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

Fri Oct 26 11:11:05 CEST 2012

```Close - but it's evaluating on 'first date' AND 'last date' - I'll be considering groups defined by 'first diagnosis' and groups defined by 'last diagnosis' completely separately, so I need it to run considering the first date (to produce e.g. INCLUDE.FIRST), then on a separate run to consider the last date (to produce e.g. INCLUDE.LAST).

-----Original Message-----
From: arun [mailto:smartpink111 at yahoo.com]
Sent: 25 October 2012 12:32
Cc: R help; Petr PIKAL
Subject: Re: [r] How to pick colums from a ragged array?

Hi Stuart,

So, I guess my result (below) serves the purpose!
A.K.

----- Original Message -----
To: arun <smartpink111 at yahoo.com>
Cc:
Sent: Thursday, October 25, 2012 3:13 AM
Subject: RE: [r] How to pick colums from a ragged array?

Even confusing myself now, serves me right for replying late at night!

** If DGs are the same, then the first (or last) diagnosis is unambiguous even if date is duplicated - so I can use the data.**

Consider we want INCLUDE.FIRST to look at first dates.
Duplicate dates: 167, 323,814, 841, 910 1019 AND This dup is the first date: 167, 841, 1019 AND This dup has different DGs: 841 1019 = give all rows of 841 and 1019  FALSE.
(All other rows TRUE)

Now consider we want INCLUDE.LAST to look at last dates.
Duplicate dates: 167, 323,814, 841, 910 1019 AND This dup is the last date: 167, 323, 814 AND This dup has different DGs: 323 = give all rows of 323 FALSE.
(All others TRUE)

Of course, I'm happy to run a function twice, either one with a 'first/last' switch, or one that assumes initial order of sort by DATE determines whether you end up with first or last date duplicates.

-----Original Message-----
From: arun [mailto:smartpink111 at yahoo.com]
Sent: 24 October 2012 22:59
Subject: Re: [r] How to pick colums from a ragged array?

Hi Stuart,
So, 167 should be FALSE eventhough DG is same because it comes under earliest/first date, but TRUE for 814 because it comes under latest/last date.  167 comes under both cases.
Let me try to make sense of that:

I am just pasting my earlier solution and its results again to see whether we are on the same page:
res2<-id.d[id.d[,1]%in%names(res1[res1\$flag==TRUE,])&(duplicated(id.d[,1:2])|duplicated(id.d[,1:2],fromLast=TRUE)),]
res3<-res2[!res2\$ID%in% res2[duplicated(res2)|duplicated(res2,fromLast=TRUE),]\$ID,]
id.d1<-id.d
res4\$INCLUDE[is.na(res4\$INCLUDE)]<-TRUE
res4
ID     DATE DG INCLUDE
1    58 20060821  1    TRUE
2    58 20061207  2    TRUE
3    58 20080102  1    TRUE
4    58 20090904  1    TRUE
5   167 20040205  4    TRUE
6   167 20040205  4    TRUE
7   323 20051111  3   FALSE
8   323 20060111  2   FALSE
9   323 20071119  3   FALSE
10  323 20080107  2   FALSE
11  323 20080407  1   FALSE
12  323 20080521  2   FALSE
13  323 20080521  3   FALSE
14  547 20041005  2    TRUE
15  794 20070905  1    TRUE
16  814 20020814  2    TRUE
17  814 20021125  2    TRUE
18  814 20040429  2    TRUE
19  814 20040429  2    TRUE
20  814 20071205  2    TRUE
21  814 20071205  2    TRUE
22  841 20050421  1   FALSE
23  841 20050421  2   FALSE
24  841 20060428  1   FALSE
25  841 20060602  1   FALSE
26  841 20060816  1   FALSE
27  841 20061025  1   FALSE
28  841 20061129  1   FALSE
29  841 20070112  1   FALSE
30  841 20070514  4   FALSE
31  910 19870508  3    TRUE
32  910 20040205  3    TRUE
33  910 20040205  3    TRUE
34  910 20080521  3    TRUE
35  910 20080521  4    TRUE
36  910 20091224  2    TRUE
37  999 20050503  2    TRUE
38 1019 19870508  1   FALSE
39 1019 19870508  2   FALSE
40 1019 19880330  1   FALSE
A.K.

----- Original Message -----
Cc: R help <r-help at r-project.org>
Sent: Wednesday, October 24, 2012 5:40 PM
Subject: RE: [r] How to pick colums from a ragged array?

