Hi all,
I'm having a little bit of trouble with some date conversions and
am hoping someone can help me out. Thanks in advance.
OK, I have two sources of data that provide date info in a csv file
differently. I've attached a small zipped file with two text files
that illustrate both. (Is it
On Sat, Jul 11, 2009 at 10:20 AM, Gabor
Grothendieckggrothendi...@gmail.com wrote:
No attachment appeared. I don't think the list allows zip files
as attachments. Try plain text.
On Sat, Jul 11, 2009 at 1:10 PM, Mark Knechtmarkkne...@gmail.com wrote:
Hi all,
I'm having a little bit of
No attachment appeared. I don't think the list allows zip files
as attachments. Try plain text.
On Sat, Jul 11, 2009 at 1:10 PM, Mark Knechtmarkkne...@gmail.com wrote:
Hi all,
I'm having a little bit of trouble with some date conversions and
am hoping someone can help me out. Thanks in
You want %Y, not %y.
You might also want to look at the zoo package:
library(zoo)
z - read.zoo(Date1.txt, header = TRUE, sep = ,, format = %m/%d/%Y)
or using chron:
library(zoo)
library(chron)
z - read.zoo(Date1.txt, header = TRUE, sep = ,, FUN = as.chron)
There are three vignettes that come
On Sat, Jul 11, 2009 at 12:05 PM, Gabor
Grothendieckggrothendi...@gmail.com wrote:
You want %Y, not %y.
You might also want to look at the zoo package:
library(zoo)
z - read.zoo(Date1.txt, header = TRUE, sep = ,, format = %m/%d/%Y)
or using chron:
library(zoo)
library(chron)
z -
Try:
format(d, %a) or format(d, %A) or as.POSIXlt(d)$wday
There is also day.of.week in chron.
On Sat, Jul 11, 2009 at 3:42 PM, Mark Knecht markkne...@gmail.com wrote:
On Sat, Jul 11, 2009 at 12:05 PM, Gabor
Grothendieckggrothendi...@gmail.com wrote:
You want %Y, not %y.
You might also
Thanks. Will do.
Cheers,
Mark
On Sat, Jul 11, 2009 at 12:49 PM, Gabor
Grothendieckggrothendi...@gmail.com wrote:
Try:
format(d, %a) or format(d, %A) or as.POSIXlt(d)$wday
There is also day.of.week in chron.
On Sat, Jul 11, 2009 at 3:42 PM, Mark Knecht markkne...@gmail.com wrote:
On Sat,
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