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13 events
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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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| Jul 30, 2013 at 15:46 | history | edited | Mr.Wizard | CC BY-SA 3.0 |
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| May 15, 2012 at 16:25 | comment | added | rcollyer |
Alongside Internal`PartitionRagged we need Developer`ParitionMapRagged. It would be very useful.
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| May 8, 2012 at 22:06 | vote | accept | John | ||
| May 8, 2012 at 20:13 | history | edited | Mr.Wizard | CC BY-SA 3.0 |
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| May 8, 2012 at 20:06 | comment | added | Mr.Wizard | @Oleksandr thank you! I should take better notes. | |
| May 8, 2012 at 18:15 | comment | added | Oleksandr R. |
@Mr.Wizard you're probably thinking of Internal`PartitionRagged, which for version 8 is indeed equivalent to your dynP. Actually I've never had cause to use it personally, but Andy Ross mentioned it here.
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| May 8, 2012 at 5:34 | comment | added | Mr.Wizard |
@rcollyer as I recall someone (Oleksandr?) posted a version-8 Internal or Developer function that does what my dynP function does. Do remember what it is? Perhaps we can use that instead and just leave a link to the Toolbag post for prior version users.
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| May 8, 2012 at 3:06 | comment | added | rcollyer | Actually, it would be preferable if the explanation were here as the ToolBag has been closed, and is on SO. That said, when I replied, I did not notice the link. Sorry. :) | |
| May 8, 2012 at 2:55 | comment | added | rcollyer |
Better, but I was trying to understand how dynP works. From what I read Accumulate@p creates a list of partition points, specifically the end points of the sublists. {0}~Join~Most@# shifts the list of points to the right, adding 1 to which creates the start points. Then it is just a matter of supplying those values to Part.
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| May 8, 2012 at 2:33 | history | edited | Mr.Wizard | CC BY-SA 3.0 |
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| May 8, 2012 at 0:33 | comment | added | rcollyer | You have my +1, but could you go into some detail about the individual steps involved, as I have no idea how you came up with that function. I see how it works, but I think additional exposition would help myself, and the newbies, understand it better. | |
| May 7, 2012 at 23:47 | history | answered | Mr.Wizard | CC BY-SA 3.0 |