I mis-typed, missing an if. I think you've got it, but let me try again:

"The function should:
-  put FALSE in a column for every instance of an ID IF ( that ID has a first (or last) DATE duplicated ) AND IF (the DGs for the duplicated dates are different)."

So for the earliest/first date function, INCLUDE should be TRUE, apart from FALSE for _all_ the instances of IDs 167, 841 and 1019 For the latest/last date function, INCLUDE should be TRUE, apart from FALSE for all the instances of ID  323.

Stuart

-----Original Message-----
From: arun [mailto:smartpink111 at yahoo.com]
Sent: 24 October 2012 21:30
Subject: Re: [r] How to pick colums from a ragged array?

Hi,

According to the OP "So the function should only exclude an ID, having identified a first (or last) DATE duplicate, the DGs for these two dates are different."
Rui:
By running your modified function (using dte <- tapply(x[,2], x[,1], FUN = function(x) duplicated(fun(x, 2),fromLast = TRUE))),

id.d\$INCLUDE <- !(rm1 | rm2)
#     ID     DATE DG INCLUDE
#1    58 20060821  1    TRUE
#2    58 20061207  2    TRUE
#3    58 20080102  1    TRUE
#4    58 20090904  1    TRUE
#5   167 20040205  4   FALSE
#6   167 20040205  4   FALSE

For #167, DGs are same.  Not sure whether to exclude it or not.

My modified solution is similar but I am excluding 167 and 814.

fun1<-function(dat){
res1last<- data.frame(flag=tapply(dat[,2],dat[,1],FUN=function(x) tail(duplicated(x)|duplicated(x,fromLast=TRUE),1)))
res2first<-dat[dat[,1]%in%names(res1first[res1first\$flag==TRUE,])&(duplicated(dat[,1:2])|duplicated(dat[,1:2],fromLast=TRUE)),]
res2last<-dat[dat[,1]%in%names(res1last[res1last\$flag==TRUE,])&(duplicated(dat[,1:2])|duplicated(dat[,1:2],fromLast=TRUE)),]
res3first<-res2first[!res2first\$ID%in% res2first[duplicated(res2first)|duplicated(res2first,fromLast=TRUE),]\$ID,]
res3last<-res2last[!res2last\$ID%in% res2last[duplicated(res2last)|duplicated(res2last,fromLast=TRUE),]\$ID,]
res3firstsubset\$INCLUDE<-FALSE
res3lastsubset<-do.call(rbind,lapply(split(res3last,res3last\$ID),tail,1))
res3lastsubset\$INCLUDE<-FALSE
res4<-merge(dat,merge(res3first,merge(res3firstsubset,merge(res3lastsubset,res3last,all=TRUE),all=TRUE),all=TRUE),all=TRUE)
res4\$INCLUDE[is.na(res4\$INCLUDE)]<-TRUE
res4
}

tail(fun1(id.d))
#     ID     DATE DG INCLUDE
#35  910 20080521  4    TRUE
#36  910 20091224  2    TRUE
#37  999 20050503  2    TRUE
#38 1019 19870508  1    TRUE
#39 1019 19870508  2   FALSE
#40 1019 19880330  1    TRUE

A.K.

----- Original Message -----
To: arun <smartpink111 at yahoo.com>
Sent: Wednesday, October 24, 2012 2:50 PM
Subject: Re: [r] How to pick colums from a ragged array?

Hello,

Inline.
Em 24-10-2012 19:05, arun escreveu:
> Hi Rui,
>
> I think now our results are matching except in the INCLUDE column
>
> id.d[c(11:13,22:24,38:40),]
> #     ID     DATE DG INCLUDE
> #11  323 20080407  1    TRUE
> #12  323 20080521  2   FALSE
> #13  323 20080521  3    TRUE
> #22  841 20050421  1    TRUE
> #23  841 20050421  2   FALSE
> #24  841 20060428  1    TRUE
> #38 1019 19870508  2    TRUE
> #39 1019 19870508  1   FALSE
> #40 1019 19880330  1    TRUE
>
>
> I thought all the rows with the above IDS would be FALSE

Why? Look at the last ID, 1019. The last of all must be included, the date doesn't repeat. And one of the first must also be included, if not we would be completely excluding that date. Or at least this is how I'm understanding the problem.

