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

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the cyclist
the cyclist on 21 May 2012
Edited: Stephen Cobeldick on 26 Apr 2020
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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Stephen Cobeldick
Stephen Cobeldick on 25 Apr 2020
"...the 'includenan' on min() also has strange behaviour:"
min([NaN; 2; 3; NaN],1,'includenan')
ans =
NaN
1
1
NaN
Why is that strange? You told MATLAB that you want the element-wise lowest values of the 1st or 2nd input, e.g. for the second element the minimum of 2 and 1 is 1... and for the first element the minimum of NaN and 1 is NaN (because of the includenan flag). It behaves exactly as the documentation describes. Perhaps you confused the min(A,B) syntax with specifying the dimension as the third input:
min([NaN; 2; 3; NaN],[],1,'includenan')
Bryan
Bryan on 25 Apr 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(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
Stephen Cobeldick
Stephen Cobeldick on 25 Apr 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...

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Answers (4)

Sean de Wolski
Sean de Wolski on 21 May 2012
At the bottom of the doc page:
The min function ignores NaNs

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the cyclist
the cyclist on 21 May 2012
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 on 22 May 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.

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Walter Roberson
Walter Roberson on 21 May 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.

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Daniel Shub
Daniel Shub on 22 May 2012
Given the behavior of MIN, I find it odd that there is a NANMIN function.

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Jonathan Sullivan
Jonathan Sullivan on 21 Jun 2012
That is really interesting. If you look inside nanmin, it has one line.:
[varargout{1:nargout}] = min(varargin{:});
per isakson
per isakson on 21 Jun 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.

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M Sohrabinia
M Sohrabinia on 21 Jun 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.

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