i=x(1:end);
 j=y(1:end);

M=[i,j];

s.x=[];
s.y=[];
c=repmat(s,12,1);

c(1).x=M(1,1);
c(1).y=M(1,2);
c(2).x=M(2,1);
c(2).y=M(2,2);
c(3).x=M(3,1);
c(3).y=M(3,2);
c(4).x=M(4,1);
c(4).y=M(4,2);
c(5).x=M(5,1);
c(5).y=M(5,2);
c(6).x=M(6,1);
c(6).y=M(6,2);
c(7).x=M(7,1);
c(7).y=M(7,2);
c(8).x=M(8,1);
c(8).y=M(8,2);
c(9).x=M(9,1);
c(9).y=M(9,2);
c(10).x=M(10,1);
c(10).y=M(10,2);
c(11).x=M(11,1);
c(11).y=M(11,2);
c(12).x=M(12,1);
c(12).y=M(12,2); 



for i=1:27
    j=c(2);
    k=c(4);
    l=c(7);
    o=c(10);
    e3(i,:)=[j;k;l];
    
    for i=1:3:25
        o=c(10);
        e1(i,:)=o;
    end
    
    for i=2:3:26
        o=c(11);
        e1(i,:)=o;
        
    end
    
    for i=3:3:27
        o=c(12);
        e1(i,:)=o;
    end
    
    
    for i=4:6
        k=c(5);
        e3(i,:)=[j;k;l];
    end
    for i=7:9
        k=c(6);
        e3(i,:)=[j;k;l];
        
    end
    
    for i=10:18
        l=c(8);
        e3(i,:)=[j;k;l];
    end
    
    for i=13:15
        k=c(5);
        e3(i,:)=[j;k;l];
        
    end
    
    for i=16:18 %#ok<*FXSET>
        k=c(6);
        d3(i,:)=[j;k;l];
        
    end
    for i=19:27
        l=c(9);
        e3(i,:)=[j;k;l];
        
        
        for i=22:24
            k=c(5);
            e3(i,:)=[j;k;l];
        end
        
        for i=25:27
            k=c(6);
            e3(i,:)=[j;k;l];
        end
        
        
    end
    e=[e3,e1];
  end

[Rik]

Those variables are not a struct, and your layout makes the code difficult to read. Please edit your question to fix this.

You are also reusing your loop variable several times. This also makes it difficult to determine what size the two arrays should be.

[Hamed Bolandi]

That was part of bigger script. I uploaded the first part which shows those are struct.

[Rik]

Try to write some comments in your code what it is doing. You are overwriting entries several times without any clear reason. The answer below (zeros(27,3)) will work, but is likely not your problem.

[Hamed Bolandi]

Thanks for your reply. Below are provided some cooments in my code.

% The objective of this code is to create 81 quadrilaterals with different
% gemoetries. Each quadrilateral has four corners (n=4) and  each corner
% has 3 nodes(m=3). Each node from one corner will be connected to
% one node in other 3 corners to creaete a unique quadrilateral. Finally,
% there would be m^n(3^4=81) quadrilaterals. 

clc;
clear;
close all; 
x=[-30;-25;-20;20;25;30;30;25;20;-20;-25;-30]; % x cordinates of quadrilaterals
y=[-25;-25;-25;-25;-25;-25;25;25;25;25;25;25]; % y cordinates of quadrilaterals


 i=x(1:end);  
 j=y(1:end);

