Hi All,
I'm trying to figure out how in my data set to add a column including a
count of unique events based on date. Here is a part of my data set:
trialno event date time
3 11301 pm_intake 2010-11-24 19:00
4 11301 am_intake 2010-11-25 07:00
5 11301 pk1 2010-11-25
10:30
6 11301 pm_intake 2010-12-22 19:00
7 11301 am_intake 2010-12-23 07:00
8 11301 pk1 2010-12-23
09:54
9 11301 pk2 2010-12-23
13:07
10 11301 pm_intake 2011-02-02 19:00
11 11301 am_intake 2011-02-03 07:00
12 11301 pk1 2011-02-03 11:30
Basically each date within each patient would indicate a new occasion. If
patient has just drug administration - it's one occasion but if patient had
drug administration and two measurements on the same day, they all count as
the same occasion. The data set does not have a regular patters (each
patient has a different number of events on each date and events in total).
What I'm trying to achieve is:
trialno event date time
OCC
3 11301 pm_intake 2010-11-24 19:00 1
4 11301 am_intake 2010-11-25 07:00 2
5 11301 pk1 2010-11-25
10:30 2
6 11301 pm_intake 2010-12-22 19:00 3
7 11301 am_intake 2010-12-23 07:00 4
8 11301 pk1 2010-12-23
09:54 4
9 11301 pk2 2010-12-23
13:07 4
10 11301 pm_intake 2011-02-02 19:00 5
11 11301 am_intake 2011-02-03 07:00 6
12 11301 pk1 2011-02-03 11:30
6
I think I should apply some kind of a loop to identify within each patient
unique dates and count them...
I thought about splitting the whole data set into patients using split
function:
splitData<- split(data, data$trialno)
And applying lapply and transform to add a new column OCC (occasion) but I
don't know how to count those as integers...
I was thinking:
splitData<- lapply(splitData, function(df) {
transform(df, OCC= ??????????????? )}
do.call ("rbind", splitData)
I know how to do it in Excell:
=IF(D5=D4, E4,E4+1)
(if the cell value in neighbouring cell is same as in the cell above, then
value in my cell is same as in one above, else it's one greater)-this way
first cell in E column has to be 1 and the others are integers of new date
events.
Help much appreciated!
Andrzej
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