# Any idea why all([]) is true while any([]) is false

2 visualizzazioni (ultimi 30 giorni)
Khaled Hamed il 29 Lug 2012
>> all([])
ans =
1
>> any([])
ans =
0
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Ryan il 29 Lug 2012
It's written into the documentation as such, but no explanation is given.
Khaled Hamed il 29 Lug 2012
I have just noticed the same with Nan! more confusing
>> all(nan)
ans =
1
>> any(nan)
ans =
0

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### Risposta accettata

Daniel Shub il 29 Lug 2012
This post by Loren lead to some comments that address the issue, especially the one by Matt Fig.
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### Più risposte (1)

the cyclist il 29 Lug 2012
I can't say I know definitively, but I expect that one reason is for consistency when taking the union of sets with the empty set. For example, one would want
all(union(true,[]))
to be true, and also
any(union(false,[]))
to be false. The definitions in your question make sense in that context.
##### 1 CommentoMostra -1 commenti meno recentiNascondi -1 commenti meno recenti
Khaled Hamed il 29 Lug 2012
I have just noticed the same with NaN! more confusing
>> all(nan)
ans =
1
>> any(nan)
ans =
0

Accedi per commentare.

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