delete the row where is a NaN

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Saad
Saad 2013-1-14
Dear all,
I am stuck in a silly programming problem and I would like to ask for some help please.
I have two cells a and b:
a=num2cell(1:10); b=num2cell(1:10);
in each cell, I have 10 column vectors (n x 1). The cell "a" has a NaN that I would like to delete. I also would like to multiply the matrices in each cell. However if I delete some row in a{i} then the matrix size will change so I have to delete the same row in b{i} so that the matrices a{i} and b{i} have the same size. To illustrate what I am trying to achieve, here the structure of my code:
for i=1:10
if a{i}(isnan(a{i}))
%% complete here
end
a{i}'*b{i};
end
I would greatly appreciate if you could help me complete the code. Basically I would like to tell the code "if there is a NaN in a{i}, please delete the row, and show me the row number. Then I will delete the same row in b{i} and finally multiply a{i} and b{i}.
Thanks a lot
Kind Regards
S

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采纳的回答

Matt J
Matt J 2013-1-14
编辑：Matt J 2013-1-14
Looping will be faster than proposals based on cellfun or arrayfun
for i=1:10
badrows=any(isnan(a{i}),2);
a{i}(badrows,:)=[];
b{i}(badrows,:)=[];
a{i}'*b{i};
end
  1 个评论
Saad
Saad 2013-1-14
Matt
Thanks a lot Matt. Your code was the simplest to understand.

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更多回答（2 个）

Shashank Prasanna
Shashank Prasanna 2013-1-14
编辑：Shashank Prasanna 2013-1-14
Is there a requirement to be using cells? if no you can do this quickly without loops, if yes, you can always convert them into cells after you are done.
Let
a = [1 2 nan 4 6 7 nan 7 3 nan 10]
b = [1 2 3 4 6 7 7 7 3 7 10]
b(isnan(a))=[]
a(isnan(a))=[]
c = a.*b
% If you would like:
c = num2cell(c)
Azzi Abdelmalek
Azzi Abdelmalek 2013-1-14
% ------Your data-------------
n=4
a1=randi(9,n,10);
b1=randi(9,n,10);
a1( randperm(10*n,10))=nan
a=num2cell(a1)
b=num2cell(b1)
% ------the code---------------
aa=cell2mat(a);
bb=cell2mat(b);
idx=~cellfun(@isnan,a)
out=arrayfun(@(x) aa(idx(:,x)).*bb(idx(:,x)),1:size(aa,2),'un',0)

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