Displaying 1 result from an estimated 1 matches for "houseyear".
2009 Jun 04
3
Getting a column of values from a list - think I'm doing it the hard way
...quot;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"...