store the index of nanvalues

Hi, So i got a matrix, 5 observations for 2 variables

A = 1 2
    3 4
    3 NaN
    Nan 5
    Nan 7

I wanna store the position when the value of A is a number. so for variable 1, i got values in row 1,2 and 3. for variable 2, i got values in rows 1 2 4 5

so the result of what I want is:

Answer = 1 1 
         2 2 
         3 NaN (because A(3,2) is not a number)
         NaN 4
         Nan 5
or 

Answer 2 = 1 1
           2 2
           3 4
           - 5

Answer 1

[Jos (10584)]

% data
A = [ 1     2
      3     4
      3   NaN
    NaN     5
    NaN     7 ]

% engines  
  tf = ~isnan(A)
  Answer1 = nan(size(A))
  [Answer1(tf), ~] = find(tf)
  Answer2 = arrayfun(@(k) find(A(:,k)), 1:size(A,2), 'un', 0)

[Tiago Dias]

what is the result of your answer2? because it's not the same of what i described in the question, since i think i would prefer answer2 over answer1 since for my main data, i have lot of variables with nans, so i would like to make a reorder, so i see the nums firstly

[Jos (10584)]

Answer2 is a cell array. Each cell contains a vector with numbers. The first cell corresponds to the first column of A, and holds the rows of that column that are not NaN. Since the amount of non-NanS varies between columns, the lengths of the vectors differ between cells.

Answer 2

[KL]

Just an approach with cellarrays.

A = [1 2
    3 4
    3 NaN
    NaN 5
    NaN 7];
res = cellfun(@(x) find(~isnan(x)),num2cell(A,1),'uni',0)

result:

res{1} =
     1
     2
     3

res{2} = 
     1
     2
     4
     5

[Jos (10584)]

?? you could simply do this:

res = ~isnan(A)

[KL]

Yes Jos, you're right! Looks like I need some coffee. (facepalm indeed)

[Tiago Dias]

huh, what Jos said only return a matriz of 1 or 0, not what i wanted

[Tiago Dias]

KL code does whata i prefer over the answer1 of Jos. thanks mate.

[Tiago Dias]

@KL i tried to do

C = cell2table(res)
so i could get 
1 1
2 2
3 4
Nan 5

but i get an error because the matrice is not consistent, could i do that in another way?, because the 1st one has 3 numbers, and the 2nd one has 4

I also tried this, but it only saves for the last j

[n,m] = size(A);
for j = 1:m
  table = res{j};
end