three level indexing and nested if statements

I have an index with three conditions, one condition in each column. I want to create a master index with one column for classifications which incorporate all three columns of the previous index. I did this successfully with two conditions but can't figure out how to add the third. I include the code that works and my attempt to change it.

for i = 1:length(DEFcomboIDX)
    if DEFcomboIDX(i,1) % Type 1 SN        %If comboIDX column 1 = 1 then Type 1
        if ~DEFcomboIDX(i,2)               % If comboIDX column 2 =0 then SN
            DEFindex(i,1) = 1;             % mark it with a 1
        elseif DEFcomboIDX(i,2) % Type 1 VTA 
            DEFindex(i,1) = 2;             % if comboIDX column 2 = 1 then VTa; Mark with a 2
        end
    elseif ~DEFcomboIDX(i,1) % Type 2 SN  
        if ~DEFcomboIDX(i,2)
            DEFindex(i,1) = 3;
        elseif DEFcomboIDX(i,2) % Type 2 VTA 
            DEFindex(i,1) = 4;
        end
    end
end

and what i'm trying now:

for i = 1:length(DEFcomboIDX)
    if DEFcomboIDX(i,1) % Type 1 SN        %If comboIDX column 1 = 1 then Type 1
        if ~DEFcomboIDX(i,2)                % If comboIDX column 2 =0 then SN
            if DEFcomboIDX(i,3)
                DEFindex(i,1) = 1;             % mark it with a 1
        elseif DEFcomboIDX(i,2) % Type 1 VTA 
            if DEFcomboIDX(i,3)
            DEFindex(i,1) = 2;             % if comboIDX column 2 = 1 then VTa; Mark with a 2
        end
    elseif ~DEFcomboIDX(i,1) % Type 2 SN  
        if ~DEFcomboIDX(i,2)
           if DEFindex(i,1) = 3;
        elseif DEFcomboIDX(i,2) % Type 2 VTA 
            DEFindex(i,1) = 4;
        end
            end
        end
        end
    end
end

[Stephen23]

@whynotwegs: Using a truth table would be much simpler than this code. You could do this in just a few lines of neater and more robust code.

Answer 1

[Walter Roberson]

In your first code, the elseif for Type 2 SN is matched with the if for Type 1 SN . In your second code, the elseif for Type 2 SN is matched with the if for Type 1 VTA. I think you are missing an "end"

[whynotwegs]

you're right I needed the ends in the middle of the statement instead of the end. here's the corrected code if anyone is interested

for i = 1:length(DEFcomboIDX)
    if DEFcomboIDX(i,1) % Type 1 SN        %If comboIDX column 1 = 1 then Type 1
        if DEFcomboIDX(i,3)                 % ANd if combo IDX column 3 =1 then meets baseline requirements
        if ~DEFcomboIDX(i,2)                % If comboIDX column 2 =0 then SN         
                DEFindex(i,1) = 1;             % mark it with a 1
        elseif DEFcomboIDX(i,2) % Type 1 VTA 
            if DEFcomboIDX(i,3)
            DEFindex(i,1) = 2;             % if comboIDX column 2 = 1 then VTa; Mark with a 2
            end
        end
        end
    elseif ~DEFcomboIDX(i,1) % Type 2 SN  
        if ~DEFcomboIDX(i,2)
           DEFindex(i,1) = 3;
        elseif DEFcomboIDX(i,2) % Type 2 VTA 
            DEFindex(i,1) = 4;
        end
            end
        end

[Walter Roberson]

You would normally get a warning showing up on the right hand side of the editor about the if or endif or end not being aligned.

Also in a case like this I would routinely select the entire block and click for "smart indent", just in case I was overlooking any mismatch on alignment.

[whynotwegs]

Actually this still doesnt work. The rows that meet the first two conditions and not the third aren't being classified. I tried to add more elseifs to reroute them but it didn't work. I made sure to match the proper if with the proper elseif and match up my ends

for i = 1:length(DEFcomboIDX)
    if DEFcomboIDX(i,1) % Type 1 SN        %If comboIDX column 1 = 1 then Type 1
        if DEFcomboIDX(i,3)                 % ANd if combo IDX column 3 =1 then meets baseline requirements    
            if ~DEFcomboIDX(i,2)                % If comboIDX column 2 =0 then SN         
                DEFindex(i,1) = 1; 
            elseif DEFcomboIDX(i,2) % Type 1 VTA
                DEFindex(i,1) = 2;             % if comboIDX column 2 = 1 then VTa; Mark with a 2

              end

          elseif ~DEFcomboIDX(1,3)
              if ~DEFcomboIDX(i,2)
              DEFindex(i,1) = 3;
              elseif DEFcomboIDX(1,2)
                  DEFindex(i,1) = 4;
              end
          end
      elseif ~DEFcomboIDX(i,1) % Type 2 SN
          if ~DEFcomboIDX(i,2)
             DEFindex(i,1) = 3;
          elseif DEFcomboIDX(i,2) % Type 2 VTA
              DEFindex(i,1) = 4;
          end
              end
          end

[Walter Roberson]

You coded

           elseif ~DEFcomboIDX(1,3)

instead of

           elseif ~DEFcomboIDX(i,3)

Might I suggest you build a truth table?

1  3 ~2 => value 1
1  3  2 => value 2
1 ~3 ~2 => value 3
1 ~3  2 => value 4
~1 ~2    => value 3
~1  2    => value 4

Is that the complete table? If it is, then

0 0 0 => value 3
0 0 1 => value 3
0 1 0 => value 4
0 1 1 => value 4
1 0 0 => value 3
1 0 1 => value 1
1 1 0 => value 4
1 1 1 => value 2

And that can be expressed as:

value_table = [3 3 4 4 3 1 4 2];
value_table( (DEFcomboIDX(i,:) * [4; 2; 1]) + 1 )

The * [4; 2; 1] converts the triple from binary into decimal 0 to 7, add 1 to get an index into the table.