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cameron

2010-Nov-09 16:49 UTC

[R] Help make this simpler – count business day

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


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Gabor Grothendieck

2010-Nov-10 01:49 UTC

head link

[R] Help make this simpler – count business day

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)



-- 
Statistics & Software Consulting
GKX Group, GKX Associates Inc.
tel: 1-877-GKX-GROUP
email: ggrothendieck at gmail.com

cameron

2010-Nov-10 13:08 UTC

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[R] Help make this simpler – count business day

thanks Gabor

There is no way i can come up with your code.  I guess if i am using loop in
R, I am doing it wrong.


-- 
View this message in context:
http://r.789695.n4.nabble.com/Help-make-this-simpler-count-business-day-tp3034621p3036042.html
Sent from the R help mailing list archive at Nabble.com.

Gabor Grothendieck

2010-Nov-10 13:55 UTC

head link

[R] Help make this simpler – count business day

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)


-- 
Statistics & Software Consulting
GKX Group, GKX Associates Inc.
tel: 1-877-GKX-GROUP
email: ggrothendieck at gmail.com

R help - Nov 2010 - Help make this simpler – count business day

[R] Help make this simpler – count business day

[R] Help make this simpler – count business day

[R] Help make this simpler – count business day

[R] Help make this simpler – count business day