[R] [r] How to pick colums from a ragged array?

Rui Barradas ruipbarradas at sapo.pt
Tue Oct 23 11:59:53 CEST 2012


I'm not sure I understand it well, in the solution below the only 
returned value is ID == 814 but it's not the first nor the last DATE.

how.many <- ave(id.d[,1], id.d[,1], id.d[,2], FUN = length)
id.d[how.many > 1, ]

See the help page for ?ave if the repetition of id.d[,1] is confusing. 
The first is the vector to average (to apply FUN to) and the second is 
one of thw two vectors defining the groups.

Hope this helps,

Rui Barradas
Em 23-10-2012 10:37, Stuart Leask escreveu:
> I have a large dataset (~1 million rows) of three variables: ID (patient's name), DATE (of appointment) and DIAGNOSIS (given on that date).
> Patients may have been assigned more than one diagnosis at any one appointment - leading to two rows, same ID and DATE but different DIAGNOSIS.
> The diagnoses may change between appointments.
> I want to subset the data in two ways:
> -          define groups of patients by the first diagnosis given
> -          define groups of patients by the last diagnosis given.
> The problem:
> Unfortunately, a small number of patients have been given more than one diagnosis at their first (or last) appointment. These individuals I need to identify and remove, as it's not possible to say uniquely what their first (or last) diagnosis was. So I need to identify and remove these individuals which have pairs of rows with the same ID and (lowest or highest) DATE. The size of the dataset precludes the option of doing this by eye.
> I suspect there is a very elegant way of doing this in R.
> This is what I've come up with:
> -          Sort by DATE then ID
> -          Make a ragged array of DATE by ID
> -          Remove IDs that only occur once.
> -          Subtract the first and second DATEs. Remove IDs for which this = zero, as this will only be true for IDs for which the appointment is recorded twice (because there were two diagnoses recorded on this date).
> -          (Then do the same to get the 'last appointment' duplicates, by reversing the initial sort by DATE.)
> I am stuck at the 'Subtract dates' step: I would like to get the data out of the ragged array by columns (so e.g. I end up with a matrix of ID, 1st DATE, 2nd DATE). But I can't get the dates out by column from the ragged array.
> I hope someone can help. My ugly code is below, with some data for testing.
> Stuart
> Dr Stuart John Leask DM FRCPsych MB BChir MA
> Clinical Senior Lecturer and Honorary Consultant Pychiatrist
> Institute of Mental Health, Innovation Park
> Triumph Road, Nottingham, Notts. NG7 2TU. UK
> Tel. +44 115 82 30419 stuart.leask at nottingham.ac.uk<mailto:stuart.leask at nottingham.ac.uk>
> Google 'Dr Stuart Leask'
> ID <- c(58,58,58,58,167,167,323,323,323,323,323,323,323
> ,547,794,814,814,814,814,814,814,841,841,841,841,841
> ,841,841,841,841,910,910,910,910,910,910,999,1019,1019
> ,1019)
> DATE <- c(20060821,20061207,20080102,20090904,20040205,20040323,20051111
> ,20060111,20071119,20080107,20080407,20080521,20080711,20041005
> ,20070905,20020814,20021125,20040429,20040429,20071205,20080227
> ,20050421,20060130,20060428,20060602,20060816,20061025,20061129
> ,20070112,20070514,20091105,20091117,20091119,20091120,20091210
> ,20091224,20050503,19870508,19880223,19880330)
> id.d <- cbind (ID,DATE )
> rag.a  <-  split ( id.d [ ,2 ], id.d [ ,1])               # create ragged array, 1-n DATES for every NAME
> # Inelegant attempt to remove IDs that only have one entry:
> rag.s <-tapply  (id.d [ ,2], id.d [ ,1], sum)             #add up the dates per row
> # Since DATE is in 'year mo da', if there's only one date, sum will be less than 2100000:
> rag.t <- rag.s [ rag.s > 21000000 ]
> multi.dates <- rownames ( rag.t )                         # all the IDs with >1 date
> rag.am <- rag.a [ multi.dates ]                           # rag.am only has IDs with > 1 Date
> # But now I'm stuck.
> # Each row of the array is rag.am$ID.
> # So I can't pick columns of DATEs from the ragged array.
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