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

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

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Akzeptierte Antwort

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

Weitere Antworten (1)

the cyclist
the cyclist am 29 Jul. 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 Kommentar
Khaled Hamed
Khaled Hamed am 29 Jul. 2012
I have just noticed the same with NaN! more confusing
>> all(nan)
ans =
1
>> any(nan)
ans =
0

Melden Sie sich an, um zu kommentieren.

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