Effacer les filtres
Effacer les filtres

How to fill empty matrix elements based on surrounding values

35 vues (au cours des 30 derniers jours)
Simon Allosserie
Simon Allosserie le 6 Avr 2022
I have a matrix with a lot of empty values and some numbers. A column could look like this (simplified for the sake of this question):
[1 0 0 0 5 0 0 0 0 0 8 0 3]'
The first and last element of the row is always non-zero.
Now I need to fill the matrix column per column, such that the zeros are replaced by linear interpolation values between the surrounding values as such:
[1 0 0 0 5 0 0 0 0 0 8 0 3]' %inital column
[1 2 3 4 5 5.5 6 6.5 7 7.5 8 5.5 3]' %filled column with linear interpolation
The way I would do it now is with loops and iterations: find a non-zero element, find the next non-zero element, replace the elements between those with linear interpolation values. Do this for all values in the column, and for each column.
This certainly is not the most optimal way of doing it. I don't see however how I could change this into a matrix operation, nor did I find a specific function or tool to do this. So I was wondering if anyone has suggestions on how optimise this, or knows a function that performs this operation?

Réponse acceptée

Stephen23 le 6 Avr 2022
Modifié(e) : Stephen23 le 6 Avr 2022
V = [1;0;0;0;5;0;0;0;0;0;8;0;3];
The new approach:
Z = fillmissing(V + 0./(V~=0), 'linear', 1)
Z = 13×1
1.0000 2.0000 3.0000 4.0000 5.0000 5.5000 6.0000 6.5000 7.0000 7.5000
The old approach (repeat for each column if they have zeros in different locations):
Xi = find(V);
Xq = 1:numel(V);
Z = interp1(Xi,V(Xi),Xq(:))
Z = 13×1
1.0000 2.0000 3.0000 4.0000 5.0000 5.5000 6.0000 6.5000 7.0000 7.5000
  1 commentaire
Simon Allosserie
Simon Allosserie le 6 Avr 2022
Hi Stephen, this is exactly what I was looking for. Unfortunately my googling terms did not show up this useful function. Thanks!

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Plus de réponses (1)

Image Analyst
Image Analyst le 6 Avr 2022
You tagged it as "Image Processing". For an image, you might look at the function regionfill().
  1 commentaire
Simon Allosserie
Simon Allosserie le 7 Avr 2022
Indeed, because in the end it is a grayscale image that I'm edititing. However, the filling must be happening in columns (and not in regions) so the solution proposed by Stephen proved to be the fitting one here. regionfill is not giving the results I need at this moment. Thanks for your input though!

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