M=[i,j];  % Make a matrix including all x and y coordinates

s.x=[];
s.y=[];   % Make a empty struct

c=repmat(s,12,1);  % create an empty 2 fields struct 

c(1).x=M(1,1);      % Assigning x and y values to the struct. Nodes are labeled from 1 to 12.
c(1).y=M(1,2);
c(2).x=M(2,1);
c(2).y=M(2,2);
c(3).x=M(3,1);
c(3).y=M(3,2);
c(4).x=M(4,1);
c(4).y=M(4,2);
c(5).x=M(5,1);
c(5).y=M(5,2);
c(6).x=M(6,1);
c(6).y=M(6,2);
c(7).x=M(7,1);
c(7).y=M(7,2);
c(8).x=M(8,1);
c(8).y=M(8,2);
c(9).x=M(9,1);
c(9).y=M(9,2);
c(10).x=M(10,1);
c(10).y=M(10,2);
c(11).x=M(11,1);
c(11).y=M(11,2);
c(12).x=M(12,1);
c(12).y=M(12,2);

% Making 3 matrix with dimension of 27X4 in three main loop to make a main matrix with
% dimension of 81x4
 
% Main loop 1

 for i=1:27  % This loop creates the first d=27X4 struct. Breaking struct d into two d3=27x3 and d1=27x1 struct
             % and finally combined d1 and d3 to make a sighe 27X4 struct. I break d struct to d3 and d1 since
             % d1 and d3 elements are changing in each step. In struct d all the quadrilaterals are starting from
             % node number 1(c(1))
            
j=c(1);
k=c(4);
l=c(7);  
o=c(10);

d3(i,:)=[j;k;l];
  

for i=1:3:25    % Assigning c(10) which  x=-20 and y=25 to d1 every 3 row starting from row 1 to row 25
    o=c(10);
    d1(i,:)=o;  
end
     
for i=2:3:26    % Assigning c(11) which  x=-25 and y=25 to d1 every 3 row starting from row 2 to row 26
    o=c(11);
    d1(i,:)=o; 
                  
end                           
                   
for i=3:3:27     % Assigning c(12) which  x=-30 and y=25 to d1 every 3 row starting from row 3 to row 27           
    o=c(12);  
    d1(i,:)=o;   
end           
        
         
    
     
for i=4:6    % Assigning c(5) which x=25 and y=-25 to the 2nd column of d3 starting from row 4 to row 6
     k=c(5);
     d3(i,:)=[j;k;l];
end       

for i=7:9   % Assigning c(6) which x=30 and y=-25 to the 2nd column of d3 starting from row 7 to row 9
    k=c(6);
    d3(i,:)=[j;k;l];     
         
end   

              
for i=10:18  % Assigning c(8) which x=25 and y=25 to the 3rd column of d3 starting from row 10 to row 18
       l=c(8);  
      d3(i,:)=[j;k;l];
end
                     
for  i=13:15 % Assigning c(5) which x=25 and y=-25 to the 2nd column of d3 starting from row 13 to row 15
     k=c(5);
     d3(i,:)=[j;k;l];
                         
end
                     
for  i=16:18  % Assigning c(6) which x=30 and y=-25 to the 2nd column of d3 starting from row 16 to row 18
     k=c(6);
     d3(i,:)=[j;k;l];
                         
end
                     
for i=19:27 % Assigning c(9) which x=20 and y=25 to the 2nd column of d3  starting from row 19 to row 27
    l=c(9);
    d3(i,:)=[j;k;l];
                             
                     
for i=22:24  % Assigning c(5) which x=25 and y=-25 to the 2nd column of d3  starting from row 22 to row 24
    k=c(5);
    d3(i,:)=[j;k;l];
end
                     
for  i=25:27 % Assigning c(6) which x=30 and y=-25 to the 2nd column of d3  starting from row 25 to row 27
     k=c(6);
     d3(i,:)=[j;k;l];
end
                                      
                     
end
d=[d3,d1];  % Combinig d3 and d1 to make a struct d with dimension of 27X4
end   
 
%Repeat all the steps of above loop(main loop 1) for the below loop( main
%loop 2) to make the 2nd 27X4 struct which is named e. In struct e all the
%quadrilaterals are starting from node number 2(c(2))

Answer 1

[Sulaymon Eshkabilov]

Did you mean memory pre-allocation? if so, then

e1=zeros(27,1); 
e3=ones(27,3);