Given the matrix A, return B in which all the rows that have one or more NaNs have been removed. So for
A = [ 1 5 8
-3 NaN 14
0 6 NaN ];then
B = [ 1 5 8 ]
Solution Stats
Problem Comments
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17 Comments
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14 older comments
Rednas
on 3 Jan 2013
It's a good problem!
gaurav parihar
on 17 Oct 2014
very good problem
Shrabya
on 1 Nov 2014
ya.. but not so difficult though
Alexander Razmyslov
on 30 Jan 2015
nice trick with the " ' "))
Padmanabh Deshmukh
on 23 Jun 2015
very good problem
GYStif
on 10 Feb 2016
I really like this problem. It makes me to improve my solutions again and again. But it seems that I still use too many loops....
Amninder Dhillon
on 19 Aug 2016
lol test case with transform was good one!
SERGII DENEGA
on 30 Sep 2016
do not understand how one get so efficient solution on this (and to be honest other problems as well)
JINWOO LEE
on 7 Nov 2016
thanks you very much.
Engineer_Josh
on 7 Dec 2016
Why doesn't this work? It works when I use it on my computer...
function B = remove_nan_rows(A)
B = A;
B(any(isnan(A')),:) = [];
end
Noriko Hounoki
on 24 Dec 2016
MATLABotic!
Cody Colclough
on 10 Feb 2017
These types of problems are giving me so many useful functions
Amol Koli
on 19 Jul 2017
A(~sum(isnan(A),2),:) this workde for me
André Nóbrega
on 9 Aug 2018
Cool!
Arthur Dimitri
on 10 Aug 2018
Very good problem!
dymaxwell
on 20 Aug 2018
A easy problem if you know fuction isnan.
Divya Yerraguntla
on 27 Feb 2019
Nice one!
Solution Comments
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1 Comment
Xiaoyi Chin
on 19 Sep 2017
really cool!
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2 Comments
Vishvaksenan A.
on 19 Apr 2017
I think that test 4 does not comply to the question.
Amol Koli
on 19 Jul 2017
A(~sum(isnan(A),2),:) this worked for me
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1 Comment
Yoshimasa Nakatani
on 16 Dec 2016
O(^-^) o
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1 Comment
Engineer_Josh
on 7 Dec 2016
function B = remove_nan_rows(A)
B = A;
B(any(isnan(A')),:) = [];
end
Why isn't this working?
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1 Comment
YUSUF AKGUNDUZ
on 15 Oct 2016
I like the solution
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1 Comment
yao gang
on 13 Oct 2016
a = a(all(a==a,2),:); % a = a(~any(isnan(a),2),:);
a(any(a~=a,2),:) = []; % a(any(isnan(a),2),:) = [];
来自知乎
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3 Comments
Ramoflaple
on 4 Jul 2015
If I can't view a smaller solution, how can I improve myself ??
I really want to view the better answers.
Matthew Dvorsky
on 25 Mar 2016
The only improvement you can really make is making it one line
B = A( not( any(isnan(A), 2 ), :)
zefeng chen
on 20 Jun 2017
really?
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2 Comments
Nicholas Hortance
on 30 Jun 2015
I really liked this problem. I know that I took a very long approach to it but in time I hope to improve it. I have had a lot of practice with for and while loops which is why I have taken this approach.
ANAND KUMAR
on 9 Oct 2015
i am unable to compare it with inf. how can i do that??
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2 Comments
Mark Orloff
on 29 Jun 2015
Can anyone tell me why matrix B is not being returned correctly? If I put it one 'end' further into the loop it returns the matrix each time a row is taken out and the tests fail
Ramoflaple
on 4 Jul 2015
when you delete one row of B, the size of B changes.
your solution works with an assumption that the size of B won't change.
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1 Comment
Simon
on 23 Nov 2014
2 points better than my previous solution with ~any(isnan(
Passes the defined tests, but would fail if Inf was in the test matrices.
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3 Comments
Karthik Raja Arumugam
on 12 Feb 2014
Can anyone please tell me how to reduce this size?
Christopher Creutzig
on 24 Jul 2014
You already use vectorized assignment. Carry that further and get rid of the for loop.
Brian Butler
on 14 Nov 2015
The functions sum() and prod() can reduce row-wise to bring out NaN values. Thus, this express produces logicals for whether each row is in or out...
~isnan(prod(A,2))
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2 Comments
Karthik Raja Arumugam
on 12 Feb 2014
Can anyone tell me, how it can coded in size of 10???
Tim
on 12 Feb 2014
Using regexp, you can put all the code into a character string, and the character string only has a size of 1.
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1 Comment
Rahul
on 29 Oct 2013
Good one...
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3 Comments
Tim
on 26 Jan 2012
Would fail if test suite contained any Infs.
David Smith
on 31 Jan 2012
^ Yes, this happens to work on the test inputs, but is technically not doing what the problem asks.
Brett Shoelson
on 1 May 2012
On the other hand, B = A(all(A==A,2),:) works.
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