How to get date in month-year format?

Question

I got the data from someone which has following types of dates:

"Jan-20", "Feb-19", "May-18"

I want to convert them into following format:

"01-20", "02-19", "05-18"

But whenever I try to convert the column to as.date it returns NA values.

eg

df<- data.frame(MonthYear = c("Jan-20", "Feb-19", "May-18"))
df$MonthYear<-as.Date(df$MonthYear, format = "%b-%y")

Output:

MonthYear
1      <NA>
2      <NA>
3      <NA>

So I used the following code and added day as well. But I don't want the date:

df<- data.frame(MonthYear = c("Jan-20", "Feb-19", "May-18"))
df$MonthYear<-paste(as.character(df$MonthYear))
df$MonthYear<-paste("01-",as.character(df$MonthYear))
df$MonthYear<-as.Date(df$MonthYear, format = "%d- %b-%y")

Output:

MonthYear
1 2020-01-01
2 2019-02-01
3 2018-05-01

But my required output is as follows:

MonthYear
1 2020-01
2 2019-02
3 2018-05

Answer 1

Using as.yearmon() from the zoo package (and the magrittr pipe):

library(zoo)
library(magrittr)
as.yearmon(df$MonthYear, "%b-%y") %>%
 format(., "%Y-%m")


[1] "2020-01" "2019-02" "2018-05"

Can also be done without the '.' used as a placeholder for the left hand side of the pipe. It was left in as these functions aren't typical tidyverse piping functions.

as.yearmon(df$MonthYear, "%b-%y") %>%
 format("%Y-%m")

Or without piping at all, and using nested functions (as pointed out by @Sotos). I find them harder to read, and usually have the tidyverse (and therefore %>% pipes) loaded anyway.

format(as.yearmon(df$MonthYear, "%b-%y"), "%Y-%m")

Answer 2

You could paste an arbitrary date, convert to Date and then use format

format(as.Date(paste0("01-",df$MonthYear), "%d-%b-%y"), "%Y-%m")
#[1] "2020-01" "2019-02" "2018-05"

Answer 3

Maybe you can try the following using gsub() to keep the year and month only, ie,

format(gsub("(.*)-\\d+","\\1",df$MonthYear),format = "%Y-%m")

or just

gsub("(.*)-\\d+","\\1",df$MonthYear)

such that

[1] "2020-01" "2019-02" "2018-05"

DATA

df <- structure(list(MonthYear = structure(c(18262, 17928, 17652), class = "Date")), class = "data.frame", row.names = c(NA, 
-3L))

Answer 4

x <- c("Jan-20", "Feb-19", "May-18")
for (i in seq_along(month.abb)) x <- sub(month.abb[i], sprintf("%02d", i), x) 
# Can add fixed = TRUE for speed

x
# [1] "01-20" "02-19" "05-18"