Or even:
> library(chron)
> ch <- chron(HouseDates)
> years(ch)
[1] 1990 1991 1992 1993 1994 1995 1996
Levels: 1990 < 1991 < 1992 < 1993 < 1994 < 1995 < 1996
> # or
> with(month.day.year(ch), year)
[1] 1990 1991 1992 1993 1994 1995 1996
On Thu, Jun 4, 2009 at 4:02 AM, ONKELINX, Thierry
<thierry.onkel...@inbo.be> wrote:
> Dear Jason,
>
> Have a look at years() from the chron package.
>
> library(chron)
> HouseDates <- c("02/27/90", "02/27/91", "01/14/92", "02/28/93",
> "02/01/94", "02/01/95", "02/01/96")
> HouseDatesFormatted<-as.Date(HouseDates, "%m/%d/%y")
> years(HouseDates)
>
> HTH,
>
> Thierry
> ------------------------------------------------------------------------
> ----
> ir. Thierry Onkelinx
> Instituut voor natuur- en bosonderzoek / Research Institute for Nature
> and Forest
> Cel biometrie, methodologie en kwaliteitszorg / Section biometrics,
> methodology and quality assurance
> Gaverstraat 4
> 9500 Geraardsbergen
> Belgium
> tel. + 32 54/436 185
> thierry.onkel...@inbo.be
> www.inbo.be
>
> To call in the statistician after the experiment is done may be no more
> than asking him to perform a post-mortem examination: he may be able to
> say what the experiment died of.
> ~ Sir Ronald Aylmer Fisher
>
> The plural of anecdote is not data.
> ~ Roger Brinner
>
> The combination of some data and an aching desire for an answer does not
> ensure that a reasonable answer can be extracted from a given body of
> data.
> ~ John Tukey
>
> -----Oorspronkelijk bericht-----
> Van: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
> Namens Jason Rupert
> Verzonden: donderdag 4 juni 2009 4:45
> Aan: R-help@r-project.org
> Onderwerp: [R] Getting a column of values from a list - think I'm doing
> it thehard way
>
>
> Example code it shown below.
>
> I think I am doing this the hard way. I'm just trying to get the full
> year value from an array of dates. An example array is shown below.
> Right now, I'm using a "for" loop to pull the year out of a list where
> the dates were split up into their individual components.
>
> This seems to work, but just wondering if there is an easier way.
>
> Thanks for any insights.
>
> #*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~*~
> HouseDates <- c("02/27/90", "02/27/91", "01/14/92", "02/28/93",
> "02/01/94", "02/01/95", "02/01/96")
>
> # ?as.Date
> HouseDatesFormatted<-as.Date(HouseDates, "%m/%d/%y")
>
> HouseDatesFormatted
>
> HouseDatesList<-strsplit(as.character(HouseDatesFormatted), "-",
> fixed=TRUE)
>
> HouseYear_array<-NULL
> length_array<-length(HouseDatesList)
> for(ii in 1:length_array)
> {
> HouseYear<-HouseDatesList[[ii]][1]
>
> HouseYear_array<-c(HouseYear_array, HouseYear) }
>
> as.character(HouseYear_array)
>
> # Desired:
> # [1] "1990" "1991" "1992" "1993" "1994" "1995" "1996"
>
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