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Handling Dates in R
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| # Sample dates | |
| dates <- c("09/27/99", "03/15/08") | |
| newDates <- as.Date(dates, format = "%m/%d/%y") # <-- Must change the format based on the format of what's being referenced | |
| newDates # R output will be ISO standard dates in the format "%Y-%m-%d" | |
| # Similar example to the last, but must be handled differently | |
| dates <- c("09/27/1999", "03/15/2008") | |
| newDates <- as.Date(dates, format = "%m/%d/%Y") # <-- the only difference here is the "Y" is capitalized | |
| newDates # R output will be ISO standard dates in the format "%Y-%m-%d" | |
| # Another example | |
| dates <- c("Nov 7 1941", "Apr 16 2012") | |
| newDates <- as.Date(dates, format = "%B %d %Y") | |
| newDates | |
| # from Excel | |
| dates <- c(41567, 42001) | |
| newDates <- as.Date(dates, origin = "1899-12-30") | |
| newDates | |
| # Now, change the default output of the new dates | |
| format(newDates, "%a %b %d %Y") | |
| # Just for fun...Build function to test if input year is a leap year | |
| is_leap_year <- function(year){ | |
| if (year %% 400 == 0) {output = "TRUE" | |
| } | |
| else if (year %% 100 == 0) {output = "FALSE" | |
| } | |
| else if (year %% 4 == 0) {output = "TRUE" | |
| } | |
| else {output = "FALSE" | |
| } | |
| return(output) | |
| } | |
| # Return logical for current year | |
| is_leap_year(as.numeric(format(Sys.time(),"%Y"))) |
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