I too had a parsimonious solution that was also fooled by IDs that had a
duplicate date that wasn't the first date, but was the same as another ID's
duplicate+first.
The right answer
>From this data:
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 )
is:
167, 841 and 1019 - correct.
814 910 - incorrect. Although there are duplicate dates, they are not the first
date.
-----Original Message-----
From: arun [mailto:smartpink...@yahoo.com]
Sent: 23 October 2012 14:29
To: Stuart Leask
Cc: R help
Subject: Re: [R] [r] How to pick colums from a ragged array?
Hi,
Also one more thing:
This should get the dates which are duplicated. In my first reply, I was
looking for the duplicated rows. Sorry for that!
id.d<-data.frame(ID,DATE)
new1<-id.d[duplicated(id.d$DATE)|duplicated(id.d$DATE,fromLast=TRUE),]
new2<-new1[order(new1$ID,new1$DATE),]
tapply(new2$ID,new2$DATE,head,1)
#19870508 20040205 20040429 20050421
# 910 167 814 841
But, still the result is not that you wanted, because 910's date is the
earliest date when compared to 1019.
new1[order(new1$ID,new1$DATE),]
# ID DATE
#5 167 20040205
#6 167 20040205
#18 814 20040429
#19 814 20040429
#22 841 20050421
#23 841 20050421
#31 910 19870508
#32 910 20040205
#33 910 20040205
#38 1019 19870508
#39 1019 19870508
A.K.
----- Original Message -----
From: Stuart Leask <stuart.le...@nottingham.ac.uk>
To: arun <smartpink...@yahoo.com>
Cc: Petr PIKAL <petr.pi...@precheza.cz>
Sent: Tuesday, October 23, 2012 9:15 AM
Subject: RE: [R] [r] How to pick colums from a ragged array?
Sorry Arun, but when I run it I get an error:
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 )
new1<-id.d[duplicated(id.d)|duplicated(id.d,fromLast=TRUE),]
tapply(new1$ID,new1$DATE,head,1)
Error in new1$DATE : $ operator is invalid for atomic vectors
-----Original Message-----
From: arun [mailto:smartpink...@yahoo.com]
Sent: 23 October 2012 14:05
To: Stuart Leask
Cc: R help; Petr PIKAL
Subject: Re: [R] [r] How to pick colums from a ragged array?
HI,
I was not following the thread.
May be this is what you are looking for:
new1<-id.d[duplicated(id.d)|duplicated(id.d,fromLast=TRUE),]
tapply(new1$ID,new1$DATE,head,1)
#19870508 20040205 20040429 20050421
# 1019 167 814 841
A.K.
----- Original Message -----
From: Stuart Leask <stuart.le...@nottingham.ac.uk>
To: PIKAL Petr <petr.pi...@precheza.cz>; "r-help@r-project.org"
<r-help@r-project.org>
Cc:
Sent: Tuesday, October 23, 2012 8:28 AM
Subject: Re: [R] [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.
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.pi...@precheza.cz]
Sent: 23 October 2012 13:15
To: Stuart Leask; r-help@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-boun...@r-project.org [mailto:r-help-bounces@r-
project.org] On Behalf Of PIKAL Petr
Sent: Tuesday, October 23, 2012 1:49 PM
To: Stuart Leask; r-help@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-boun...@r-project.org [mailto:r-help-bounces@r-
project.org] On Behalf Of Stuart Leask
Sent: Tuesday, October 23, 2012 11:38 AM
To: r-help@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.le...@nottingham.ac.uk<mailto:stuart.le...@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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______________________________________________
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
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.
This message has been checked for viruses but the contents of an attachment may
still contain software viruses which could damage your computer system:
you are advised to perform your own checks. Email communications with the
University of Nottingham may be monitored as permitted by UK legislation.
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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