delete the row where is a NaN

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

Answer 1

[Matt J]

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

[Saad]

Matt

Thanks a lot Matt. Your code was the simplest to understand.

Answer 2

[Shashank Prasanna]

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)

[Saad]

Thank you your help SP.

Answer 3

[Azzi Abdelmalek]

% ------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)

[Saad]

Thank you Azzi.