On Tue, Nov 9, 2010 at 11:49 AM, cameron <raymond...@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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