Using cumsum with 'group by' ?

Hi,
No problem.
One more method if you wanted to try:
library(data.table)
dat2<-data.table(dat1)
dat2[,list(x,time,Cumsum=cumsum(x)),list(id)]
?# ? id? x? time Cumsum
?#1:? 1? 5 12:01????? 5
?#2:? 1 14 12:02???? 19
?#3:? 1? 6 12:03???? 25
?#4:? 1? 3 12:04???? 28
?#5:? 2 98 12:01???? 98
?#6:? 2 23 12:02??? 121
?#7:? 2? 1 12:03??? 122
?#8:? 2? 4 12:04??? 126
?#9:? 3? 5 12:01????? 5
#10:? 3 65 12:02???? 70
#11:? 3 23 12:03???? 93
#12:? 3 23 12:04??? 116


A.K.



----- Original Message -----
From: TheRealJimShady <james.david.smith at gmail.com>
To: r-help at r-project.org
Cc: 
Sent: Thursday, November 22, 2012 12:27 PM
Subject: Re: [R] Using cumsum with 'group by' ?

Thank you very much, I will try these tomorrow morning.

On 22 November 2012 17:25, arun kirshna [via R]

HI,
You can do this in many ways:
dat1<-read.table(text="
id? ? time? ? x
1?  12:01? ? 5
1?  12:02?  14
1?  12:03?  6
1?  12:04?  3
2?  12:01?  98
2?  12:02?  23
2?  12:03?  1
2?  12:04?  4
3?  12:01?  5
3?  12:02?  65
3?  12:03?  23
3?  12:04?  23
",sep="",header=TRUE,stringsAsFactors=FALSE)
? dat1$Cumsum<-ave(dat1$x,dat1$id,FUN=cumsum)
#or
? unlist(tapply(dat1$x,dat1$id,FUN=cumsum),use.names=FALSE)
# [1]?  5? 19? 25? 28? 98 121 122 126?  5? 70? 93 116
#or
library(plyr)
? ddply(dat1,.(id),function(x) cumsum(x[3]))[,2]
# [1]?  5? 19? 25? 28? 98 121 122 126?  5? 70? 93 116
head(dat1)
#? id? time? x Cumsum
#1? 1 12:01? 5? ? ? 5
#2? 1 12:02 14? ?  19
#3? 1 12:03? 6? ?  25
#4? 1 12:04? 3? ?  28
#5? 2 12:01 98? ?  98
#6? 2 12:02 23? ? 121
A.K.

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