Why does argmin index of NaN array have a value of 1 and not NaN?

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Michael 2023-12-11

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评论： Matt J 2023-12-12

采纳的回答： Matt J

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[M, I] = min([NaN, NaN])

produces

M =
   NaN
I =
     1
     

Why? When there is no minimum (M=NaN) there should be no index returned for the minimum (the output variable I should also be NaN). This seems odd behaviour.

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Dyuman Joshi 2023-12-11

Because that's how it is defined.

From the documentation of min - "If all elements in the operating dimension are missing, then the corresponding element in M is missing."

M is the output array here.

Michael 2023-12-11

In this case, I'm interested in the I output of the min function:

Index, returned as a scalar, vector, matrix, multidimensional array, or table. I is the same size as the first output.

When "linear" is not specified, I is the index into the operating dimension. When "linear" is specified, I contains the linear indices of A corresponding to the minimum values.

If the smallest element occurs more than once, then I contains the index to the first occurrence of the value.

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Answer 1

Matt J 2023-12-11

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编辑：Matt J 2023-12-11

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I imagine it is so that the logical test tf (below) will always return true. If the I output were NaN, an error message would result.

A=[NaN, NaN];
[M, I] = min(A);
tf=isequaln(M, A(I)) %We want this to be true, always
ans = logical
   1

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Answer 2

Fangjun Jiang 2023-12-11

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编辑：Fangjun Jiang 2023-12-11

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Maybe the index is indeed returned by matching the min value to the input vector. Why does it matter? What is the significance in the case of min([nan nan])?

[M, I]=min([])
M =

     []


I =

     []
[M, I]=min([inf,nan])
M = Inf
I = 1
[M, I]=min([nan,inf])
M = Inf
I = 2
[M,I]=min([nan nan])
M = NaN
I = 1
[M,I]=min([inf inf])
M = Inf
I = 1

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Walter Roberson 2023-12-12

The IEEE 754 committee that defined NaN explicitly created two kinds of NaN: "signalling NaN" and "quiet NaN".

Signalling NaN are intended to trigger a processor exception as soon as they are detected. They are for cases that "Something has gone wrong in the code and the code needs to know right away!". The processor itself is responsible for raising an interrupt when a Signalling NaN is detected.

Quiet NaN are for cases where "Something unusual has happened in the code, but as long as it is marked, the code will take care of detecting and any repairs afterwards."

MATLAB choose not to implement signalling NaNs -- with the limited exception of "dbstop if naninf" (which uses a different detection mechanism)

When IEEE 754 was defined, at least

different NaN were defined for each of "quiet" and "signalling". The intent was that code libraries could use different NaN bit patterns to indicate different causes for the NaN; there was no standardization at all about particular patterns having specific meaning, so meanings were not inherently compatible between library implementations... but it was expected that standardization might come in time if it was going to be useful.

MATLAB treats all NaN the same way. It does not make any effort to preserve NaN bit patterns.

So, what happened in practice with signalling NaNs? Well, they have limited use, but it turned out that in practice the hardware handling of signalling NaNs gums up high speed computation. Hardware makes use of parallel execution, and of speculative execution, and of out-of-order execution, but when a signalling NaN is encounted, the state needs to be unwound and reconstructed as if the instructions had been done in serial in strict sequential order. Just keeping track of everything is non-trivial overhead that slows down computation whenver the processor bits are set to permit Signalling NaN to interrupt.

Because of this, there are some very specialized programmes where it is important to detect the exception ASAP that might use signalling NaN -- but that it turns out that for the great majority of programs, that inserting a quiet NaN and doing fix-ups later is fine.

Michael 2023-12-12

@Torsten, in the case of my application, the NaNs originally arise because a NaN array is assigned using NaN(). Then the array is partially filled in with values where values should be. It is an intentional part of the code design. Thanks for your note of caution and for commenting on the question.

@Walter Roberson thanks for the historical background, this is interesting to know!

Torsten 2023-12-12

in the case of my application, the NaNs originally arise because a NaN array is assigned using NaN(). Then the array is partially filled in with values where values should be. It is an intentional part of the code design.

If you have "control" over your NaN values, I apologize for my provocative comments.

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Answer 3

Steven Lord 2023-12-12

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If the second output from min in the case where the input is all NaN values were NaN, every single call to min that wanted to use that second output as an index into the input would have to guard themselves against the all-NaN input case using isnan. With the current behavior, that second output is always* usable as an index.

x = [NaN, NaN]
x = 1×2
   NaN   NaN
[minvalue, minindex] = min(x)
minvalue = NaN
minindex = 1
x(minindex) % Current behavior
ans = NaN
x(NaN) % Your proposed behavior
Array indices must be positive integers or logical values.

Why is minindex equal to 1? Well, since all the elements are the same 1 is as good as any other index. And there are other places in MATLAB where we default to the first dimension / element / etc. (functions that accept a dim argument and get passed a scalar and no dimension as input, for example.)

* There may be a case where it's not, involving very tall sparse matrices and linear indices. But I don't remember off the top of my head what that does; I'd have to double-check. That might just throw an error.

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Steven Lord 2023-12-12

I was speaking colloquially. I should have been more precise. All the elements are NaN. There's no inherent reason based on the value of the NaN elements to favor one over another. We could have chosen to return the index of the first element, the last element, the middle element, or any of the elements in the array. I suspect that it was Cleve's decision to keep it simple and just use the first element.

Matt J 2023-12-12

Why is minindex equal to 1? Well, since all the elements are the same 1 is as good as any other index. And there are other places in MATLAB where we default to the first dimension

It's also the most efficient choice. Why update a register when you don't have to?

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Why does argmin index of NaN array have a value of 1 and not NaN?

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