Hi Anna,
In addition, you may also try:
library(data.table)
dt1 <- data.table(my_dat, key=c('sr', 'bond'))
dt2 <- dt1[,list(cashflow_total=sum(cashflow)), by=c('sr','bond')]
A.K.
On Wednesday, May 21, 2014 6:46 PM, Jim Lemon <j...@bitwrit.com.au> wrote:
On Wed, 21 May 2014 03:59:26 PM Anna Carter wrote:
> Dear R forum,
>
> I have following data.frame -
>
> my_dat = data.frame(sr = c(0,0,0,0, 1, 1, 1, 1, 2, 2, 2, 2), bond =
c("A",
> "B", "B", "B", "A", "B", "B", "B", "A", "B", "B", "B"), cashflow = c(1000,
> 2000, 2000, 4000, 10, 200, 300, 100, 80, 40, 60, 120))
> > my_dat
>
> sr bond cashflow
> 1 0 A 1000
> 2 0 B 2000
> 3 0 B 2000
> 4 0 B 4000
> 5 1 A 10
> 6 1 B 200
> 7 1 B 300
> 8 1 B 100
> 9 2 A 80
> 10 2 B 40
> 11 2 B 60
> 12 2 B 120
>
> The above data.frame is just an example. My original data is bit
large. My
> requirement is for given sr no, I need to add the cashflow values, for
each
> bond. Thus, I need the output (as a data.frame) as -
>
>
> # OUTPUT
>
> sr bond cashflow_total
> 1 0 A 1000
> 2 0 B 8000
> 3 1 A 10
> 4 1 B 600
> 5 2 A 80
> 6 2 B 220
>
> My - code
>
> my_dat$key = paste(my_dat$bond, my_dat$sr, sep = "_")
> cashflow_total = tapply(my_dat$cashflow, my_dat$key, sum)
>
> > cashflow_total
>
> A_0 A_1 A_2 B_0 B_1 B_2
> 1000 10 80 8000 600 220
>
> How do maintain the original order of sr and bond as in my_dat
data.frame
> and obtain the
>
> above OUTPUT?
>
Hi Anna,
Try this:
aggregate(my_dat$cashflow,my_dat[c("bond","sr")],sum)
Jim
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