Dataframe cust has Date-type column open.date. I wish to set up another
column, with (first day of) the quarter of open.date.
To be comprehensive (of course, improvement suggestions are welcome),
month = function(date)
{
return(as.numeric(format(date,"%m")))
}
first.day.of.month = function(date)
{
return(date + 1 - as.numeric(format(date,"%d")))
}
first.day.of.quarter = function(date)
{
t = seq.Date(first.day.of.month(date), by = "-1 month", length
month(date) %% 3)
return(t[length(t)])
}
Now the main part,
> cust$open.quarter = apply(cust$open.date, 1, FUN = first.day.of.quarter)
Error in apply(cust$open.date, 1, FUN = first.day.of.quarter) :
dim(X) must have a positive length
> cust$open.quarter = tapply(cust$open.date, FUN = first.day.of.quarter)
Error in tapply(cust$open.date, FUN = first.day.of.quarter) :
element 1 is empty;
the part of the args list of 'is.list' being evaluated was:
(INDEX)
> cust$open.quarter = lapply(cust$open.date, FUN = first.day.of.quarter)
Error in prettyNum(.Internal(format(x, trim, digits, nsmall, width, 3L, :
invalid 'trim' argument
Can anyone suggest the right syntax?
Thank you.
--
View this message in context:
http://n4.nabble.com/Newbie-woes-with-apply-tp1555149p1555149.html
Sent from the R help mailing list archive at Nabble.com.
Bug fix:
first.day.of.quarter = function(date)
{
t = first.day.of.month(date)
l = month(date) %% 3
if (l == 0) return(t)
t = seq.Date(t, by = "-1 month", length = l)
return(t[length(t)])
}
But the *apply part still does not work.
--
View this message in context:
http://n4.nabble.com/Newbie-woes-with-apply-tp1555149p1555167.html
Sent from the R help mailing list archive at Nabble.com.
Try library(zoo) x <- as.Date(seq(1, 500, 50)) # test data as.Date(as.yearmon(x)) as.Date(as.yearqtr(x)) On Sun, Feb 14, 2010 at 8:59 AM, Dimitri Shvorob <dimitri.shvorob at gmail.com> wrote:> > Dataframe cust has Date-type column open.date. I wish to set up another > column, with (first day of) the quarter of open.date. > > To be comprehensive (of course, improvement suggestions are welcome), > > month = function(date) > { > ?return(as.numeric(format(date,"%m"))) > } > > first.day.of.month = function(date) > { > ?return(date + 1 - as.numeric(format(date,"%d"))) > } > > first.day.of.quarter = function(date) > { > ?t = seq.Date(first.day.of.month(date), by = "-1 month", length > month(date) %% 3) > ?return(t[length(t)]) > } > > Now the main part, > >> cust$open.quarter ?= apply(cust$open.date, 1, FUN = first.day.of.quarter) > Error in apply(cust$open.date, 1, FUN = first.day.of.quarter) : > ?dim(X) must have a positive length > >> cust$open.quarter ?= tapply(cust$open.date, FUN = first.day.of.quarter) > Error in tapply(cust$open.date, FUN = first.day.of.quarter) : > ?element 1 is empty; > ? the part of the args list of 'is.list' being evaluated was: > ? (INDEX) > >> cust$open.quarter ?= lapply(cust$open.date, FUN = first.day.of.quarter) > Error in prettyNum(.Internal(format(x, trim, digits, nsmall, width, 3L, ?: > ?invalid 'trim' argument > > Can anyone suggest the right syntax? > > Thank you. > > -- > View this message in context: http://n4.nabble.com/Newbie-woes-with-apply-tp1555149p1555149.html > Sent from the R help mailing list archive at Nabble.com. > > ______________________________________________ > R-help at r-project.org mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. >
Many thanks, but my focus is actually on *apply usage. -- View this message in context: http://n4.nabble.com/Newbie-woes-with-apply-tp1555149p1555329.html Sent from the R help mailing list archive at Nabble.com.
Your function is not receiving what you may think its receiving. Try this:> x <- Sys.Date() + 1:3 > junk <- lapply(x, print)[1] 14655 [1] 14656 [1] 14657 So add this as the first statement in the body of each of your functions: date <- as.Date(date, "1970-01-01") and then using x from above try these: lapply(x, first.day.of.quarter) xx <- x; xx[] <- sapply(x, first.day.of.quarter) On Sun, Feb 14, 2010 at 8:59 AM, Dimitri Shvorob <dimitri.shvorob at gmail.com> wrote:> > Dataframe cust has Date-type column open.date. I wish to set up another > column, with (first day of) the quarter of open.date. > > To be comprehensive (of course, improvement suggestions are welcome), > > month = function(date) > { > ?return(as.numeric(format(date,"%m"))) > } > > first.day.of.month = function(date) > { > ?return(date + 1 - as.numeric(format(date,"%d"))) > } > > first.day.of.quarter = function(date) > { > ?t = seq.Date(first.day.of.month(date), by = "-1 month", length > month(date) %% 3) > ?return(t[length(t)]) > } > > Now the main part, > >> cust$open.quarter ?= apply(cust$open.date, 1, FUN = first.day.of.quarter) > Error in apply(cust$open.date, 1, FUN = first.day.of.quarter) : > ?dim(X) must have a positive length > >> cust$open.quarter ?= tapply(cust$open.date, FUN = first.day.of.quarter) > Error in tapply(cust$open.date, FUN = first.day.of.quarter) : > ?element 1 is empty; > ? the part of the args list of 'is.list' being evaluated was: > ? (INDEX) > >> cust$open.quarter ?= lapply(cust$open.date, FUN = first.day.of.quarter) > Error in prettyNum(.Internal(format(x, trim, digits, nsmall, width, 3L, ?: > ?invalid 'trim' argument > > Can anyone suggest the right syntax? > > Thank you. > > -- > View this message in context: http://n4.nabble.com/Newbie-woes-with-apply-tp1555149p1555149.html > Sent from the R help mailing list archive at Nabble.com. > > ______________________________________________ > R-help at r-project.org mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. >