Michael, thanks a lot!
Really appreciate it - what wasn't hard for you would be for me!
Dimitri
On Thu, Aug 4, 2011 at 2:20 PM, R. Michael Weylandt
wrote:
> You are getting 105 because the default behavior of findInterval is such
> that v[N+1] := + Inf (as noted in ? findInterval); that is, the
You are getting 105 because the default behavior of findInterval is such
that v[N+1] := + Inf (as noted in ? findInterval); that is, the last
interval is actually taken to stretch from the final element of sbwl.dates
unto eternity. It shouldn't be hard to write a catch line to set that to
whatever
Sorry for renewing the topoic. I thought it worked but now I've run
into a little problem:
# My data frame with dates for week starts (Mondays)
y<-data.frame(week=seq(as.Date("2009-12-28"),
as.Date("2011-12-26"),by="week") )
# I have a vector of super bowl dates (including the future one for 201
Thanks a lot, everyone!
Dimitri
On Tue, Aug 2, 2011 at 12:34 PM, Dennis Murphy wrote:
> Hi:
>
> You could try the lubridate package:
>
> library(lubridate)
> week(weekly$week)
> week(july4)
> [1] 27 27
>
>> week
> function (x)
> yday(x)%/%7 + 1
>
>
> which is essentially Gabor's code :)
>
> HTH,
Hi:
You could try the lubridate package:
library(lubridate)
week(weekly$week)
week(july4)
[1] 27 27
> week
function (x)
yday(x)%/%7 + 1
which is essentially Gabor's code :)
HTH,
Dennis
On Tue, Aug 2, 2011 at 7:36 AM, Dimitri Liakhovitski
wrote:
> Hello!
>
> I have dates for the beginning of
On Tue, 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 intere
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
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