Timeline for How to make lists the same size?
Current License: CC BY-SA 3.0
13 events
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| Mar 22, 2018 at 7:18 | history | edited | kglr | CC BY-SA 3.0 |
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| Mar 21, 2018 at 20:52 | comment | added | Mlo27 | Perfect thanks. I'll give that a try tomorrow morning :D | |
| Mar 21, 2018 at 20:23 | comment | added | kglr |
@Mlo27, "exact code" to get the average of 19 realizations is Mean[RESULTS[[All, ;; Min[Length /@RESULTS]]]. If you also need the totals you can use Total[RESULTS[[All, ;; Min[Length /@RESULTS]]].
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| Mar 21, 2018 at 20:19 | comment | added | Mlo27 | Thanks for getting back to me. I'm still slightly confused. Can I not just trim my results using "trimmedlists = lists[[All, ;; Min[Length /@ lists]]];". And then define "nthrealisation[n_]:=trimmedresults[[n]]". Then I can do "Table[nthrealisation[n],{n,19}]" and then add each table together and then divide through? If not, is it possible for you to provide the exact coding I need to use in order to find the average of the 19 realisation? Thanks in advance @kglr | |
| Mar 21, 2018 at 19:40 | comment | added | kglr | @Mlo27, please see the update. | |
| Mar 21, 2018 at 19:37 | history | edited | kglr | CC BY-SA 3.0 |
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| Mar 21, 2018 at 18:03 | comment | added | Mlo27 | I have attached my coding for you to see. @kglr | |
| Mar 21, 2018 at 17:42 | comment | added | Mlo27 | Still doesn't seem to work. May I add that my lists "nthrealisation[n]" contains approx 6000 elements. | |
| Mar 21, 2018 at 17:41 | history | edited | kglr | CC BY-SA 3.0 |
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| Mar 21, 2018 at 17:35 | comment | added | kglr |
@Mlo27, welcome to mma.se. You can replace lists with your RESULTS.
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| Mar 21, 2018 at 17:33 | history | edited | kglr | CC BY-SA 3.0 |
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| Mar 21, 2018 at 17:29 | comment | added | Mlo27 | I have a very basic understanding of Mathematica, so am unsure of how to implement this into my coding | |
| Mar 21, 2018 at 17:27 | history | answered | kglr | CC BY-SA 3.0 |