Why won't cells populate via if-else statements?

I am taking the transpose of the last three rows of matrix "con". What I want to do is search each newly transposed column for specific numbers using find() (i.e. 1,2,3,4,5,6,7,8,9). I want each of these specific numbers to correspond to a 2x1 matrix, each of which is to be inserted into a cell structure. Then I want to use cell2array to form a matrix. For example [1;2;3;4;5;6] maps to [1;2;3;4;5;6;7;8;9;10;11;12]. My issue is that when I run the code T is displaying as an empty matrix (T=[ ]). I have attached the code down below. Thank you in advance!

clc
clear all
close all

con=[1 2 3 4 5 6;                                                           %Connectivity matrix
    1 2 3 4 5 6;
    3 5 4 8 7 9;
    4 5 6 2 5 7];

Con_Transpose=con(2:end,:).'                                                %Transpose of last three rows of connectivity matrix


width=1:1:size(Con_Transpose,2);
for i=1:length(width)
    
    a={};
    if Con_Transpose(:,width(i))==1;
     
        a{find(Con_Transpose==1,1),1}=[1;2]
        
    end
    
    if Con_Transpose(:,width(i))==2;

        a{find(Con_Transpose==2,1),1}=[3;4]
        
    end
        
    if Con_Transpose(:,width(i))==3;
        
        a{find(Con_Transpose==3,1),1}=[5;6]
        
    end
    
    if Con_Transpose(:,width(i))==4;
  
        a{find(Con_Transpose==4,1),1}=[7;8]
        
    end
    
    if Con_Transpose(:,width(i))==5;
        
        a{find(Con_Transpose==5,1),1}=[9;10]
        
    end

    if Con_Transpose(:,width(i))==6;
        
        a{find(Con_Transpose==6,1),1}=[11;12]
        
    end
    
    if Con_Transpose(:,width(i))==7;
        
        a{find(Con_Transpose==7,1),1}=[13;14]
        
    end
    
    if Con_Transpose(:,width(i))==8;
        
        a{find(Con_Transpose==8,1),1}=[15;16]
        
    end
    
    if Con_Transpose(:,width(i))==9;
      
        a{find(Con_Transpose==9,1),1}=[17;18]
        
    end
    
    T=cell2mat(a)
    
end

Answer 1

[J. Alex Lee]

The if statements are probably your problem...you are comparing a column vector with scalars, which off the top of my head I don't know the behavior of.

I won't attempt to fix your code, but consider this strategy that's based on indexing, doesn't need to loop, and doesn't touch cell arrays...does it do what you want:

% generate your "map" from 1->[1;2], 2->[3;4], 4->[7;8], etc.
rmap = reshape(1:18,2,[])

rmap = 2×9

1 3 5 7 9 11 13 15 17 2 4 6 8 10 12 14 16 18

% test the mechanism
x = [1 4 6]

x = 1×3

1 4 6

a = reshape(rmap(:,x),[],1)

a = 6×1

1 2 7 8 11 12

% now solve your problem
con=[1 2 3 4 5 6;                                                           %Connectivity matrix
    1 2 3 4 5 6;
    3 5 4 8 7 9;
    4 5 6 2 5 7];

Con_Transpose=con(2:end,:).'

Con_Transpose = 6×3

1 3 4 2 5 5 3 4 6 4 8 2 5 7 5 6 9 7

sz = [2,1].*size(Con_Transpose)

sz = 1×2

12 3

T = reshape(rmap(:,Con_Transpose),sz)

T = 12×3

1 5 7 2 6 8 3 9 9 4 10 10 5 7 11 6 8 12 7 15 3 8 16 4 9 13 9 10 14 10

[Kyle Lyman]

Hey J. Alex Lee! Thank you for the prompt response. The method that you presented certainly works for matrices that have an even number of elements (i.e. size(con,2)*(size(con,1)-1)=18 in the example you replied to). However, when I change the matrix "con" to have an odd number of elements, I receive the error:

"Error using reshape

Product of known dimensions, 2, not divisible into total number of elements, 9.

Error in DegreeOfFreedom (line 8)

rmap = reshape(1:range,2,[])"

The code that reproduces this error is listed below:

con=[1 2 3;                         %The first line is local degrees of freedom, thus we focus on the second row through the end row, which sums to 9 total elements.
    6 7 9;
    4 1 6;
    9 7 2]

Con_Transpose=con(2:end,:).'

range=numel(con)-size(con,2);

% generate your "map" from 1->[1;2], 2->[3;4], 4->[7;8], etc.
rmap = reshape(1:range,2,[])

x=[con(2:end,:)];

a = reshape(rmap(:,x),[],1);

sz = [2,1].*size(Con_Transpose);

T = reshape(rmap(:,Con_Transpose),sz)

[J. Alex Lee]

So what does your conversion map look like...

1->1,2

2->3,4

3->5,6

4->7,8

5->9

6->?

7->?

8->?

9->?