lubridate - How to extract Month from date in R

Question

I am using the lubridate package and applying the month function to extract month from date. I ran the str command on date field and I got

Factor w/ 9498 levels "01/01/1979","01/01/1980",..: 5305 1 1 1 1 1 1 1 1 1 ...

> v1$Date<-month(v1$Date)
Error in as.POSIXlt.character(as.character(x), ...) : 
character string is not in a standard unambiguous format

Here is an example of my data frame

https://drive.google.com/file/d/0B6cqWmwsEk20Q2dHblhXZi14Wk0/edit?usp=sharing

I don't know what I am doing wrong.

question from:https://stackoverflow.com/questions/22603847/how-to-extract-month-from-date-in-r

Answer 1

?month states:

Date-time must be a POSIXct, POSIXlt, Date, Period, chron, yearmon, yearqtr, zoo, zooreg, timeDate, xts, its, ti, jul, timeSeries, and fts objects.

Your object is a factor, not even a character vector (presumably because of stringsAsFactors = TRUE). You have to convert your vector to some datetime class, for instance to POSIXlt:

library(lubridate)
some_date <- c("01/02/1979", "03/04/1980")
month(as.POSIXlt(some_date, format="%d/%m/%Y"))
[1] 2 4

There's also a convenience function dmy, that can do the same (tip proposed by @Henrik):

month(dmy(some_date))
[1] 2 4

Going even further, @IShouldBuyABoat gives another hint that dd/mm/yyyy character formats are accepted without any explicit casting:

month(some_date)
[1] 2 4

For a list of formats, see ?strptime. You'll find that "standard unambiguous format" stands for

The default formats follow the rules of the ISO 8601 international standard which expresses a day as "2001-02-28" and a time as "14:01:02" using leading zeroes as here.