Timeline for Calculate Cumulative Mean and Standard Deviation of a list without FOR loops [duplicate]
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
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| Apr 11, 2016 at 13:55 | vote | accept | Minh N | ||
| Apr 11, 2016 at 13:32 | history | closed |
Kuba xzczd♦ xyz Martin Ender MarcoB |
Duplicate of Any built-in function to generate successive sublists from a list? | |
| Apr 11, 2016 at 12:38 | answer | added | kirma | timeline score: 3 | |
| Apr 11, 2016 at 11:56 | comment | added | Minh N | @kirma Yup It's very convenient. I'd say I'll stick to this one. | |
| Apr 11, 2016 at 11:51 | comment | added | Minh N | @Kuba: I think it is safe to say that. | |
| Apr 11, 2016 at 11:49 | comment | added | Minh N | @MartinBüttner I would say it is still acceptable, at least for me. :-) | |
| Apr 11, 2016 at 11:34 | comment | added | kirma |
In the lines of @TomD, you could define a helper function for this: ClearAll[MapAccumulate]; MapAccumulate[f_, list_List] := f@Take[list, #] & /@ Range@Length@list; ... MapAccumulate[f, mylist]
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| Apr 11, 2016 at 11:09 | review | Close votes | |||
| Apr 11, 2016 at 13:32 | |||||
| Apr 11, 2016 at 11:05 | answer | added | wolfies | timeline score: 2 | |
| Apr 11, 2016 at 10:54 | comment | added | user1066 |
mylist = {x1, x2, x3, xN}. f@Take[mylist, #] & /@ Range[4] or Mean@Take[mylist, #] & /@ Range[4], maybe
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| Apr 11, 2016 at 10:23 | comment | added | Martin Ender |
Would using Table be an acceptable solution or would that count as a simple loop? While Mathematica has Accumulate which performs the operation you look for for Plus, I don't think there's a way to do a generalised accumulation with an arbitrary function like Mean. Neither is there a built-in to get all the prefixes of a list, so I guess using Table or Array might be the best you can get. (E.g. Table[Mean@myList[[1 ;; i]], {i, Length@myList}])
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| Apr 11, 2016 at 10:16 | history | asked | Minh N | CC BY-SA 3.0 |