>   (from my solution):
>
> res4[c(11:13,22:24,38:40),]
>       ID     DATE DG INCLUDE
> #11  323 20080407  1   FALSE
> #12  323 20080521  2   FALSE
> #13  323 20080521  3   FALSE
> #22  841 20050421  1   FALSE
> #23  841 20050421  2   FALSE
> #24  841 20060428  1   FALSE
> #38 1019 19870508  1   FALSE
> #39 1019 19870508  2   FALSE
> #40 1019 19880330  1   FALSE
>
> A.K.
>
>
>
>
> ----- Original Message -----
> Cc: "arun (smartpink111 at yahoo.com)" <smartpink111 at yahoo.com>; PIKAL
> Petr <petr.pikal at precheza.cz>; r-help <r-help at r-project.org>
> Sent: Wednesday, October 24, 2012 1:41 PM
> Subject: Re: [r] How to pick colums from a ragged array?
>
> Hello,
>
> Using one of Arun's ideas, some post ago, this new function returns a
> logical index into id.d of the rows that should be _removed_, hence
> rm1 and rm2. I think
>
>
>
> getRepLogical <- function(x, first = TRUE){
>       fun <- if(first) head else tail
>       dte <- tapply(x[,2], x[,1], FUN = function(x) duplicated(fun(x,
>2)))
>       len <- tapply(x[,2], x[,1], FUN = length)
>       lst <- lapply(seq_along(dte), function(i) c(dte[[i]], rep(FALSE,
>if(len[[i]] > 2) len[[i]] - 2 else 0)))
>       lst <- if(first) lst else lapply(lst, rev)
>       i1 <- unlist(lst)
>       dg <- tapply(x[,3], x[,1], FUN = function(x) !duplicated(fun(x,
>2)))
>       lst <- lapply(seq_along(dte), function(i) c(dg[[i]], rep(FALSE,
>if(len[[i]] > 2) len[[i]] - 2 else 0)))
>       lst <- if(first) lst else lapply(lst, rev)
>       i2 <- unlist(lst)
>       i1 & i2
> }
>
> rm1 <- getRepLogical(id.d)
> rm2 <- getRepLogical(id.d, first = FALSE)
>
> id.d[rm1, ]
> id.d[rm2, ]
>
> id.d\$INCLUDE <- !(rm1 | rm2)
>
>
> Hope this helps,
>
> Em 24-10-2012 16:41, Stuart Leask escreveu:
>> (And, considering  the real application, the functions ideally should
>> probably output a variable INCLUDE, the same length as the original
>> data, with TRUE and FALSE for whether or not that row should be
>> included...)
>>
>> -----Original Message-----
>> Sent: 24 October 2012 16:25
>> To: arun (smartpink111 at yahoo.com); 'PIKAL Petr'; Rui Barradas
>> Subject: RE: [r] How to pick colums from a ragged array?
>>
>> Arun, Petr, Rui, many thanks for your help, and the functions you have written.
>>
>> You'll recall I wanted to remove these first (or last) duplicates, because they represented instances where two different diagnoses (in this case, variable DG, value 1, 2, 3, 4 or 5) had been recorded on the same day - so I can't say which was 'first' (or 'last').
>>
>> Your functions have revealed something I wasn't expecting: In some cases, the diagnoses recorded on the duplicated DATEs are the same!
>> This is a surprise to me, but probably reflects someone going to two different departments in a clinic, and both departments submit data. I have to say that crazy things like this are often a feature of real data, which I'm sure you've come across yourselves.
>>
>> Of course, I don't want to remove records in which I can determine an unambiguous 'first diagnosis'.
>>
>> You have all put in so much effort on my behalf, I'm ashamed to ask,
>> but I wonder if any of the functions you've written could do this
>> with a little more Indexing and the 'duplicate' function So the function should only exclude an ID, having identified a first (or last) DATE duplicate, the DGs for these two dates are different.
>>
>> Test dataset:
>>
>> 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(20060821,20061207,20080102,20090904,20040205,20040205,20051111
>>     ,20060111,20071119,20080107,20080407,20080521,20080521,20041005
>>     ,20070905,20020814,20021125,20040429,20040429,20071205,20071205
>>     ,20050421,20050421,20060428,20060602,20060816,20061025,20061129
>>     ,20070112,20070514, 19870508,20040205,20040205, 20080521,20080521
>>     ,20091224,20050503,19870508,19870508,19880330)
>>
>> DG<-
>> c(1,2,1,1,4,4,3,2,3,2,1,2,3,2,1,2,2,2,2,2,2,1,2,1,1,1,1,1,1,4,3,3,3,4
>> ,3,2,2,2,1,1)
>>
>> id.d<-data.frame(ID,DATE,DG)
>> id.d
>>
>> # Considering Ruis  getRepeat function:
>>
>> g.r<-getRepeat(id.d)    # defaults to first = TRUE getRepeat(id.d,
>> first = FALSE)  to get the last ones g.rr<-do.call(rbind, g.r) # put
>> the data into a matrix
>>
>> # I can remove the date duplicates with:
>> g.rr[rep(!duplicated(g.rr)[(1:(dim(g.rr)[1]/2))*2],each=2),]
>>
>> I'm not sure how to add this to your suggestions, Arun & Petr...
