Extract year from date

Question

How can I remove the first elements from a variable  especially if this variable has a special characters  For example  I have the following column   Date 01 01 2009 01 01 2010 01 01 2011 01 01 2012   I need to have a new column like the following   Date 2009 2010 2011 2012

User · Answer

If you are using the date package  this can be done fairly easily     library date  Date  lt - c  01 01 2009    01 01 2010    01 01 2011    01 01 2012   Date  lt - as date Date  Date    1  1Jan2009 1Jan2010 1Jan2011 1Jan2012 date mdy Date  year    1  2009 2010 2011 2012     be aware that these are now integers and thus different methods may be invoked  str date mdy Date  year    int  1 4  2009 2010 2011 2012 summary Date        First      Last       1Jan2009   1Jan2012   summary date mdy Date  year       Min  1st Qu   Median    Mean 3rd Qu     Max        2009    2010    2010    2010    2011    2012

User · Answer

if all your dates are the same width  you can put the dates in a vector and use substring  Date a  lt - c  01 01 2009    01 01 2010     01 01 2011   substring a 7 10   This takes string and only keeps the characters beginning in position 7 to position 10   output   1   2009   2010   2011

User · Answer

library lubridate  a mdy b  year a    https   cran r-project org web packages lubridate vignettes lubridate html http   vita had co nz papers lubridate pdf

User · Answer

When you convert your variable to Date   date  lt -  as Date  10 30 2018    m  d  Y     you can then cut out the elements you want and make new variables  like year   year  lt - as numeric format date   Y      or month   month  lt - as numeric format date   m

User · Answer

For some time now  you can also only rely on the data table package and its IDate class plus associated functions  Check  as IDate     So  no need to additionally install lubridate  require data table   a  lt - c  quot 01 01 2009 quot    quot 01 01 2010 quot     quot 01 01 2011 quot   year as IDate a    d  m  Y      all data table functions

User · Answer

As discussed in the comments  this can be achieved by converting the entry into Date format and extracting the year  for instance like this   format as Date df1 Date  format   d  m  Y     Y

User · Answer

This is more advice than a specific answer  but my suggestion is to convert dates to date variables immediately  rather than keeping them as strings  This way you can use date  and time  functions on them  rather than trying to use very troublesome workarounds   As pointed out  the lubridate package has nice extraction functions    For some projects  I have found that piecing dates out from the start is helpful  create year  month  day  of month  and day  of week  variables to start with  This can simplify summaries  tables and graphs  because the extraction code is separate from the summary table graph code  and because if you need to change it  you don t have to roll out those changes in multiple spots

[r] Extract year from date

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