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Removing NaN frm matrix

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Mayank Lakhani
Mayank Lakhani 2016-3-11
编辑： Stephen23 2016-3-11
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 个评论
Stephen23
Stephen23 2016-3-11
编辑：Stephen23 2016-3-11
See Guillaume's answer for the neatest and fastest way to solve this.

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采纳的回答

KSSV
KSSV 2016-3-11
a = [NaN NaN NaN
NaN NaN NaN
4 5 6
7 8 9];
k = find(isnan(a)) ;
a(k) = [] ;
reshape(a,[2,3])
  3 个评论
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KSSV
KSSV 2016-3-11
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 2016-3-11
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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更多回答（1 个）

Guillaume
Guillaume 2016-3-11
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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