Find first non-NaN in each column of array & combine into one vector

10 vues (au cours des 30 derniers jours)
mfclose
mfclose le 19 Juil 2019
Modifié(e) : Adam Danz le 19 Juil 2019
I am trying to:
1) Identify the first non-NaN value of each column in an array, and convert the rest to NaN
2) Identify the first non-NaN value of each row in the array, and convert the rest to NaN
3) Combine these numbers back into one column vector
I have created something that looks roughly like this (although much larger):
X =
2 NaN NaN NaN
4 6 NaN NaN
NaN NaN NaN NaN
5 4 7 8
After the first step (keeping only the first number in each column), I would hope to have this:
Xa =
2 NaN NaN NaN
NaN 6 NaN NaN
NaN NaN NaN NaN
NaN NaN 7 8
After the second step (keeping only the first number in each row), I would hope to have this:
Xb =
2 NaN NaN NaN
NaN 6 NaN NaN
NaN NaN NaN NaN
NaN NaN 7 NaN
Lastly, I would want to turn this into one column (there should only be NaN's or 1 non-NaN value in each row - I would want it to keep whichever is there)
Xc =
2
6
NaN
7
I apologize for not having any attempted code for this. I appear to have gotten stuck in the doorway of this one...

Réponse acceptée

Bruno Luong
Bruno Luong le 19 Juil 2019
Modifié(e) : Bruno Luong le 19 Juil 2019
X = [2 NaN NaN NaN;
4 6 NaN NaN;
NaN NaN NaN NaN;
5 4 7 8 ]
[m,n] = size(X);
[I,J] = ndgrid(1:m,1:n);
Xa = X;
[~,r]=max(isfinite(Xa),[],1);
Xa(I>r) = NaN
Xb = Xa;
[~,c]=max(isfinite(Xb),[],2);
Xb(J>c) = NaN
[~,c]=max(isfinite(Xb),[],2);
Xc = Xb(sub2ind([m,n],(1:m)',c))
  2 commentaires
mfclose
mfclose le 19 Juil 2019
Thank you, this did exactly what I needed!
Adam Danz
Adam Danz le 19 Juil 2019
Modifié(e) : Adam Danz le 19 Juil 2019
Be sure to add semicolons to the end of 3 lines that are missing them.
Also, use ~isnan() rather than isfinite().

Connectez-vous pour commenter.

Plus de réponses (2)

Adam Danz
Adam Danz le 19 Juil 2019
Modifié(e) : Adam Danz le 19 Juil 2019
X = [2 NaN NaN NaN
4 6 NaN NaN
NaN NaN NaN NaN
5 4 7 8];
X(cumsum(cumsum(~isnan(X)))~=1) = NaN; % your Xa
X(cumsum(cumsum(~isnan(X),2),2) ~=1) = NaN; % your Xb
Xc = NaN(size(X,1),1);
Xc(any(~isnan(X),2)) = X(~isnan(X)); % your Xc
  2 commentaires
mfclose
mfclose le 19 Juil 2019
Thank you this also worked!
Adam Danz
Adam Danz le 19 Juil 2019
Modifié(e) : Adam Danz le 19 Juil 2019
Glad I could help! Just FYI, this solution and the max(infinite()) solution are both quite fast but this one is 4.5x faster (10,000 iterations, p<0.0001 Wicoxon signed rank) with less overhead. Also, instead of isfinite(), you should use ~isnan().

Connectez-vous pour commenter.


Mario Chiappelli
Mario Chiappelli le 19 Juil 2019
Check out this question asked earlier on the forum, I think it asks and answers what you want.
  3 commentaires
Mario Chiappelli
Mario Chiappelli le 19 Juil 2019
Try this:
x = [2,NaN,NaN,NaN;4,6,NaN,NaN;NaN,NaN,NaN,NaN;5,4,7,8];
rows = length(x(:,1));
columns = length(x(1,:));
% This finds the first non NaN in the columns
columnVector = double(columns);
for i = 1:columns
found = 0;
for j = 1:rows
if ~isnan(x(j,i)) && found == 0
found = 1;
columnVector(i) = x(j,i);
end
end
if found == 0
columnVector(i) = NaN;
end
end
% This finds the first non NaN in the rows
rowVector = double(rows);
for i = 1:rows
found = 0;
for j = 1:columns
if ~isnan(x(i,j)) && found == 0
found = 1;
rowVector(i) = x(i,j);
end
end
if found == 0
rowVector(i) = NaN;
end
end
I could include some more functionality to delete any NaN values in the resulting columnVector and rowVector arrays. This does what you want, however it skips over the middle steps. I did not know if that was a necessary step for you or just your way of visualizing what needed to be done.
Hope this helps :)
mfclose
mfclose le 19 Juil 2019
This was very close, but I did need the middle step getting rid of the other non-NaN values in a column. The code below fixed this. Thank you so much for helping!

Connectez-vous pour commenter.

Catégories

En savoir plus sur Logical dans Help Center et File Exchange

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by