Extract Month and Year From Date in R

Solution 1:

This will add a new column to your data.frame with the specified format.

df$Month_Yr <- format(as.Date(df$Date), "%Y-%m")

df
#>   ID       Date Month_Yr
#> 1  1 2004-02-06  2004-02
#> 2  2 2006-03-14  2006-03
#> 3  3 2007-07-16  2007-07

# your data sample
  df <- data.frame( ID=1:3,Date = c("2004-02-06" , "2006-03-14" , "2007-07-16") )

a simple example:

dates <- "2004-02-06"

format(as.Date(dates), "%Y-%m")
> "2004-02"

side note: the data.table approach can be quite faster in case you're working with a big dataset.

library(data.table)
setDT(df)[, Month_Yr := format(as.Date(Date), "%Y-%m") ]

Solution 2:

Use substring?

d = "2004-02-06"
substr(d,0,7)
>"2004-02"

Solution 3:

Here's another solution using a package solely dedicated to working with dates and times in R:

library(tidyverse)
library(lubridate)

(df <- tibble(ID = 1:3, Date = c("2004-02-06" , "2006-03-14", "2007-07-16")))
#> # A tibble: 3 x 2
#>      ID Date      
#>   <int> <chr>     
#> 1     1 2004-02-06
#> 2     2 2006-03-14
#> 3     3 2007-07-16

df %>%
  mutate(
    Date = ymd(Date),
    Month_Yr = format_ISO8601(Date, precision = "ym")
  )
#> # A tibble: 3 x 3
#>      ID Date       Month_Yr
#>   <int> <date>     <chr>   
#> 1     1 2004-02-06 2004-02 
#> 2     2 2006-03-14 2006-03 
#> 3     3 2007-07-16 2007-07

^{Created on 2020-09-01 by the reprex package (v0.3.0)}

Solution 4:

The zoo package has the function of as.yearmon can help to convert.

require(zoo)

df$ym <- as.yearmon(df$date, "%Y %m")