How to efficiently replace NAN with Numirical value a reference vector

Dear User,

As per the title, may I know how to make the following code much compact and efficient. I wonder if the number of FOR-Loops can be reduced further?

Thanks in advance.

Refr=[1 20 1 4 5 2];

WithNan=[1 NaN 1 3 2 2;
    1 2 1 NaN 50 2;
    NaN NaN 4 9 NaN NaN;
    NaN NaN NaN 9 NaN NaN];

NoNan=zeros(size(WithNan,1),size(WithNan,2));
for f_x=1:size(WithNan,1)
    SelctCase=WithNan(f_x,:);
    NaNLoc=find (isnan(SelctCase));
    RefForNaN=Refr(NaNLoc);
    
    for f_xx=1:size(NaNLoc,2)
    SelctCase(NaNLoc(f_xx))=RefForNaN(f_xx);
    end
    
    NoNan(f_x,:)=SelctCase;

Answer 1

[Guillaume]

Certainly, the inner for loop is unnecessary (and the find, use logical indexing).

Refr=[1 20 1 4 5 2];
WithNan=[1 NaN 1 3 2 2;
    1 2 1 NaN 50 2;
    NaN NaN 4 9 NaN NaN;
    NaN NaN NaN 9 NaN NaN];

NoNan = WithNan;
for row = 1:size(WithNan, 1)
    toreplace = isnan(WithNan(row, :)); 
    NoNan(row, toreplace) = Refr(toreplace);
end

But loops are not needed at all:

NoNan = WithNan;
filler = repmat(Refr, size(NoNan, 1), 1);
NoNan(isnan(NoNan)) = filler(isnan(NoNan));

Answer 2

[balandong]

Thanks for the quick response. While your approach look elegant, but it consume to much memory if the array become larger. As for my case, WithNan (50 * 500). However, I did not specify about the dimension issue in the original question. Anyhow, II really appreciate for your response.