The only problem with this is that Chris's unique individuals are a
combination of Type and ID, as I understand it. So Type=A, ID=1 is a
different individual from Type=B,ID=1. So we need to create a unique
identifier per person, simplistically by uniqueID=paste(Type, ID,
sep=''). Then, using this new identifier, everything follows.
On 08/24/2010 01:53 PM, David Winsemius wrote:
On Aug 24, 2010, at 1:19 PM, Chris Beeley wrote:
Hello-
A basic question which has nonetheless floored me entirely. I have a
dataset which looks like this:
Type ID Date Value
A 1 16/09/2020 8
A 1 23/09/2010 9
B 3 18/8/2010 7
B 1 13/5/2010 6
There are two Types, which correspond to different individuals in
different conditions, and loads of ID labels (1:50) corresponding to
the different individuals in each condition, and measurements at
different times (from 1 to 10 measurements) for each individual.
I want to perform the following operations:
1) Delete all individuals for whom only one measurement is available.
In the dataset above, you can see that I want to delete the row Type B
ID 3, and Type B ID 1, but without deleting the Type A ID 1 data
because there is more than one measurement for Type A ID 1 (but not
for Type B ID1)
2) Produce difference scores for each of the Dates, so each individual
(Type A ID1 and all the others for whom more than one measurement
exists) starts at Date "1" and goes up in integers according to how
many days have elapsed.
I just know there's some incredibly cunning R-ish way of doing this
but after many hours of fiddling I have had to admit defeat.
Not sure about terribly cunning. Let's assume your dataframe was read
in with stringsAsFactors=FALSE and is called txt.df:
> txt.df$dt2 <- as.Date(txt.df$Date, format="%d/%m/%Y")
> txt.df
Type ID Date Value dt2
1 A 1 16/09/2020 8 2020-09-16
2 A 1 23/09/2010 9 2010-09-23
3 B 3 18/8/2010 7 2010-08-18
4 B 1 13/5/2010 6 2010-05-13
> txt.df$nn <- ave(txt.df$ID,txt.df$ID, FUN=length)
> txt.df
Type ID Date Value dt2 nn
1 A 1 16/09/2020 8 2020-09-16 3
2 A 1 23/09/2010 9 2010-09-23 3
3 B 3 18/8/2010 7 2010-08-18 1
4 B 1 13/5/2010 6 2010-05-13 3
> txt.df[ -which( txt.df$nn <=1), ]
Type ID Date Value dt2 nn
1 A 1 16/09/2020 8 2020-09-16 3
2 A 1 23/09/2010 9 2010-09-23 3
4 B 1 13/5/2010 6 2010-05-13 3
# Task #1 accomplished
> tapply(txt.df$dt2, txt.df$ID, function(x) x[1] -x)
$`1`
Time differences in days
[1] 0 3646 3779
$`3`
Time difference of 0 days
> unlist( tapply(txt.df$dt2, txt.df$ID, function(x) x[1] -x) )
11 12 13 3
0 3646 3779 0
> txt.df$diffdays <- unlist( tapply(txt.df$dt2, txt.df$ID, function(x)
x[1] -x) )
> txt.df
Type ID Date Value dt2 nn diffdays
1 A 1 16/09/2020 8 2020-09-16 3 0
2 A 1 23/09/2010 9 2010-09-23 3 3646
3 B 3 18/8/2010 7 2010-08-18 1 3779
4 B 1 13/5/2010 6 2010-05-13 3 0
>
I would be very grateful for any words of advice.
Many thanks,
Chris Beeley,
Institute of Mental Health, UK
______________________________________________
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PLEASE do read the posting guide
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and provide commented, minimal, self-contained, reproducible code.
David Winsemius, MD
West Hartford, CT
______________________________________________
R-help@r-project.org mailing list
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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.
--
Abhijit Dasgupta, PhD
Director and Principal Statistician
ARAASTAT
Ph: 301.385.3067
E: adasgu...@araastat.com
W: http://www.araastat.com
______________________________________________
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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.
