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

2 visualizzazioni (ultimi 30 giorni)
>> all([])
ans =
1
>> any([])
ans =
0
  2 Commenti
Ryan
Ryan il 29 Lug 2012
It's written into the documentation as such, but no explanation is given.
Khaled Hamed
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.

Risposta accettata

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

Più risposte (1)

the cyclist
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 Commento
Khaled Hamed
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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