[R] Help make this simpler – count business day

[Gabor Grothendieck]

On Tue, Nov 9, 2010 at 11:49 AM, cameron <raymond.fu at invesco.com> wrote:
>
> Help make this simpler – count business day
>
>
> I am a beginner in R and this is my first post
>
> Want to count the day in month.  For example
>
>                 Day
> 2010-09-01   1      Wed
> 2010-09-02   2  Thurs
> 2010-09-03   3  Friday
> 2010-09-07   4  Tuesday
> 2010-09-08   5  Wed
> 2010-09-09   6  Thursday
> 2010-09-10   7  Friday
>
> #-------------------------
> library(tseries)
>
> msft <- get.hist.quote(instrument="MSFT", start="1986-03-31",
> end="2008-09-10", quote=c("O","H","L","C","A","V"), provider="yahoo",
> retclass="zoo")
>
> # tail(msft)
> #            Open  High   Low Close AdjClose   Volume
> #2008-09-03 27.00 27.18 26.84 26.90    25.73 57127700
> #2008-09-04 26.74 26.89 26.35 26.35    25.21 66141900
> #2008-09-05 26.03 26.22 25.63 25.65    24.54 82305200
> #2008-09-08 26.21 26.33 25.67 26.12    24.99 62110800
> #2008-09-09 26.20 26.60 26.05 26.10    24.97 85735700
> #2008-09-10 26.52 26.86 26.25 26.44    25.29 75064900
>
>
> countday<-function(z)
> {
>        z <- cbind(z,rep(0,times=nrow(z)))
>
>        rng <- range(time(z))
>        StartDate <- rng[1]
>        EndDate   <- rng[2]
>
>        starty <- as.numeric(format(StartDate, "%Y"))
>        endy   <- as.numeric(format(EndDate, "%Y"))
>
>        year <- starty
>
>        for (year in starty:endy){
>                for (month in 1:12){
>                        rows <- which(as.numeric(format(time(z),"%m")) == month &
> as.numeric(format(time(z),"%Y")) == year )
>                        temp <- z[rows,]
>                        n <- 1:nrow(temp)
>                        z[rows,ncol(z)] <- n
>                }
>        }
>        colnames(z) <- c(colnames(z[,1:(ncol(z)-1)]),"Day")
>        return(z)
> }
>
> msft <- countday(msft)
>
>>msft
> #            Open  High   Low Close AdjClose   Volume        Day
> #2008-09-03 27.00 27.18 26.84 26.90    25.73 57127700   2
> #2008-09-04 26.74 26.89 26.35 26.35    25.21 66141900   3
> #2008-09-05 26.03 26.22 25.63 25.65    24.54 82305200   4
> #2008-09-08 26.21 26.33 25.67 26.12    24.99 62110800   5
> #2008-09-09 26.20 26.60 26.05 26.10    24.97 85735700   6
> #2008-09-10 26.52 26.86 26.25 26.44    25.29 75064900   7

Try this:

msft$Day <- ave(1:nrow(msft), as.yearmon(time(msft)), FUN = seq_along)



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