array summation with matrices of different size

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Saad
Saad on 3 Jan 2013
Dear All,
I would appreciate some help on this one please.
I am summing matrix elements in different arrays. Say we have 2 arrays:
Price=num2cell(1:10); Coupon=num2cell(1:10);
Price is a cell composed of different matrices. The same for Coupon.
Now the size of the matrices inside "Price" and "Coupon" are not always equal. Of course if I run a code similar to this...
for i=1:10
Price{1,i}+ coupon{1,i};
end
...I get an error which what you would expect because the first matrix of "Price" and "Coupon" are of different size, ....
Now I would like to run an If statement to say: if matrix sizes of Price and Coupon are similar then sum them, if they are not equal then drop that extra elements (at the end) and just sum the first elements. How can I do that please?
example to illustrate: size(Price{1,1})=size(Coupon{1,1})= (1000,1) size(Price{1,2})=(995,1) < size(Coupon{1,2})= (1000,1)
I would like to ignore the extra elements (at the end) of Coupon{1,2} and then sum the remaining elements (i.e. from 1:995) with Price{1,2}? How Can I do that please?
Thanks a lot

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Accepted Answer

Walter Roberson
Walter Roberson on 3 Jan 2013
for K = 1:10
s1 = size(Price{K});
s2 = size(Coupon{K}};
common = min(s1,s2);
rc = common(1);
cc = common(2);
Price{K}(1:rc, 1:cc) = Price{K}(1:rc, 1:cc} + Coupon{K}(1:rc, 1:cc);
end

More Answers (1)

Wayne King
Wayne King on 3 Jan 2013
Edited: Wayne King on 3 Jan 2013
When you say "matrices" are they are always really Nx1 column vectors as your example shows? If so you can just do this
Price = cell(10,1);
Coupon = cell(10,1);
Price{1} = randn(995,1);
Coupon{1} = randn(1000,1);
minval = min(length(Price{1}),length(Coupon{1}));
Price{1} = Price{1}(1:minval);
Coupon{1} = Coupon{1}(1:minval);
summ = Price{1}+Coupon{1};
Obviously, you can embed the above in a for loop
for ii = 1:10
minval = min(length(Price{ii}),length(Coupon{ii}));
Price{ii} = Price{ii}(1:minval);
Coupon{ii} = Coupon{ii}(1:minval);
summz{ii} = Price{ii}+Coupon{ii};
end

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