>>
>>
>> Stuart
>>
>>
>> -----Original Message-----
>> From: PIKAL Petr [mailto:petr.pikal at precheza.cz]
>> Sent: 23 October 2012 15:24
>> Subject: RE: [r] How to pick colums from a ragged array?
>>
>> Hi
>>
>> I assumed that id.d is data frame
>>
>> id.d <- data.frame (ID,DATE )
>>
>> and
>>
>> fff(id.d)
>>
>> works for me
>>
>> Petr
>>
>>
>>> -----Original Message-----
>>> Sent: Tuesday, October 23, 2012 3:13 PM
>>> To: PIKAL Petr
>>> Subject: RE: [r] How to pick colums from a ragged array?
>>>
>>> Hi Petr.
>>> I see what you mean it should do, but when I run it I get an error
>>> (see below).
>>> Stuart
>>>
>>>
>>>> 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(20060821,20061207,20080102,20090904,20040205,20040205,20051111
>>> +  ,20060111,20071119,20080107,20080407,20080521,20080711,20041005
>>> +  ,20070905,20020814,20021125,20040429,20040429,20071205,20080227
>>> +  ,20050421,20050421,20060428,20060602,20060816,20061025,20061129
>>> +  ,20070112,20070514, 19870508,20040205,20040205, 20091120,20091210
>>> +  ,20091224,20050503,19870508,19870508,19880330)
>>>>     id.d <- cbind (ID,DATE )
>>>> fff<-function(data, first=TRUE, remove=FALSE) {
>>> +
>>> + testfirst <- function(x) x[1,2]==x[2,2] testlast <- function(x)
>>> + x[nrow(x),2]==x[nrow(x)-1,2]
>>> +
>>> + if(first) sel <- as.numeric(names(which(unlist(sapply(split(data,
>>> + data[,1]), testfirst))))) else sel <-
>>> + as.numeric(names(which(unlist(sapply(split(data, data[,1]),
>>> + testlast)))))
>>> +
>>> + if (remove) data[!data[,1] %in% sel,] else data[data[,1] %in%
>>> + sel,] }
>>>> fff(id.d)
>>> Error in x[1, 2] : incorrect number of dimensions -----Original
>>> Message-----
>>> From: PIKAL Petr [mailto:petr.pikal at precheza.cz]
>>> Sent: 23 October 2012 13:51
>>> To: Stuart Leask; r-help at r-project.org
>>> Subject: RE: [r] How to pick colums from a ragged array?
>>>
>>> Hi
>>>
>>>> -----Original Message-----
>>>> Sent: Tuesday, October 23, 2012 2:29 PM
>>>> To: PIKAL Petr; r-help at r-project.org
>>>> Subject: RE: [r] How to pick colums from a ragged array?
>>>>
>>>> Hi there.
>>>>
>>>> Not sure I follow what you are doing.
>>>>
>>>> I want a list of all the IDs that have duplicate DATE entries, only
>>>> when the DATE is the earliest (or last) date for that ID.
