R help - Aug 2011 - identifying weeks (dates) that certain days (dates) fall into

Hello!

I have dates for the beginning of each week, e.g.:
weekly<-data.frame(week=seq(as.Date("2010-04-01"),
as.Date("2011-12-26"),by="week"))
week  # each week starts on a Monday

I also have a vector of dates I am interested in, e.g.:
july4<-as.Date(c("2010-07-04","2011-07-04"))

I would like to flag the weeks in my weekly$week that contain those 2
individual dates.
I can only think of a very clumsy way of doing it:

myrows<-c(which(weekly$week==weekly$week[weekly$week>july4[1]][1]-7),
	which(weekly$week==weekly$week[weekly$week>july4[2]][1]-7))
weekly$flag<-0
weekly$flag[myrows]<-1

It's clumsy - because actually, my vector of dates of interest (july4
above) is much longer.
Is there maybe a more elegant way of doing it?
Thank you!
-- 
Dimitri Liakhovitski
marketfusionanalytics.com

The findInterval function should surely be tried in some form or  
another.

On Aug 2, 2011, at 10:36 AM, Dimitri Liakhovitski wrote:
> Hello!
>
> I have dates for the beginning of each week, e.g.:
> weekly<-data.frame(week=seq(as.Date("2010-04-01"),
> as.Date("2011-12-26"),by="week"))
> week  # each week starts on a Monday
>
> I also have a vector of dates I am interested in, e.g.:
> july4<-as.Date(c("2010-07-04","2011-07-04"))
>
> I would like to flag the weeks in my weekly$week that contain those 2
> individual dates.
 > findInterval(july4, weekly$week)
[1] 14 66   # works "out of the box"

Provides an index you cna use with weekly$week
> I can only think of a very clumsy way of doing it:
>
> myrows<-c(which(weekly$week==weekly$week[weekly$week>july4[1]][1]-7),
> 	which(weekly$week==weekly$week[weekly$week>july4[2]][1]-7))
> weekly$flag<-0
> weekly$flag[myrows]<-1
>
> It's clumsy - because actually, my vector of dates of interest (july4
> above) is much longer.
> Is there maybe a more elegant way of doing it?
-- 

David Winsemius, MD
West Hartford, CT

On Tue, Aug 2, 2011 at 10:36 AM, Dimitri Liakhovitski
<dimitri.liakhovitski at gmail.com> wrote:
> Hello!
>
> I have dates for the beginning of each week, e.g.:
> weekly<-data.frame(week=seq(as.Date("2010-04-01"),
> as.Date("2011-12-26"),by="week"))
> week ?# each week starts on a Monday
>
> I also have a vector of dates I am interested in, e.g.:
> july4<-as.Date(c("2010-07-04","2011-07-04"))
>
> I would like to flag the weeks in my weekly$week that contain those 2
> individual dates.
> I can only think of a very clumsy way of doing it:
>
> myrows<-c(which(weekly$week==weekly$week[weekly$week>july4[1]][1]-7),
> ? ? ? ?which(weekly$week==weekly$week[weekly$week>july4[2]][1]-7))
> weekly$flag<-0
> weekly$flag[myrows]<-1
>
> It's clumsy - because actually, my vector of dates of interest (july4
> above) is much longer.
> Is there maybe a more elegant way of doing it?
> Thank you!
This gives myrows:

   as.numeric(july4 - weekly[1,1]) %/% 7 + 1

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

Hi:

You could try the lubridate package:

library(lubridate)
week(weekly$week)
week(july4)
[1] 27 27
> week
function (x)
yday(x)%/%7 + 1
<environment: namespace:lubridate>

which is essentially Gabor's code :)

HTH,
Dennis

On Tue, Aug 2, 2011 at 7:36 AM, Dimitri Liakhovitski
<dimitri.liakhovitski at gmail.com> wrote:
> Hello!
>
> I have dates for the beginning of each week, e.g.:
> weekly<-data.frame(week=seq(as.Date("2010-04-01"),
> as.Date("2011-12-26"),by="week"))
> week ?# each week starts on a Monday
>
> I also have a vector of dates I am interested in, e.g.:
> july4<-as.Date(c("2010-07-04","2011-07-04"))
>
> I would like to flag the weeks in my weekly$week that contain those 2
> individual dates.
> I can only think of a very clumsy way of doing it:
>
> myrows<-c(which(weekly$week==weekly$week[weekly$week>july4[1]][1]-7),
> ? ? ? ?which(weekly$week==weekly$week[weekly$week>july4[2]][1]-7))
> weekly$flag<-0
> weekly$flag[myrows]<-1
>
> It's clumsy - because actually, my vector of dates of interest (july4
> above) is much longer.
> Is there maybe a more elegant way of doing it?
> Thank you!
> --
> Dimitri Liakhovitski
> marketfusionanalytics.com
>
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