[R] merging-binning data

Boris Steipe boris.steipe at utoronto.ca
Wed Nov 4 14:09:48 CET 2015


I don't understand: 
 - where does the "label" come from? (It's not an element of your data that I see.)
 - what do you want to do with this "label" i.e. how does it need to be associated with the data?


B.



On Nov 4, 2015, at 7:57 AM, Alaios <alaios at yahoo.com> wrote:

> Thanks it works great and gives me group numbers as integers and thus I can with which group the elements as needed (which (groups== 2))
> 
> Question though is how to keep also the labels for each group. For example that my first group is the [13,206)
> 
> Regards
> Alex
> 
> 
> 
> On Wednesday, November 4, 2015 1:00 PM, Boris Steipe <boris.steipe at utoronto.ca> wrote:
> 
> 
> I would transform the original numbers into integers which you can use as group labels. The row numbers of the group labels are the indexes of your values.
> 
> Example: assume your input vector is dBin
> 
> nGroups <- 5  # number of groups
> groups <- (dBin - min(dBin)) / (max(dBin) - min(dBin)) # rescale to the range [0,1]
> groups <- floor(groups * nGroups) + 1  # discretize to nGroups integers
> 
> Now you can eg. get the indices for group 2
> 
> groups[groups == 2]
> 
> Depending on the nature of your input data, it may be better to keep these groups in a column adjacent to your values, rather than in a separate vector, or even better to just calculate the groups on the fly in your downstream analysis with the approach given above in a function, rather than storing them at all. These are simple operations that should not add perceptibly to execution time.
> 
> Cheers,
> Boris
> 
> 
> 
> 
> 
> 
> On Nov 4, 2015, at 6:40 AM, Alaios via R-help <r-help at r-project.org> wrote:
> 
> > Thanks for the answer. Split does not give me the indexes though but only in which group they fall in. I also need the index of the group. Is the first, the second .. group?Alex
> > 
> > 
> > 
> >    On Tuesday, November 3, 2015 5:05 PM, Ista Zahn <istazahn at gmail.com> wrote:
> > 
> > 
> > Probably
> > 
> > split(binDistance, test).
> > 
> > Best,
> > Ista
> > 
> > On Tue, Nov 3, 2015 at 10:47 AM, Alaios via R-help <r-help at r-project.org> wrote:
> >> Dear all,I am not exactly sure on what is the proper name of what I am trying to do.
> >> I have a vector that looks like
> >>  binDistance
> >>            [,1]
> >>  [1,] 238.95162
> >>  [2,] 143.08590
> >>  [3,]  88.50923
> >>  [4,] 177.67884
> >>  [5,] 277.54116
> >>  [6,] 342.94689
> >>  [7,] 241.60905
> >>  [8,] 177.81969
> >>  [9,] 211.25559
> >> [10,] 279.72702
> >> [11,] 381.95738
> >> [12,] 483.76363
> >> [13,] 480.98841
> >> [14,] 369.75241
> >> [15,] 267.73650
> >> [16,] 138.55959
> >> [17,] 137.93181
> >> [18,] 184.75200
> >> [19,] 254.64359
> >> [20,] 328.87785
> >> [21,] 273.15577
> >> [22,] 252.52830
> >> [23,] 252.52830
> >> [24,] 252.52830
> >> [25,] 262.20084
> >> [26,] 314.93064
> >> [27,] 366.02996
> >> [28,] 442.77467
> >> [29,] 521.20323
> >> [30,] 465.33071
> >> [31,] 366.60582
> >> [32,]  13.69540
> >> so numbers that start from 13 and go up to maximum 522 (I have also many other similar sets).I want to put these numbers into 5 categories and thus I have tried cut
> >> 
> >> 
> >> Browse[2]> test<-cut(binDistance,seq(min(binDistance)-0.00001,max(binDistance),length.out=scaleLength+1))
> >> Browse[2]> test
> >>  [1] (217,318]  (115,217]  (13.7,115] (115,217]  (217,318]  (318,420]
> >>  [7] (217,318]  (115,217]  (115,217]  (217,318]  (318,420]  (420,521]
> >> [13] (420,521]  (318,420]  (217,318]  (115,217]  (115,217]  (115,217]
> >> [19] (217,318]  (318,420]  (217,318]  (217,318]  (217,318]  (217,318]
> >> [25] (217,318]  (217,318]  (318,420]  (420,521]  (420,521]  (420,521]
> >> [31] (318,420]  (13.7,115]
> >> Levels: (13.7,115] (115,217] (217,318] (318,420] (420,521]
> >> 
> >> 
> >> I want then for the numbers of my initial vector that fall within the same "category" lets say the (318,420] to be collected on a vector.I rephrase it the indexes of my initial vector that have a value between 318 to 420 to be put in a same vector that I can process then as I want.
> >> How I can do that effectively in R?
> >> I would like to thank you for your replyRegardsAlex
> >> 
> >>        [[alternative HTML version deleted]]
> >> 
> >> ______________________________________________
> >> R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see
> >> https://stat.ethz.ch/mailman/listinfo/r-help
> >> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> >> and provide commented, minimal, self-contained, reproducible code.
> > 
> > 
> >     [[alternative HTML version deleted]]
> > 
> > ______________________________________________
> > R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see
> > https://stat.ethz.ch/mailman/listinfo/r-help
> > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> > and provide commented, minimal, self-contained, reproducible code.
> 
> 



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