Changing date format in R

Question

I have some very simple data in R that needs to have its date format changed    date midpoint 1   31 08 2011   0 8378 2   31 07 2011   0 8457 3   30 06 2011   0 8147 4   31 05 2011   0 7970 5   30 04 2011   0 7877 6   31 03 2011   0 7411 7   28 02 2011   0 7624 8   31 01 2011   0 7665 9   31 12 2010   0 7500 10  30 11 2010   0 7734 11  31 10 2010   0 7511 12  30 09 2010   0 7263 13  31 08 2010   0 7158 14  31 07 2010   0 7110 15  30 06 2010   0 6921 16  31 05 2010   0 7005 17  30 04 2010   0 7113 18  31 03 2010   0 7027 19  28 02 2010   0 6973 20  31 01 2010   0 7260 21  31 12 2009   0 7154 22  30 11 2009   0 7287 23  31 10 2009   0 7375   Rather than  d  m  Y  I would like it in the standard R format of  Y- m- d  How can I make this change  I have tried   nzd date  lt - format as Date nzd date     Y  m  d     But that just cut off the year and added zeros to the day     1   0031 08 20   0031 07 20   0030 06 20   0031 05 20   0030 04 20    6   0031 03 20   0028 02 20   0031 01 20   0031 12 20   0030 11 20    11   0031 10 20   0030 09 20   0031 08 20   0031 07 20   0030 06 20    16   0031 05 20   0030 04 20   0031 03 20   0028 02 20   0031 01 20    21   0031 12 20   0030 11 20   0031 10 20   0030 09 20   0031 08 20    26   0031 07 20   0030 06 20   0031 05 20   0030 04 20   0031 03 20    31   0028 02 20   0031 01 20   0031 12 20   0030 11 20   0031 10 20    36   0030 09 20   0031 08 20   0031 07 20   0030 06 20   0031 05 20    Thanks

User · Answer

You could also use the parse date time function from the lubridate package   library lubridate  day lt - 31 08 2011  as Date parse date time day  dmy     1   2011-08-31    parse date time returns a POSIXct object  so we use as Date to get a date object  The first argument of parse date time specifies a date vector  the second argument specifies the order in which your format occurs  The orders argument makes parse date time very flexible

User · Answer

nzd date  lt - format as Date nzd date     Y  m  d     In the above piece of code  there are two mistakes  First of all  when you are reading nzd date inside as Date you are not mentioning in what format you are feeding it the date  So  it tries it s default set format to read it  If you see the help doc   as Date you will see      format       A character string  If not specified  it will try   Y- m- d        then   Y  m  d  on the first non-NA element  and give an error       if neither works  Otherwise  the processing is via strptime   The second mistake is  even though you would like to read it in  Y- m- d format  inside format you wrote   Y  m  d     Now  the correct way of doing it is     gt  nzd  lt - data frame date c  31 08 2011    31 07 2011    30 06 2011                                             mid c 0 8378 0 8457 0 8147    gt  nzd         date    mid 1 31 08 2011 0 8378 2 31 07 2011 0 8457 3 30 06 2011 0 8147  gt  nzd date  lt - format as Date nzd date  format     d  m  Y      Y- m- d    gt  head nzd          date    mid 1 2011-08-31 0 8378 2 2011-07-31 0 8457 3 2011-06-30 0 8147

User · Answer

This is really easy using package lubridate  All you have to do is tell R what format your date is already in  It then converts it into the standard format  nzd date  lt - dmy nzd date   that s it

User · Answer

After reading your data in via a textConnection  the following seems to work   dat  lt - read table textConnection txt   header   TRUE  dat date  lt - strptime dat date  format    d  m  Y   format dat date  format   Y- m- d     gt  format dat date  format   Y- m- d     1   2011-08-31   2011-07-31   2011-06-30   2011-05-31   2011-04-30   2011-03-31    7   2011-02-28   2011-01-31   2010-12-31   2010-11-30   2010-10-31   2010-09-30   13   2010-08-31   2010-07-31   2010-06-30   2010-05-31   2010-04-30   2010-03-31   19   2010-02-28   2010-01-31   2009-12-31   2009-11-30   2009-10-31    gt  str dat   data frame     23 obs  of  2 variables     date      POSIXlt  format   2011-08-31   2011-07-31   2011-06-30         midpoint  num  0 838 0 846 0 815 0 797 0 788

User · Answer

I believe that  nzd date  lt - as Date nzd date  format     d  m  Y     is sufficient

User · Answer

There are two steps here    Parse the data  Your example is not fully reproducible  is the data in a file  or the variable in a text or factor variable   Let us assume the latter  then if you data frame is called X  you can do     X newdate  lt - strptime as character X date     d  m  Y      Now the newdate column should be of type Date    Format the data   That is a matter of calling format   or strftime        format X newdate    Y- m- d      A more complete example   R gt  nzd  lt - data frame date c  31 08 2011    31 07 2011    30 06 2011                          mid c 0 8378 0 8457 0 8147   R gt  nzd         date    mid 1 31 08 2011 0 8378 2 31 07 2011 0 8457 3 30 06 2011 0 8147 R gt  nzd newdate  lt - strptime as character nzd date     d  m  Y   R gt  nzd txtdate  lt - format nzd newdate    Y- m- d   R gt  nzd         date    mid    newdate    txtdate 1 31 08 2011 0 8378 2011-08-31 2011-08-31 2 31 07 2011 0 8457 2011-07-31 2011-07-31 3 30 06 2011 0 8147 2011-06-30 2011-06-30 R gt     The difference between columns three and four is the type  newdate is of class Date whereas txtdate is character

User · Answer

Using one line to convert the dates to preferred format   nzd date  lt - format as Date nzd date  format   d  m  Y     Y  m  d

[r] Changing date format in R

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