Re: [R] Newbie woes with *apply
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.
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Re: [R] Newbie woes with *apply
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.shvo...@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.
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Re: [R] Newbie woes with *apply
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. __ R-help@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.
Re: [R] Newbie woes with *apply
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.shvo...@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.
--
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http://n4.nabble.com/Newbie-woes-with-apply-tp1555149p1555149.html
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__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
R-help@r-project.org mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
