What is the reasoning behind the fact that min(0,NaN) is 0?
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I know that Mathworks pays a lot of attention to this stuff, so I am wondering why the expression
>> min(0,NaN)
is 0. Returning a NaN here seems more logical to me.
4 comentarios
Bryan
el 25 de Abr. de 2020
Editada: Bryan
el 25 de Abr. de 2020
yes i noticed that very quickly (hence deleted that bit of the post). but you were in there very quick with helpful feedback, thanks for the explanation.
As a small bit of feedback, perhaps the min() documentation could mention in the very first line of the documentation that NaN values will be ignored by default, since this is unusal behaviour in the Matlab environment. The documentation here: https://se.mathworks.com/help/matlab/ref/min.html and in 'help min' states
M = min(X) is the smallest element in the vector X.
perhaps it could state
M = min(X) is the smallest non-NaN element in the vector X.
or alternatively
M = min(X) is the smallest element in the vector X, whereby NaN elements are ignored by default.
cheers
Stephen23
el 25 de Abr. de 2020
Editada: Stephen23
el 26 de Abr. de 2020
@Bryan: you should make that as an enhancement request.
Another option is to stop relying on inconsistent "default" behavior and always specify any flags, dimensions, etc. for any function that has these kind of options. Although it requires a little more typing, it has the following advantages:
- makes the intention clear
- avoids bugs, e.g. when a matrix ony has one row (and thus min returns a scalar, not a row vector)
- throws an error on versions that do not support that option, rather than silently continuing...
Respuestas (5)
Walter Roberson
el 21 de Mayo de 2012
If you initialize the result to inf, and then loop testing whether the current value is less than the result and replace the result if it is, then since NaN < any number is false, the result will never get replaced with NaN. You would have to add special code to return NaN in such a case.
0 comentarios
Sean de Wolski
el 21 de Mayo de 2012
2 comentarios
Jan
el 22 de Mayo de 2012
It depends on how you understand the MIN function. 0 < NaN replies FALSE, but NaN < 0 replies FALSE also. As long as it is well documented, both values are reasonable.
Daniel Shub
el 22 de Mayo de 2012
Given the behavior of MIN, I find it odd that there is a NANMIN function.
2 comentarios
Jonathan Sullivan
el 21 de Jun. de 2012
That is really interesting. If you look inside nanmin, it has one line.:
[varargout{1:nargout}] = min(varargin{:});
per isakson
el 21 de Jun. de 2012
MIN and MAX ignores NaN. MEAN and SUM does not. I guess NANMIN (in stat toolbox) is for people like me who cannot remember all the details when we cannot see the underlying logic.
M Sohrabinia
el 21 de Jun. de 2012
NaN is considered undefined, so undefined is ignored by most functions (0/0 will be resulted in NaN which is basically undefined but any number divided by 0, say 4/0, will result in inf). However, the question is why Matlab has decided to treat NaNs in a certain way in some functions, e.g., sort function will always arrange NaNs at the top end (A to Z mode). I guess Matlab has just decided to adopt some rules to handle exceptions.
0 comentarios
Mark vanRossum
el 3 de Jun. de 2021
I encountered this when working on arrays.
x=[1 NaN 10];
y=[5 5 5];
m=min(x,y) and m=nanmin(x,y) return [1,5,5]
In V2020, min(x,y,'includenan') returns [1, NaN,5]
Here is an ugly workaround to get the desired behaviour in older versions.
q=isnan(x)
m=min(x,y)
m(q)=NaN
0 comentarios
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