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| From | Wolfgang Rohdewald <wolfgang@rohdewald.de> |
| To | python-list@python.org |
| Subject | Re: float("nan") in set or as key |
| Date | Sun, 29 May 2011 10:27:14 +0200 |
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On Sonntag 29 Mai 2011, Tim Delaney wrote: > There's a second part the mystery - sets and dictionaries (and > I think lists) assume that identify implies equality (hence > the second result). This was recently discussed on > python-dev, and the decision was to leave things as-is. On Sonntag 29 Mai 2011, Grant Edwards wrote: > Even if they are the same nan, it's still not equal to itself. if I understand this thread correctly, they are not equal to itself as specified by IEEE but Python treats them equal in sets and dictionaries for performance reasons -- Wolfgang
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Re: float("nan") in set or as key Albert Hopkins <marduk@letterboxes.org> - 2011-05-28 20:28 -0400
Re: float("nan") in set or as key Erik Max Francis <max@alcyone.com> - 2011-05-28 17:44 -0700
Re: float("nan") in set or as key Grant Edwards <invalid@invalid.invalid> - 2011-05-29 02:25 +0000
Re: float("nan") in set or as key Wolfgang Rohdewald <wolfgang@rohdewald.de> - 2011-05-29 10:27 +0200
Re: float("nan") in set or as key Grant Edwards <invalid@invalid.invalid> - 2011-05-29 14:41 +0000
Re: float("nan") in set or as key MRAB <python@mrabarnett.plus.com> - 2011-05-29 18:44 +0100
Re: float("nan") in set or as key Steven D'Aprano <steve+comp.lang.python@pearwood.info> - 2011-05-29 18:27 +0000
Re: float("nan") in set or as key Chris Angelico <rosuav@gmail.com> - 2011-05-30 03:50 +1000
Re: float("nan") in set or as key Christian Heimes <lists@cheimes.de> - 2011-05-29 20:05 +0200
Re: float("nan") in set or as key Steven D'Aprano <steve+comp.lang.python@pearwood.info> - 2011-05-29 18:46 +0000
Re: float("nan") in set or as key Dennis Lee Bieber <wlfraed@ix.netcom.com> - 2011-05-29 20:36 -0700
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