Moving NaN elements from the last columns to the first column, iteratively

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Hello community,
I'm struggling to figure out a way to automatize a problem I'm confronting on Matlab. Here is a fictitious matrix that replicates my issue (mine is much bigger, but always has between 0 and 2 NaN observations at the end of the matrix):
A = [1 2 3 4 5 6 7 8 NaN NaN;
1 2 3 4 5 6 7 8 9 NaN;
1 2 3 4 5 6 7 8 9 10]
For those rows that have NaNs, I want to move the NaN observations to the front of the matrix, so as to have this;
A = [NaN NaN 1 2 3 4 5 6 7 8;
NaN 1 2 3 4 5 6 7 8 9;
1 2 3 4 5 6 7 8 9 10]
This is a code which works for me with a row vector (i.e A = [1 2 3 4 5 6 7 8 NaN NaN]):
while isnan(A(end))
A = A(1:end-1);
A = [NaN,A];
end
I can't seem to figure out how to broaden the loop to the general case presented above (the 3x10 matrix).
Thank you!

Respuesta aceptada

TADA
TADA el 10 de Jun. de 2019
Editada: TADA el 11 de Jun. de 2019
The simplest way would be to loop through the rows:
i = isnan(A);
for j = 1:size(A,1)
A(j,:) = [A(j,i(j,:)) A(j,~i(j,:))];
end
Another approach:
a1 = A';
[~, ord] = sort(~isnan(a1));
ord2 = ord + repmat(0:size(a1,1):numel(a1)-1, size(a1,1),1);
a2 = a1(ord2)'
  3 comentarios
gd94
gd94 el 11 de Jun. de 2019
Works.... great! Thank you so much TADA

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Más respuestas (1)

Bob Thompson
Bob Thompson el 10 de Jun. de 2019
Do you want your numeric values in ascending order? If so you should be able to do the following:
A = sort(A,2,'MissingPlacement','first');
  1 comentario
gd94
gd94 el 11 de Jun. de 2019
No, not necessarily. The 1 2 3 ..... 9 were just examples of numbers, but they need not be in ascending order. The order should just be maintained when moving the NaNs to the front.

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