three level indexing and nested if statements

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whynotwegs
whynotwegs 2017-1-13
编辑： Stephen23 2017-1-13
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
  1 个评论
Stephen23
Stephen23 2017-1-13
编辑：Stephen23 2017-1-13
@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.

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Walter Roberson
Walter Roberson 2017-1-13
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"
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whynotwegs
whynotwegs 2017-1-13
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
Walter Roberson 2017-1-13
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.

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