Extract Month and Year from a Zoo::Yearmon Object

Extract month and year from a zoo::yearmon object

Use the format() method for objects of class "yearmon". Here is your example date (properly created!)

date1 <- as.yearmon("Mar 2012", "%b %Y")

Then we can extract the date parts as required:

> format(date1, "%b") ## Month, char, abbreviated
[1] "Mar"
> format(date1, "%Y") ## Year with century
[1] "2012"
> format(date1, "%m") ## numeric month
[1] "03"

These are returned as characters. Where appropriate, wrap in as.numeric() if you want the year or numeric month as a numeric variable, e.g.

> as.numeric(format(date1, "%m"))
[1] 3
> as.numeric(format(date1, "%Y"))
[1] 2012

See ?yearmon and ?strftime for details - the latter explains the placeholder characters you can use.

Identify fiscal year for yearmon object in R

We assume that the fiscal year ends in September so that the fiscal year corresponding to October, November and December is the following calendar year and for other months the fiscal year is the same as the calendar year.

Push the input forward by three months by adding 3/12 to the input yearmon object so that October, November and December get pushed into the next calendar year but no other month is and then format:

library(zoo)
ym <- yearmon(2020 + 0:11/12) # test data: Jan '20, Feb '20, ..., Dec '20

format(ym + 3/12, "FY%y")
##  [1] "FY20" "FY20" "FY20" "FY20" "FY20" "FY20" "FY20" "FY20" "FY20" "FY21"
## [11] "FY21" "FY21"

Extract Month and Year From Date in R

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") ]

Extract month and year from datetime in R

lubridate month and year will work.

as.data.frame(Order.Date) %>%
  mutate(Month = lubridate::month(Order.Date, label = FALSE),
         Year = lubridate::year(Order.Date))

  Order.Date Month Year
1 2011-10-20    10 2011
2 2011-12-25    12 2011
3 2012-04-15     4 2012
4 2012-08-23     8 2012
5 2013-09-25     9 2013

If you want month format as Jan, use month.abb and as January, use month.name

as.data.frame(Order.Date) %>%
  mutate(Month = month.abb[lubridate::month(Order.Date, label = TRUE)],
         Year = lubridate::year(Order.Date))

  Order.Date Month Year
1 2011-10-20   Oct 2011
2 2011-12-25   Dec 2011
3 2012-04-15   Apr 2012
4 2012-08-23   Aug 2012
5 2013-09-25   Sep 2013

as.data.frame(Order.Date) %>%
  mutate(Month = month.name[lubridate::month(Order.Date, label = TRUE)],
         Year = lubridate::year(Order.Date))

  Order.Date     Month Year
1 2011-10-20   October 2011
2 2011-12-25  December 2011
3 2012-04-15     April 2012
4 2012-08-23    August 2012
5 2013-09-25 September 2013

Converting yearmon column to last date of the month in R

If the Date variable is an actual yearmon class vector, from the zoo package, the as.Date.yearmon method can do what you want via its argument frac.

Using your data, and assuming that the Date was originally a character vector

library("zoo")
df <- data.frame(Date = c("2014-07", "2014-08", "2014-09"),
                 Arrivals = c(100, 150, 200))

I convert this to a yearmon vector:

df <- transform(df, Date2 = as.yearmon(Date))

Assuming this is what you have, then you can achieve what you want using as.Date() with frac = 1:

df <- transform(df, Date3 = as.Date(Date2, frac = 1))

which gives:

> df
     Date Arrivals    Date2      Date3
1 2014-07      100 Jul 2014 2014-07-31
2 2014-08      150 Aug 2014 2014-08-31
3 2014-09      200 Sep 2014 2014-09-30

That shows the individual steps. If you only want the final Date this is a one-liner

## assuming `Date` is a `yearmon` object
df <- transform(df, Date = as.Date(Date, frac = 1))
## or if not a `yearmon`
df <- transform(df, Date = as.Date(as.yearmon(Date), frac = 1))

The argument frac in the fraction of the month to assign to the resulting dates when converting from yearmon objects to Date objects. Hence, to get the first day of the month, rather than convert to a character and paste on "-01" as your Question showed, it's better to coerce to a Date object with frac = 0.

If the Date in your df is not a yearmon class object, then you can solve your problem by converting it to one and then using the as.Date() method as described above.

Extract Month from Week and Year number

This extracts the month of the first day of the week, you could add a constant or something if you wanted the 2nd or 3rd day of the week instead.

df %>% 
  mutate(Month = month(ymd(Year * 10000 + 0101) + Week * 7))

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