Here are some alternatives. They all return numeric years but if you really need a string starting with FY then use paste0(“FY”, result) where result is any of the results below. They all support vector input, i.e. the input dates can be a vector. 1) zoo::as.yearmon The zoo package has a “yearmon” class which represents … Read more
You can use parse_date_time from lubridate: some.dates <- c(“23/11/12”, “20/10/2012”, “22/10/2012″ ,”23/11/12”) parse_date_time(some.dates,c(‘dmy’)) [1] “2012-11-23 UTC” “2012-10-20 UTC” “2012-10-22 UTC” “2012-11-23 UTC” But , Note that the order of format is important : some.dates <- c(“20/10/2012″,”23/11/12”, “22/10/2012″ ,”23/11/12”) parse_date_time(some.dates,c(‘dmY’,’dmy’)) [1] “2012-10-20 UTC” “2012-11-23 UTC” “2012-10-22 UTC” “2012-11-23 UTC” EDIT Internally parse_date_time is using guess_formats (which … Read more
Here is a function that allows you to do this: library(lubridate) x <- mdy(c(“1/2/54″,”1/2/68″,”1/2/69″,”1/2/99″,”1/2/04”)) foo <- function(x, year=1968){ m <- year(x) %% 100 year(x) <- ifelse(m > year %% 100, 1900+m, 2000+m) x } Try it out: x [1] “2054-01-02 UTC” “2068-01-02 UTC” “1969-01-02 UTC” “1999-01-02 UTC” [5] “2004-01-02 UTC” foo(x) [1] “2054-01-02 UTC” “2068-01-02 … Read more
Yes, you found the correct trick: going back a day from the first of the next month. Here is as a one-liner in base R: R> seq(as.Date(“2010-02-01″), length=24, by=”1 month”) – 1 [1] “2010-01-31” “2010-02-28” “2010-03-31” “2010-04-30” “2010-05-31” [6] “2010-06-30” “2010-07-31” “2010-08-31” “2010-09-30” “2010-10-31” [11] “2010-11-30” “2010-12-31” “2011-01-31” “2011-02-28” “2011-03-31” [16] “2011-04-30” “2011-05-31” “2011-06-30” “2011-07-31” … Read more
Here’s a possible solution. The basic idea here is to compare lagged start date with the maximum end date “until now” using the cummax function and create an index that will separate the data into groups data %>% arrange(ID, start) %>% # as suggested by @Jonno in case the data is unsorted group_by(ID) %>% mutate(indx … Read more
You don’t need to use lubridate for this, the base function as.Date handles this type of conversion nicely. The trick is that you have to provide the origin, which in Excel is December 30, 1899. as.Date(42705, origin = “1899-12-30”) # [1] “2016-12-01” If you want to preserve your column types, you can try using the … Read more
The easiest way is to use lubridate: library(lubridate) prods.all$Date2 <- mdy(prods.all$Date2) This function automatically returns objects of class POSIXct and will work with either factors or characters.
Before converting year-week to a date you have to specify a day of the week but more importantly you have to ensure which of the different conventions is being used. Base R’s strptime() function knows 3 definitions of week of the year (but supports only 2 of them on input) and 2 definitions of weekday … Read more