How to sum values of two cell arrays based on another array?.

Hey all

How to sum values of Add1 with all elements of Add2 based on ConditionArray.

    Add1 = {1,1,1,0.6,0.6,0.3; 1.6,1,1,1,0.6,0.6}
    Add2{1,1} = {0,0.8,-1,0.2,-0.5}
    Add2{1,2} = {0,0,0.8,1,0}
    conditionArray = {[2,3,4,5];[1,3,4,5]}

Result will be based on ConditionArray. Like As ConditionArray{1,1} contains [2,3,4,5] which means that we have to take 2nd,3rd,4th and 5th value from Add1 and add up these values with Add2{1,1}. Similarly ConditionArray{2,1} contains [1,3,4,5] which means that we have to take 1st,3rd,4th and 5th value from Add1 and add up these values with Add2{1,2}. Result may look like this:

    ResultantArray{1,1} = {1,1.8,-0.4,0.8,-0.2} % {(1+0),(1+0.8),(0.6+(-1)),(0.6+0.2),(0.3+(-0.5))}
    ResultantArray{1,2} = {1.6,1,1.8,1.6,0.6}     % {(1.6+0),(1+0),(1+0.8),(0.6+1),(0.6+0)}

Thanks in advance

Answer 1

[Birdman]

%conversion
row1Add1=cell2mat(Add1(1,:));
row2Add1=cell2mat(Add1(2,:));
%taking necessary values
row1Add1=row1Add1(cell2mat(conditionArray(1,:)));
row2Add1=row2Add1(cell2mat(conditionArray(2,:)));
%resultant array
ResultantArray{1,1}=[zeros(1,numel(cell2mat(Add2{1,1}))-numel(row1Add1)) row1Add1]+cell2mat(Add2{1,1})
ResultantArray{1,2}=[zeros(1,numel(cell2mat(Add2{1,2}))-numel(row2Add1)) row2Add1]+cell2mat(Add2{1,2})

[lucksBi]

Thank You so much.

One thing i want to ask is that when we apply

    %taking necessary values
row1Add1=row1Add1(cell2mat(conditionArray(1,:)));
row2Add1=row2Add1(cell2mat(conditionArray(2,:)));

Why it eliminated the last column value? Like row1Add1 gives following

    row1Add1 =[1,1,0.6,0.6]
Why not this :     row1Add1 =[1,1,0.6,0.6,0.3]

[Birdman]

Because your last column value has a index of 6, but your index for first row was

[1 3 4 5]

[lucksBi]

Ok but Add1 needs to be 1 index ahead to Add2 as we have to remove 1 values from Add1 based on conditionalArray. So that indexes left behind are equal in number

[Birdman]

If you want to include that number as well, then change the index to

[1 3 4 5 6]