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" ______________________________________________ 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. Dit bericht en eventuele bijlagen geven enkel de visie van de schrijver weer en binden het INBO onder geen enkel beding, zolang dit bericht niet bevestigd is door een geldig ondertekend document. The views expressed in this message and any annex are purely those of the writer and may not be regarded as stating an official position of INBO, as long as the message is not confirmed by a duly signed document. ______________________________________________ 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.