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Removing NaN frm matrix

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Mayank Lakhani
Mayank Lakhani el 11 de Mzo. de 2016
Editada: Stephen23 el 11 de Mzo. de 2016
a = NaN NaN NaN
NaN NaN NaN
4 5 6
7 8 9
i want t remove NaN frm matrix and want to have matrix like belw
a = 4 5 6
7 8 9
how to do it with find functin. i can do like below this but i want using find function
Result_All(isnan(Result_All(:,1)),:) = []
  1 comentario
Stephen23
Stephen23 el 11 de Mzo. de 2016
Editada: Stephen23 el 11 de Mzo. de 2016
See Guillaume's answer for the neatest and fastest way to solve this.

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Respuesta aceptada

KSSV
KSSV el 11 de Mzo. de 2016
a = [NaN NaN NaN
NaN NaN NaN
4 5 6
7 8 9];
k = find(isnan(a)) ;
a(k) = [] ;
reshape(a,[2,3])
  3 comentarios
KSSV
KSSV el 11 de Mzo. de 2016
if every time...you are removing a complete row..it would be easy to get a matrix with same columns as the original. If not getting matrix of same columns or rows of original would be tough.
PS: If it is useful. Accept the answer.
Guillaume
Guillaume el 11 de Mzo. de 2016
Note that the find in the above answer is completely unnecessary and only slows down the code. Admittedly the slowdown is negligible in this case.

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Más respuestas (1)

Guillaume
Guillaume el 11 de Mzo. de 2016
Note that you cannot remove arbitrary nans from a matrix since that would leave 'gaps'. You can only remove entire rows or columns. You can detect rows or columns that are just nan with the all function. It is much better than checking the first column as you have done.
To remove all rows that contain nans:
a = [NaN NaN NaN
NaN NaN NaN
4 5 6
7 8 9];
nanrows = all(isnan(a), 2);
a(nanrows, :) = []
This is the "dynamic" versions. find is completely unnecessary. You can of course add as many unnecessary statements as you want to your code to make slower and more efficient, but why?
a = [NaN NaN NaN
NaN NaN NaN
4 5 6
7 8 9];
nanrows = find(all(isnan(a), 2)); %<- completely unnecessary find as you asked
nanrows = nanrows * 1 + 0; %<- completely unnecessary addition and multiplication, while we're at it
a(nanrows, :) = [];

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