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Re: sumskipnan(nan) = 0 ?
From: |
gail |
Subject: |
Re: sumskipnan(nan) = 0 ? |
Date: |
Sun, 09 Sep 2007 21:04:58 +0200 |
On Fri Sep 7 21:43 , Thomas Shores sent:
>> >> sumskipnan(nan) = 0 ?
>> >
>> >I'm guessing that the sum of the input with all the NaN's removed
>> > is zero. That is the sum of no numbers is considered 0.
>>
>> That seems logical. Take the set of arguments, remove NaNs, get
>> an empty set. I think that the sum of an empty set of numbers
>> should be 0. Why should it be different?
What is so special with 0 here? - Why, if you decide to choose a number, not 1
or
any other number?
>
>Actually, it's not only logical, it's the only choice. Look at it
>this way: the sum of elements over the union of disjoint index sets
>A and B should be the sum of the sum over indices in A plus the sum
>over indices in B. Now if B is the empty set, the union of the sets
>is just A, and A and B are disjoint. Hence the sum over A union B
>should be the sum over A. Hence the sum over B must be zero.
sum(A) + sum(empty) = sum(A) + nan = sum(A). No?
To me, the only reasonable choice for the sum of nothing is nothing, that is:
sum(nan) = nan.
G.