>>> And that is what the function (with 3 small modifications) does
>>>
>>>
>>> fff<-function(data, first=TRUE, remove=FALSE) {
>>>
>>> testfirst <- function(x) x[1,2]==x[2,2] testlast <- function(x)
>>> x[nrow(x),2]==x[nrow(x)-1,2]
>>>
>>> if(first) sel <- as.numeric(names(which(unlist(sapply(split(data,
>>> data[,1]), testfirst))))) else sel <-
>>> as.numeric(names(which(unlist(sapply(split(data, data[,1]),
>>> testlast)))))
>>>
>>> if (remove) data[!data[,1] %in% sel,] else data[data[,1] %in% sel,]
>>> }
>>>
>>> See the result of your refined data
>>>
>>> fff(id.d)
>>>         ID       DATE
>>> 5   167 2004-02-05
>>> 6   167 2004-02-05
>>> 22  841 2005-04-21
>>> 23  841 2005-04-21
>>> 24  841 2006-04-28
>>> 25  841 2006-06-02
>>> 26  841 2006-08-16
>>> 27  841 2006-10-25
>>> 28  841 2006-11-29
>>> 29  841 2007-01-12
>>> 30  841 2007-05-14
>>> 38 1019 1987-05-08
>>> 39 1019 1987-05-08
>>> 40 1019 1988-03-30
>>>> fff(id.d, first=F)
>>>       ID       DATE
>>> 5 167 2004-02-05
>>> 6 167 2004-02-05
>>>> fff(id.d, remove=T)
>>>        ID       DATE
>>> 1   58 2006-08-21
>>> 2   58 2006-12-07
>>> 3   58 2008-01-02
>>> 4   58 2009-09-04
>>> 7  323 2005-11-11
>>> 8  323 2006-01-11
>>> 9  323 2007-11-19
>>> 10 323 2008-01-07
>>> 11 323 2008-04-07
>>> 12 323 2008-05-21
>>> 13 323 2008-07-11
>>> 14 547 2004-10-05
>>> 15 794 2007-09-05
>>> 16 814 2002-08-14
>>> 17 814 2002-11-25
>>> 18 814 2004-04-29
>>> 19 814 2004-04-29
>>> 20 814 2007-12-05
>>> 21 814 2008-02-27
>>> 31 910 1987-05-08
>>> 32 910 2004-02-05
>>> 33 910 2004-02-05
>>> 34 910 2009-11-20
>>> 35 910 2009-12-10
>>> 36 910 2009-12-24
>>> 37 999 2005-05-03
>>> You can do surgery on fff function to see what result comes from
>>>some  piece of the function e.g.
>>>
>>> sapply(split(id.d, id.d[,1]), testlast)
>>>
>>> Regards
>>> Petr
>>>
>>>> I have refined my test dataset, to include some tests (e.g. 910 has
>>>> the same dup as 1019, but for 910 it's not the earliest date):
>>>>
>>>>
>>>> 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(20060821,20061207,20080102,20090904,20040205,20040205,20051111
>>>>     ,20060111,20071119,20080107,20080407,20080521,20080711,20041005
>>>>     ,20070905,20020814,20021125,20040429,20040429,20071205,20080227
>>>>     ,20050421,20050421,20060428,20060602,20060816,20061025,20061129
>>>>     ,20070112,20070514, 19870508,20040205,20040205,
>>>>20091120,20091210
>>>>     ,20091224,20050503,19870508,19870508,19880330)
>>>>
>>>> Correct output:
>>>> "167"  "841"  "1019"
>>>>
>>>> Stuart
>>>>
>>>> -----Original Message-----
>>>> From: PIKAL Petr [mailto:petr.pikal at precheza.cz]
>>>> Sent: 23 October 2012 13:15
>>>> To: Stuart Leask; r-help at r-project.org
>>>> Subject: RE: [r] How to pick colums from a ragged array?
>>>>
>>>> Hi
>>>>
>>>> Rui's answer brought me to more elaborated solution which still
>>>> needs data frame to be ordered by date
>>>>
>>>> fff<-function(data, first=TRUE, remove=FALSE) {
>>>>
>>>> testfirst <- function(x) x[1,2]==x[2,2] testlast <- function(x)
>>>> x[length(x),2]==x[length(x)-1,2]
>>>>
>>>> if(first) sel <- as.numeric(names(which(sapply(split(data,
>>>> data[,1]),
>>>> testfirst)))) else sel <- as.numeric(names(which(sapply(split(data,
>>>> data[,1]), testlast))))
>>>>
>>>> if (remove) data[data[,1]!=sel,] else data[data[,1]==sel,] }
>>>>
>>>>
>>>>> fff(id.d)
>>>>        ID     DATE
>>>> 31 910 20091105
>>>> 32 910 20091105
>>>> 33 910 20091117
>>>> 34 910 20091119
>>>> 35 910 20091120
>>>> 36 910 20091210
>>>> 37 910 20091224
>>>> 38 910 20091224
>>>>
>>>>> fff(id.d, remove=T)
>>>>         ID     DATE
>>>> 1    58 20060821
>>>> 2    58 20061207
>>>> 3    58 20080102
>>>> 4    58 20090904
>>>> 5   167 20040205
>>>> 6   167 20040323
>>>> 7   323 20051111
>>>> 8   323 20060111
>>>> 9   323 20071119
>>>> 10  323 20080107
>>>> 11  323 20080407
>>>> 12  323 20080521
>>>> 13  323 20080711
>>>> 14  547 20041005
>>>> 15  794 20070905
>>>> 16  814 20020814
>>>> 17  814 20021125
>>>> 18  814 20040429
>>>> 19  814 20040429
>>>> 20  814 20071205
>>>> 21  814 20080227
>>>> 22  841 20050421
>>>> 23  841 20060130
>>>> 24  841 20060428
>>>> 25  841 20060602
>>>> 26  841 20060816
>>>> 27  841 20061025
>>>> 28  841 20061129
>>>> 29  841 20070112
>>>> 30  841 20070514
>>>> 39  999 20050503
>>>> 40 1019 19870508
>>>> 41 1019 19880223
>>>> 42 1019 19880330
>>>> 43 1019 19880330
>>>> Regards
>>>> Petr
>>>>
>>>>
>>>>> -----Original Message-----
>>>>> From: r-help-bounces at r-project.org [mailto:r-help-bounces at r-
>>>>> project.org] On Behalf Of PIKAL Petr
>>>>> Sent: Tuesday, October 23, 2012 1:49 PM
>>>>> To: Stuart Leask; r-help at r-project.org
>>>>> Subject: Re: [R] [r] How to pick colums from a ragged array?
