What is the reasoning behind the fact that min(0,NaN) is 0?

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the cyclist
the cyclist 2012-5-21
回答： Mark vanRossum 2021-6-3
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.
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Bryan
Bryan 2020-4-25
编辑：Bryan 2020-4-25
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(A) returns the minimum elements of an array.
M = min(X) is the smallest element in the vector X.
perhaps it could state
M = min(A) returns the minimum non-NaN elements of an array.
M = min(X) is the smallest non-NaN element in the vector X.
or alternatively
M = min(A) returns the minimum elements of an array, whereby NaN elements are ignored by default.
M = min(X) is the smallest element in the vector X, whereby NaN elements are ignored by default.
cheers
Stephen23
Stephen23 2020-4-25
编辑：Stephen23 2020-4-26
@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...

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回答（5 个）

Walter Roberson
Walter Roberson 2012-5-21
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.
Sean de Wolski
Sean de Wolski 2012-5-21
At the bottom of the doc page:
doc min
The min function ignores NaNs
  2 个评论
the cyclist
the cyclist 2012-5-21
Sean, I appreciate the reply. I realize that the documentation informs me THAT the function will ignore NaNs. That does not enlighten me as to WHY that choice was made, which is what I am trying learn.
Jan
Jan 2012-5-22
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.

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Daniel Shub
Daniel Shub 2012-5-22
Given the behavior of MIN, I find it odd that there is a NANMIN function.
  2 个评论
Jonathan Sullivan
Jonathan Sullivan 2012-6-21
That is really interesting. If you look inside nanmin, it has one line.:
[varargout{1:nargout}] = min(varargin{:});
per isakson
per isakson 2012-6-21
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.

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M Sohrabinia
M Sohrabinia 2012-6-21
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.
Mark vanRossum
Mark vanRossum 2021-6-3
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

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