[R] How to remove rows based on frequency of factor and then difference date scores

[David Winsemius]

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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> and provide commented, minimal, self-contained, reproducible code.

David Winsemius, MD
West Hartford, CT