[R] Help make this simpler – count business day
Gabor Grothendieck
ggrothendieck at gmail.com
Wed Nov 10 14:55:48 CET 2010
On Tue, Nov 9, 2010 at 8:49 PM, Gabor Grothendieck
<ggrothendieck at gmail.com> wrote:
> 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)
Note that as.yearmon is from the zoo package:
library(zoo)
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