How to substitue NaN cells in a column (or matrix) with the previous valid number?

Hi guys, pretty easy question i guess...

I have a very big matrix full of NaN values, thousands of them in each column. I need to replace them with the fist oldest value available..

Let's say that i have a column like this:

1

2

4

NaN

2

In this case i should replace NaN with 2. I have some complexities, for example i may not have a rectangular matrix and i could have some nan at the end of the column... in this case i should replace the nan with the first value "up".. Example:

1

3

2

NaN

NaN

In this case the NaN must become both 2.

I attach an example of a little .mat file ready to make some tries.

Thank you so much for the help.

Answer 1

[Joseph Cheng]

This is the quickest way I was able to come up with. Not very elegant or probably the most optimal way but here is it. I would start by finding the last non-NaN value and having it substitute the last entry in the column. This ensures that the very last entry is a number. Then replaces each NaN value with the following number until there is no more NaN values left in that column. Then moves to the next column.

iexample = Example;
for i =1:4
    temp = iexample(:,i);
    A = isnan(temp);
    A = find(A==0);
    temp(end) = temp(A(end));

    while sum(isnan(temp))~=0
       B = isnan(temp);
       B = find(B==1);
       temp(B)=temp(B+1);
    end
    iexample(:,i)=temp;
end

I was hoping i would be able to come up with a scheme instead of the while loop by substituting consecutive NaN by the first valid number after the NaN. But haven't figure that one out yet.

[Ale]

Hi Joseph, thanks a lot for the help. Your code runs perfectly and it's fast enough for what i need.

Bye