Compute monthly averages from daily data

MB123 picture MB123 · May 20, 2013 · Viewed 26.3k times · Source

I have this dataframe "df1" as example which is actually part of a much larger one (15 years):

             X1          X2
3798 2009-12-29           0
3799 2009-12-30           0
3800 2009-12-31           0 
3802 2010-01-02           0
3803 2010-01-03         2.1
3804 2010-01-04           0
3805 2010-01-05           0
3806 2010-01-06           0
3807 2010-01-07           0
3808 2010-01-08           0
3809 2010-01-09           0
3810 2010-01-10         6.8
3811 2010-01-12           0
3812 2010-01-13           0
3813 2010-01-14        17.7
3814 2010-01-16           0
3815 2010-01-17           0
3816 2010-01-18         1.5
3817 2010-01-19           0
3818 2010-01-20           0
3819 2010-01-21           0
3820 2010-01-22           0
3821 2010-01-23           0
3822 2010-01-24           0
3823 2010-01-25           0
3824 2010-01-26           0
3825 2010-01-27         4.5
3826 2010-01-28           0
3827 2010-01-29           0
3828 2010-01-31           0
3829 2010-02-01           0
3830 2010-02-03           0
3831 2010-02-04           0
3832 2010-02-05           0
3833 2010-02-07           0
3834 2010-02-08           0
3835 2010-02-09         1.2  

I want to use this dataframe to create a new one "df2" with averages per month. Does anyone know how to do this? Help would be great!

Answer

Simon O'Hanlon picture Simon O'Hanlon · May 20, 2013

One way, using base R would be to make sure your dates are of class Date or similar ( e.g. POSIXct) if you haven't already, and then to extract the months and years (as your data spans more than one year) and aggregate like so:

#  Convert to date if not already
df1$X1 <- as.Date(df1$X1)

#  Get months
df1$Month <- months(df1$X1)

#  Get years
df1$Year <- format(df1$X1,format="%y")

#  Aggregate 'X2' on months and year and get mean
aggregate( X2 ~ Month + Year , df1 , mean )
#    Month Year        X2
#1 December   09 0.0000000
#2 February   10 0.1714286
#3  January   10 1.2074074

There are quite a few ways of doing this if you have a look around.