How to pick colums from a ragged array?

-----Original Message-----
From: Stuart Leask [mailto:Stuart.Leask at nottingham.ac.uk]
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.

fff(id.d, first=F)

fff(id.d, remove=T)

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
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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