Many thanks Mikhail,
aggregate() is fine!
Il 14/02/2011 20.42, Mikhail Titov ha scritto:
It depends what would you like to get at the end. Perhaps you don't
necessary need this type of numbering. For instance, if you'd like to
calculate daily average.
london$id<- as.Date(london$id)
For sum by day you could use, let's say, this
aggregate(words~id,london,FUN=sum)
If you really want what you've asked:
london$one=1
u=unique(london$id)
z=aggregate(one~id,london,FUN=sum)
london$day=rep(seq(along.with=z$one),z$one)
Mikhail
-----Original Message-----
From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
project.org] On Behalf Of Michela Ferron
Sent: Monday, February 14, 2011 11:09 AM
To: r-help@r-project.org
Subject: [R] How to group data by day
Hi everybody,
I'm a beginner in R and I'm having a hard time grouping my data by day.
The data are in this format:
id; words
2005-07-07T09:59:56Z; 35
2005-07-07T10:01:39Z; 13
2005-07-08T10:02:22Z; 1
2005-07-09T10:03:16Z; 23
2005-07-10T10:04:23Z; 39
2005-07-10T10:04:39Z; 15
I've transformed the date strings in dates with the function:
london$id<- transform(london$id, as.Date(london$id, format="%Y-%m-
%d%T%H:%M:%S%Z"))
and it seems to work.
Now I would like to add a new "day" variable to group data by day, like
this:
id; words; day
2005-07-07T09:59:56Z; 35; 1
2005-07-07T10:01:39Z; 13; 1
2005-07-08T10:02:22Z; 1; 2
2005-07-09T10:03:16Z; 23; 3
2005-07-10T10:04:23Z; 39; 4
2005-07-10T10:04:39Z; 15; 4
How can I do that?
Many thanks!
Michela
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