Timeline for How can one write a robust ListableQ function?
Current License: CC BY-SA 3.0
16 events
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| Oct 30, 2020 at 0:23 | comment | added | Mr.Wizard | @NamNguyen That is what I meant by "In the case of Symbols this is merely a matter of checking Attributes." | |
| Oct 28, 2020 at 18:27 | comment | added | Nam Nguyen |
How about this? :) Just check the FullForm function in this list Select[Names["*"], MemberQ[Attributes[#], Listable] &]
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| Apr 13, 2017 at 12:55 | history | edited | CommunityBot |
replaced http://mathematica.stackexchange.com/ with https://mathematica.stackexchange.com/
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| Oct 31, 2013 at 21:48 | comment | added | Mr.Wizard | @Rojo I was feeling burned out. I decided to take a break before I started to dislike spending time here. I still stop by at least daily to check for fires (flags). | |
| Oct 31, 2013 at 21:46 | comment | added | Rojo | Oh, I hadn't noticed. Why is that? | |
| Oct 31, 2013 at 21:39 | comment | added | Mr.Wizard | @Rojo I didn't (and don't) have a strict definition. "... and moreover, that the question itself may be hard to answer in a complete way." What are you thinking? I should revisit this some time but as you may have noticed I'm taking a break from Mathematica. | |
| Oct 31, 2013 at 15:43 | comment | added | Rojo | Do you have a strict definition of what you call inherent listability, or you are looking for something that more or less works in the most common cases? | |
| Sep 3, 2013 at 13:25 | comment | added | Mr.Wizard | @rcollyer I know that feeling. Too well. | |
| Sep 3, 2013 at 12:45 | comment | added | rcollyer | @Mr.Wizard initially it was very clear to me when I wrote, not so much now, especially with ssch's comment. I need to see if I can recover what I intended, but my mental tape drive is faulty. :P | |
| Sep 1, 2013 at 22:38 | comment | added | Mr.Wizard | @rcollyer I don't see the problem with your example. Perhaps you have another one? | |
| Sep 1, 2013 at 15:25 | comment | added | ssch |
@rcollyer / is a listable binary operator so it works out fine since both arguments will have the same shape.
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| Sep 1, 2013 at 15:16 | history | tweeted | twitter.com/#!/StackMma/status/374189047717322752 | ||
| Sep 1, 2013 at 14:34 | comment | added | rcollyer |
What if your list is {3, 4, {1,2}}? Then, listable f, gives {f[3], f[4], {f[1], f[2]}}. So, while (3 - #)/(7 #)& is inherently listable, I don't think it is completely listable. So, ListableQ will have some limitations.
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| Sep 1, 2013 at 13:49 | answer | added | ssch | timeline score: 2 | |
| S Sep 1, 2013 at 13:31 | answer | added | Mr.Wizard | timeline score: 6 | |
| S Sep 1, 2013 at 13:31 | history | asked | Mr.Wizard | CC BY-SA 3.0 |