I've searched quite a bit for this to no avail, let alone trying to make it work myself, so without further a do:
I have this data.table :
DT = data.table(date=rep(c(as.Date("2010-01-01"),as.Date("2010-01-02")), each=3), bucket=rep(c("bucket1","bucket2","bucket3"),each=2),
kbucket=c("(0,.5]","(.5,1]","(1,1.5]","(1.5,2]","(1.5,2]","(2.5,3]"),vol=1:6,o=10:15,m=20:25)
which looks like:
date bucket kbucket vol o m
1: 2010-01-01 bucket1 (0,.5] 1 10 20
2: 2010-01-01 bucket1 (.5,1] 2 11 21
3: 2010-01-01 bucket2 (1,1.5] 3 12 22
4: 2010-01-02 bucket2 (1.5,2] 4 13 23
5: 2010-01-02 bucket3 (1.5,2] 5 14 24
6: 2010-01-02 bucket3 (2.5,3] 6 15 25
I've used ddply like so, on DF, the facsimile of DT, but which is a data frame:
out <- ddply(DF,.(date,bucket,kbucket),wrap_summarize)
where wrap_summarize is defined as:
wrap_summarize = function(x)
{
out <- summarize( x,
N = length(x$date),
sumVol = sum(x$vol),
sumO = sum(x$o),
avgM = mean(x$m,na.rm=TRUE))
}
to get
date bucket kbucket N sumVol sumO avgM
1 2010-01-01 bucket1 (.5,1] 1 2 11 21
2 2010-01-01 bucket1 (0,.5] 1 1 10 20
3 2010-01-01 bucket2 (1,1.5] 1 3 12 22
4 2010-01-02 bucket2 (1.5,2] 1 4 13 23
5 2010-01-02 bucket3 (1.5,2] 1 5 14 24
6 2010-01-02 bucket3 (2.5,3] 1 6 15 25
This is the desired result.
The actual data has this structure but is hundreds of thousands of rows. Thus the need for data.table methods. So I try this:
test <- DT[,list(N=length(DT$date),sumVol=sum(DT$vol),sumO=sum(DT$o),avgM=mean(DT$m,na.rm=T)),
by=list(date,bucket,kbucket)]
only to get, which is clearly not what is desired:
date bucket kbucket N sumVol sumO avgM
1: 2010-01-01 bucket1 (0,.5] 6 21 75 22.5
2: 2010-01-01 bucket1 (.5,1] 6 21 75 22.5
3: 2010-01-01 bucket2 (1,1.5] 6 21 75 22.5
4: 2010-01-02 bucket2 (1.5,2] 6 21 75 22.5
5: 2010-01-02 bucket3 (1.5,2] 6 21 75 22.5
6: 2010-01-02 bucket3 (2.5,3] 6 21 75 22.5
I think i need to use .SD here, but at this point, i thought it best to ask and share this problem if not get the most efficient solution. Thanks in advance!
You're looking for this:
DT[,list(
.N,
sumVol=sum(vol),
sumO=sum(o),
avgM=mean(m,na.rm=T)
),by=list(date,bucket,kbucket)]
which gives
# date bucket kbucket N sumVol sumO avgM
# 1: 2010-01-01 bucket1 (0,.5] 1 1 10 20
# 2: 2010-01-01 bucket1 (.5,1] 1 2 11 21
# 3: 2010-01-01 bucket2 (1,1.5] 1 3 12 22
# 4: 2010-01-02 bucket2 (1.5,2] 1 4 13 23
# 5: 2010-01-02 bucket3 (1.5,2] 1 5 14 24
# 6: 2010-01-02 bucket3 (2.5,3] 1 6 15 25
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