Converting characters to dates










-2















In R, it seems like this should be obvious, but I'm having trouble. I have dates formatted as 1/1/00, 12/31/00, etc., where each part is abbreviated.



When I try to convert it to a date, I get this error:



> headlines$Date <- as.Date(headlines$Date)
Error in charToDate(x) :
character string is not in a standard unambiguous format


I've also tried the below, but get all NAs:



> headlines$Date <- as.Date(headlines$Date,format="%b/%d/%y")


How should I convert this column to dates?










share|improve this question


























    -2















    In R, it seems like this should be obvious, but I'm having trouble. I have dates formatted as 1/1/00, 12/31/00, etc., where each part is abbreviated.



    When I try to convert it to a date, I get this error:



    > headlines$Date <- as.Date(headlines$Date)
    Error in charToDate(x) :
    character string is not in a standard unambiguous format


    I've also tried the below, but get all NAs:



    > headlines$Date <- as.Date(headlines$Date,format="%b/%d/%y")


    How should I convert this column to dates?










    share|improve this question
























      -2












      -2








      -2








      In R, it seems like this should be obvious, but I'm having trouble. I have dates formatted as 1/1/00, 12/31/00, etc., where each part is abbreviated.



      When I try to convert it to a date, I get this error:



      > headlines$Date <- as.Date(headlines$Date)
      Error in charToDate(x) :
      character string is not in a standard unambiguous format


      I've also tried the below, but get all NAs:



      > headlines$Date <- as.Date(headlines$Date,format="%b/%d/%y")


      How should I convert this column to dates?










      share|improve this question














      In R, it seems like this should be obvious, but I'm having trouble. I have dates formatted as 1/1/00, 12/31/00, etc., where each part is abbreviated.



      When I try to convert it to a date, I get this error:



      > headlines$Date <- as.Date(headlines$Date)
      Error in charToDate(x) :
      character string is not in a standard unambiguous format


      I've also tried the below, but get all NAs:



      > headlines$Date <- as.Date(headlines$Date,format="%b/%d/%y")


      How should I convert this column to dates?







      r date






      share|improve this question













      share|improve this question











      share|improve this question




      share|improve this question










      asked Nov 15 '18 at 1:12









      Adam_GAdam_G

      2,302114898




      2,302114898






















          2 Answers
          2






          active

          oldest

          votes


















          1














          You were on the right track by adding the format argument. I'm guessing you realized that the first error ("character string is not in a standard unambiguous format") happened because R doesn't know which of the numbers is the day, month, or the year. Say one of the values was "01/02/03"; there's no way of knowing whether it's 2 January 2003, or 1 February 2003, and so on.



          In this case I think you just need to fix what you're passing to the format argument. %b is the symbol for abbreviated month in text form, not number form (e.g. "Jan" instead of "01"). You need to use %m instead for months stored as numbers. Try this:



          headlines$Date <- as.Date(headlines$Date,format="%m/%d/%y")


          See this page for more info about date formats in R.






          share|improve this answer






























            1














            Replace format = "%b/%d/%y" with format = "%m/%d/%y".



            %b means month as in Jan, Feb, Mar and so on.



            %m is the integer equivalent (1, 2, 3 etc).



            Further reading: https://www.stat.berkeley.edu/~s133/dates.html






            share|improve this answer






















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              2 Answers
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              2 Answers
              2






              active

              oldest

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              active

              oldest

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              active

              oldest

              votes









              1














              You were on the right track by adding the format argument. I'm guessing you realized that the first error ("character string is not in a standard unambiguous format") happened because R doesn't know which of the numbers is the day, month, or the year. Say one of the values was "01/02/03"; there's no way of knowing whether it's 2 January 2003, or 1 February 2003, and so on.



              In this case I think you just need to fix what you're passing to the format argument. %b is the symbol for abbreviated month in text form, not number form (e.g. "Jan" instead of "01"). You need to use %m instead for months stored as numbers. Try this:



              headlines$Date <- as.Date(headlines$Date,format="%m/%d/%y")


              See this page for more info about date formats in R.






              share|improve this answer



























                1














                You were on the right track by adding the format argument. I'm guessing you realized that the first error ("character string is not in a standard unambiguous format") happened because R doesn't know which of the numbers is the day, month, or the year. Say one of the values was "01/02/03"; there's no way of knowing whether it's 2 January 2003, or 1 February 2003, and so on.



                In this case I think you just need to fix what you're passing to the format argument. %b is the symbol for abbreviated month in text form, not number form (e.g. "Jan" instead of "01"). You need to use %m instead for months stored as numbers. Try this:



                headlines$Date <- as.Date(headlines$Date,format="%m/%d/%y")


                See this page for more info about date formats in R.






                share|improve this answer

























                  1












                  1








                  1







                  You were on the right track by adding the format argument. I'm guessing you realized that the first error ("character string is not in a standard unambiguous format") happened because R doesn't know which of the numbers is the day, month, or the year. Say one of the values was "01/02/03"; there's no way of knowing whether it's 2 January 2003, or 1 February 2003, and so on.



                  In this case I think you just need to fix what you're passing to the format argument. %b is the symbol for abbreviated month in text form, not number form (e.g. "Jan" instead of "01"). You need to use %m instead for months stored as numbers. Try this:



                  headlines$Date <- as.Date(headlines$Date,format="%m/%d/%y")


                  See this page for more info about date formats in R.






                  share|improve this answer













                  You were on the right track by adding the format argument. I'm guessing you realized that the first error ("character string is not in a standard unambiguous format") happened because R doesn't know which of the numbers is the day, month, or the year. Say one of the values was "01/02/03"; there's no way of knowing whether it's 2 January 2003, or 1 February 2003, and so on.



                  In this case I think you just need to fix what you're passing to the format argument. %b is the symbol for abbreviated month in text form, not number form (e.g. "Jan" instead of "01"). You need to use %m instead for months stored as numbers. Try this:



                  headlines$Date <- as.Date(headlines$Date,format="%m/%d/%y")


                  See this page for more info about date formats in R.







                  share|improve this answer












                  share|improve this answer



                  share|improve this answer










                  answered Nov 15 '18 at 1:34









                  Boops BoopsBoops Boops

                  15114




                  15114























                      1














                      Replace format = "%b/%d/%y" with format = "%m/%d/%y".



                      %b means month as in Jan, Feb, Mar and so on.



                      %m is the integer equivalent (1, 2, 3 etc).



                      Further reading: https://www.stat.berkeley.edu/~s133/dates.html






                      share|improve this answer



























                        1














                        Replace format = "%b/%d/%y" with format = "%m/%d/%y".



                        %b means month as in Jan, Feb, Mar and so on.



                        %m is the integer equivalent (1, 2, 3 etc).



                        Further reading: https://www.stat.berkeley.edu/~s133/dates.html






                        share|improve this answer

























                          1












                          1








                          1







                          Replace format = "%b/%d/%y" with format = "%m/%d/%y".



                          %b means month as in Jan, Feb, Mar and so on.



                          %m is the integer equivalent (1, 2, 3 etc).



                          Further reading: https://www.stat.berkeley.edu/~s133/dates.html






                          share|improve this answer













                          Replace format = "%b/%d/%y" with format = "%m/%d/%y".



                          %b means month as in Jan, Feb, Mar and so on.



                          %m is the integer equivalent (1, 2, 3 etc).



                          Further reading: https://www.stat.berkeley.edu/~s133/dates.html







                          share|improve this answer












                          share|improve this answer



                          share|improve this answer










                          answered Nov 15 '18 at 1:17









                          12b345b6b7812b345b6b78

                          792116




                          792116



























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