[R] calculating the number of days from dates

[Bob Green]

Hello,

I gather variants of this question have been asked previously. I have
done some reading but only became more confused, as I suspect what I
am trying to do is more basic than other applications.

The following code readily calculates the difference in days between two dates:

newdays <- ISOdate(2005, 5,12) - ISOdate(2006, 12, 22)

However, I wanted to be able to deduct the dates in one variable from
the dates in another variable, resulting in a new variable - e.g the
difference in days between the two dates. Below is a sample of my
data. My questions:

1. I tried changing the data to dates via as.Date. Is this necessary
or do I need to alter the date format itself, e.g to 12/12/78 or some
other format?
2. I gather there are various packages as well as date formats.What
is the most straight forward approach to calculate the difference
between two dates, as below.

> dates <- read.csv("c:\\dates.csv",header=T)
> dates
v1 v2
1 12/12/1978 12/12/2005
2 23/01/1965 23/09/2001
3 24/12/2004 16/03/2007
4 3/03/2003 4/04/2004
5 8/11/2006 1/05/2007

> class(dates$v1)
[1] "factor"
> class(dates$v2)
[1] "factor"

> dates <- read.csv("c:\\dates.csv",header=T,
as.Date(as.character(dates) "%d/%m/%Y"))
Error: syntax error, unexpected STR_CONST, expecting ',' in "dates <-
read.csv("c:\\dates.csv",header=T, as.Date(as.character(dates) "%d/%m/%Y""
>

Any assistance is much appreciated,

Bob

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[Henrique Dallazuanna]

Try this:

dates <- read.csv("c:\\dates.csv",header=T)

dates[,1] <- as.Date(dates[,1], "%d/%m/%Y")
dates[,2] <- as.Date(dates[,2], "%d/%m/%Y")
transform(dates,
Dif=V2-V1)

--
Henrique Dallazuanna
Curitiba-Paraná-Brasil
25° 25' 40" S 49° 16' 22" O

[Knut Krueger]

Bob Green schrieb:

>
> > dates <- read.csv("c:\\dates.csv",header=T)
> > dates
> v1 v2
> 1 12/12/1978 12/12/2005
> 2 23/01/1965 23/09/2001
> 3 24/12/2004 16/03/2007
> 4 3/03/2003 4/04/2004
> 5 8/11/2006 1/05/2007
>
> > class(dates$v1)
> [1] "factor"
> > class(dates$v2)
> [1] "factor"
>
>

What about chron library:

dts <- dates(c("02/27/92", "02/27/92", "01/14/92",
"02/28/92", "02/01/92"))
dts
# [1] 02/27/92 02/27/92 01/14/92 02/28/92 02/01/92
tms <- times(c("23:03:20", "22:29:56", "01:03:30",
"18:21:03", "16:56:26"))
tms
# [1] 23:03:20 22:29:56 01:03:30 18:21:03 16:56:26
x <- chron(dates = dts, times = tms)
x
# [1] (02/27/92 23:03:19) (02/27/92 22:29:56) (01/14/92 01:03:30)
# [4] (02/28/92 18:21:03) (02/01/92 16:56:26)

# We can add or subtract scalars (representing days) to dates or
# chron objects:
c(dts[1], dts[1] + 10)
# [1] 02/27/92 03/08/92
dts[1] - 31
# [1] 01/27/92

Knut

[Martin Maechler]

>>>>> "KK" == Knut Krueger <r...@family-krueger.com>
>>>>> on Sat, 15 Dec 2007 08:39:02 +0100 writes:

KK> Bob Green schrieb:

>>
>> > dates <- read.csv("c:\\dates.csv",header=T)
>> > dates
>> v1 v2
>> 1 12/12/1978 12/12/2005
>> 2 23/01/1965 23/09/2001
>> 3 24/12/2004 16/03/2007
>> 4 3/03/2003 4/04/2004
>> 5 8/11/2006 1/05/2007
>>
>> > class(dates$v1)
>> [1] "factor"
>> > class(dates$v2)
>> [1] "factor"
>>
>>

KK> What about chron library:

it's a >> package << , not a library, please!

and as Henrique has shown it's really not needed for the question.
There's the "Date" (S3) class, and even a "difftime" one
for time *differences*
See
?as.Date
?difftime
and also note the output of
methods(class = "Date")

Martin

KK> dts <- dates(c("02/27/92", "02/27/92", "01/14/92",
KK> "02/28/92", "02/01/92"))
KK> dts
KK> # [1] 02/27/92 02/27/92 01/14/92 02/28/92 02/01/92
KK> tms <- times(c("23:03:20", "22:29:56", "01:03:30",
KK> "18:21:03", "16:56:26"))
KK> tms
KK> # [1] 23:03:20 22:29:56 01:03:30 18:21:03 16:56:26
KK> x <- chron(dates = dts, times = tms)
KK> x
KK> # [1] (02/27/92 23:03:19) (02/27/92 22:29:56) (01/14/92 01:03:30)
KK> # [4] (02/28/92 18:21:03) (02/01/92 16:56:26)

KK> # We can add or subtract scalars (representing days) to dates or
KK> # chron objects:
KK> c(dts[1], dts[1] + 10)
KK> # [1] 02/27/92 03/08/92
KK> dts[1] - 31
KK> # [1] 01/27/92

KK> Knut

KK> ______________________________________________
KK> R-h...@r-project.org mailing list
KK> https://stat.ethz.ch/mailman/listinfo/r-help
KK> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
KK> and provide commented, minimal, self-contained, reproducible code.

[Knut Krueger]

> it's a >> package << , not a library, please!
>
>

Sorry for using library instead package, but

library() is one command for using packages.

Therefore I (and it seems that i am not the only one) used library instead package.

Knut

[bogdan romocea]

> Sorry for using library instead package, but
> library() is one command for using packages.

... which is why all efforts to make folks say "package" instead of >>
"library" << are doomed to fail, IMHO. Besides, in English, "library"
also means "a collection of software or data usually reflecting a
specific theme or application" (#9 on the list from
http://dictionary.reference.com/ ). Therefore:
> "library" == "package"
[1] TRUE!
and just about the only way to clear up the "confusion" would be to
rename library() to package(), and replace "library" with "folder" or
"directory".

[Rolf Turner]

On 19/12/2007, at 6:46 AM, bogdan romocea wrote:

>> Sorry for using library instead package, but
>> library() is one command for using packages.
>
> ... which is why all efforts to make folks say "package" instead of >>
> "library" << are doomed to fail, IMHO.

Yes, but it gives so much pleasure to those who appreciate
the distinction to rail at those who don't! :-)

cheers,

Rolf Turner

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