Timeline for How to efficiently replace the repetitive sequence?
Current License: CC BY-SA 4.0
19 events
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| Apr 2, 2021 at 12:40 | vote | accept | tjx6499 | ||
| Apr 2, 2021 at 12:40 | comment | added | tjx6499 | Thank you very much @kglr. | |
| Apr 1, 2021 at 16:36 | history | edited | kglr | CC BY-SA 4.0 |
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| Apr 1, 2021 at 16:21 | comment | added | kglr | @tjx6499, please see the update. | |
| Apr 1, 2021 at 16:16 | history | edited | kglr | CC BY-SA 4.0 |
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| Apr 1, 2021 at 15:13 | comment | added | kglr | Thank you @tjx6499. | |
| Apr 1, 2021 at 15:01 | comment | added | tjx6499 | @kglr, I drew sth to add the explanation on the repetitive cases for "1", pls see my update in the question. | |
| Apr 1, 2021 at 14:22 | comment | added | kglr |
@tjx6499, I am afraid i don't understand _"The one underneath the other number will be much later" and how exactly you got {{8, 2}, {9, 6}, {5, 7}, {3}, {10, 4, 1}}. If it is not too much trouble, can you add an explanation of the steps you followed to get from {{2, 1}, {2, 2}, {1, 2}, {1}, {2, 1, 1}} to {{8, 2}, {9, 6}, {5, 7}, {3}, {10, 4, 1}}?
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| Apr 1, 2021 at 14:06 | comment | added | tjx6499 | @kglr, it is possible to get the same desired result with rules that I stated in my question={{8, 2}, {9, 6}, {5, 7}, {3}, {10, 4, 1}}? | |
| Apr 1, 2021 at 13:40 | vote | accept | tjx6499 | ||
| Apr 1, 2021 at 14:01 | |||||
| Apr 1, 2021 at 13:35 | history | edited | kglr | CC BY-SA 4.0 |
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| Apr 1, 2021 at 12:11 | comment | added | tjx6499 | @kglr, how can I get the outcome in list form={{8,2},{9,6},{4,7},{5},{10,3,1}}? | |
| Mar 31, 2021 at 12:54 | history | edited | kglr | CC BY-SA 4.0 |
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| Mar 31, 2021 at 12:44 | history | edited | kglr | CC BY-SA 4.0 |
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| Mar 31, 2021 at 11:33 | comment | added | kglr | @tjx6499, the method should work for any number of repetitions. If you have an example where the method fails, please update your question with the example input and desired output. | |
| Mar 31, 2021 at 11:13 | comment | added | tjx6499 | How can I efficiently assign the sequence for repeated cases(like randomly assigned or specific method), for example when there are 4 or even more number of no. 1 in the list, so that when I retrieve each number according to the sequences will be more effective by reducing the blocks on each targeted no. in tableform-transpose. | |
| Mar 16, 2021 at 20:51 | comment | added | Somos |
If your version of Mathematica does not have TakeList then try TakeList = If[#2 == {}, #1, Prepend[#0[Drop[#1, First@#2], Rest@#2], Take[#1, First@#2]]] &;
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| Mar 16, 2021 at 14:10 | review | Low quality posts | |||
| Mar 16, 2021 at 14:15 | |||||
| Mar 16, 2021 at 13:55 | history | answered | kglr | CC BY-SA 4.0 |