>>>>>
>>>>> Hi
>>>>>
>>>>> I did not check your code and rather followed your explanation.
>>> BTW,
>>>>> thanks for test data.
>>>>>
>>>>> small change in data frame to make DATE as Date class
>>>>>
>>>>> datum<-as.Date(as.character(DATE), format="%Y%m%d") id.d <-
>>>>> data.frame(ID,datum )
>>>>>
>>>>> ordering by date
>>>>>
>>>>> id.d<-id.d[order(id.d\$datum),]
>>>>>
>>>>>
>>>>> two functions to test if first two dates are the same or last two
>>>>> dates are the same
>>>>>
>>>>> testfirst <- function(x) x[1,2]==x[2,2] testlast <- function(x)
>>>>> x[length(x),2]==x[length(x)-1,2]
>>>>>
>>>>> change one last date in the data frame to be the same as previous
>>>>>
>>>>> id.d[35,2]<-id.d[36,2]
>>>>>
>>>>> and here are results
>>>>>
>>>>> sapply(split(id.d, id.d\$ID), testlast)
>>>>>       58   167   323   547   794   814   841   910   999  1019
>>>>>FALSE FALSE FALSE    NA    NA FALSE FALSE  TRUE    NA FALSE
>>>>>
>>>>>> sapply(split(id.d, id.d\$ID), testfirst)
>>>>>       58   167   323   547   794   814   841   910   999  1019
>>>>>FALSE FALSE FALSE    NA    NA FALSE FALSE FALSE    NA FALSE
>>>>>
>>>>> Now you can select ID which is true and remove it from your data
>>>>> which(sapply(split(id.d, id.d\$ID), testlast))
>>>>>
>>>>> and use it for your data frame to subset/remove id.d\$ID ==
>>>>> as.numeric(names(which(sapply(split(id.d, id.d\$ID), testlast))))
>>> [1]
>>>>> FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE
>>>>> FALSE [13] FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE
>>>> FALSE
>>>>> FALSE FALSE [25] FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE
>>>> FALSE
>>>>> FALSE TRUE  TRUE [37]  TRUE  TRUE  TRUE  TRUE
>>>>>
>>>>> However I am not sure if this is exactly what you want.
>>>>>
>>>>> Regards
>>>>> Petr
>>>>>
>>>>>> -----Original Message-----
>>>>>> From: r-help-bounces at r-project.org [mailto:r-help-bounces at r-
>>>>>> project.org] On Behalf Of Stuart Leask
>>>>>> Sent: Tuesday, October 23, 2012 11:38 AM
>>>>>> To: r-help at r-project.org
>>>>>> Subject: [R] [r] How to pick colums from a ragged array?
>>>>>>
>>>>>> 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
>>>>>>
>>>>>>
>>>>>>
>>>>>> 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.
>>>>>>
>>>>>> This message and any attachment are intended solely for the
>>>>>> addressee and may contain confidential information. If you have
>>>>>> received this message in error, please send it back to me, and
>>>>>> immediately delete
>>>>> it.
>>>>>> Please do not use, copy or disclose the information contained in
>>>>>> this message or in any attachment.  Any views or opinions
>>>>>> expressed by the author of this email do not necessarily reflect
>>>>>> the views of the University of Nottingham.
>>>>>>
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>>> by
>>>>>> UK legislation.
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>>>>>>
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