Timeline for How to figure out what numbers often appear together in a dataset?
Current License: CC BY-SA 3.0
8 events
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| Apr 13, 2017 at 12:44 | history | edited | CommunityBot |
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| Nov 14, 2013 at 18:55 | vote | accept | daniellopez46 | ||
| Nov 14, 2013 at 18:55 | comment | added | daniellopez46 | np. I was actually just inspecting y as you were writing your answer and thinking "I was not expecting to see 15 columns for m<-3..Should be choose(5,3) =10". But your edit fixed that. This is way cool! Thank you for very much! | |
| Nov 14, 2013 at 18:41 | comment | added | whuber♦ |
Sorry: although I checked that, I misread the output and did not catch the discrepancy. There was a typo in the code where "2" appeared but "m" should have (and R issued no warnings about a mismatch between the sizes of arrays x0 and y; it merely "recycled" the elements). I fixed that typo and pasted the new output. You might find it informative (and reassuring) to run the code and inspect the array y, which contains all the $m$-tuples for each row in x. As another check, sum(z) should equal $\binom{p}{m}$ times the number of rows in $x$: $\binom{5}{3}\times 6=60$ in the example.
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| Nov 14, 2013 at 18:38 | history | edited | whuber♦ | CC BY-SA 3.0 |
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| Nov 14, 2013 at 18:31 | comment | added | daniellopez46 | @ whuber. Thanks for your posted Answer. Why is the output of z showing a count of 5 for (3,5,10)? Based on the original matrix I was expecting to see a count of 3. This would come from rows x[c(1:2,6),]. If I enter m<-5 I would expect to see in the output of z (3,5,8,10,15) with a count of 2 and and 5 additional groups each with a count of 1. | |
| Nov 14, 2013 at 16:41 | history | edited | whuber♦ | CC BY-SA 3.0 |
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| Nov 14, 2013 at 15:44 | history | answered | whuber♦ | CC BY-SA 3.0 |