Formatting dates using a condition











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I am using "R" to format a character variable that has two different kinds of date formats (MM-DD-YYYY & YYYY-MM-DD). The second is an excel origin date.



DateVar <- c("12-07-2017", "43229", "43137", "03-27-2018")


I created vector using grepl to identify both types and then a for loop to apply the as.date function to only the "excel origin dates".



indicator <- !grepl("-", DateVar)

for(i in indicator == TRUE){
as.date(DateVar, origin = "1899-12-30")


It is not working for me however, so I am looking if someone can point me in the right direction.



Thanks.










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    up vote
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    down vote

    favorite












    I am using "R" to format a character variable that has two different kinds of date formats (MM-DD-YYYY & YYYY-MM-DD). The second is an excel origin date.



    DateVar <- c("12-07-2017", "43229", "43137", "03-27-2018")


    I created vector using grepl to identify both types and then a for loop to apply the as.date function to only the "excel origin dates".



    indicator <- !grepl("-", DateVar)

    for(i in indicator == TRUE){
    as.date(DateVar, origin = "1899-12-30")


    It is not working for me however, so I am looking if someone can point me in the right direction.



    Thanks.










    share|improve this question
























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      I am using "R" to format a character variable that has two different kinds of date formats (MM-DD-YYYY & YYYY-MM-DD). The second is an excel origin date.



      DateVar <- c("12-07-2017", "43229", "43137", "03-27-2018")


      I created vector using grepl to identify both types and then a for loop to apply the as.date function to only the "excel origin dates".



      indicator <- !grepl("-", DateVar)

      for(i in indicator == TRUE){
      as.date(DateVar, origin = "1899-12-30")


      It is not working for me however, so I am looking if someone can point me in the right direction.



      Thanks.










      share|improve this question













      I am using "R" to format a character variable that has two different kinds of date formats (MM-DD-YYYY & YYYY-MM-DD). The second is an excel origin date.



      DateVar <- c("12-07-2017", "43229", "43137", "03-27-2018")


      I created vector using grepl to identify both types and then a for loop to apply the as.date function to only the "excel origin dates".



      indicator <- !grepl("-", DateVar)

      for(i in indicator == TRUE){
      as.date(DateVar, origin = "1899-12-30")


      It is not working for me however, so I am looking if someone can point me in the right direction.



      Thanks.







      r loops date






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      asked Nov 19 at 13:11









      SteveM

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          Couple of things: The for loop is unnecessary - just subset DateVar with [indicator]. Second, it's as.Date, not as.date (note the "D"). Third, since it's a character vector, you need to pass the origin numbers through as.integer for as.Date to be able to work with them:



          as.Date(as.integer(DateVar[indicator]), origin = "1899-12-30")


          (or, without the intervening indicator assignment:



          as.Date(as.integer(DateVar[!grepl("-",DateVar)]), origin = "1899-12-30")

          [1] "2018-05-09" "2018-02-06"


          If you wish to input these dates back into DateVar, you again use the subset function:



          DateVar[indicator]<-format(as.Date(as.integer(DateVar[indicator]), origin = "1899-12-30"), "%m-%d-%Y")





          share|improve this answer























          • Thank you for your detailed reply as I am fairly new to "R". This appears to work for my case, however my followup question would be "How can I apply to this my existing variable?". I tried to use dplyr::mutate and apply the as.Date function you provided above but I get an error message.
            – SteveM
            Nov 19 at 14:07










          • You want to change DateVar so that the origin dates become class Date?
            – iod
            Nov 19 at 14:10










          • Yes - or alternatively change the other date format be be YYYY-MM-DD since that is the format the excel origin dates are in. I just want to have a consistent date format for my DateVar variable.
            – SteveM
            Nov 19 at 14:18










          • Added code above.
            – iod
            Nov 19 at 14:19










          • This worked beautifully. Thanks so much!
            – SteveM
            Nov 19 at 14:40











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          1 Answer
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          active

          oldest

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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes








          up vote
          1
          down vote



          accepted










          Couple of things: The for loop is unnecessary - just subset DateVar with [indicator]. Second, it's as.Date, not as.date (note the "D"). Third, since it's a character vector, you need to pass the origin numbers through as.integer for as.Date to be able to work with them:



          as.Date(as.integer(DateVar[indicator]), origin = "1899-12-30")


          (or, without the intervening indicator assignment:



          as.Date(as.integer(DateVar[!grepl("-",DateVar)]), origin = "1899-12-30")

          [1] "2018-05-09" "2018-02-06"


          If you wish to input these dates back into DateVar, you again use the subset function:



          DateVar[indicator]<-format(as.Date(as.integer(DateVar[indicator]), origin = "1899-12-30"), "%m-%d-%Y")





          share|improve this answer























          • Thank you for your detailed reply as I am fairly new to "R". This appears to work for my case, however my followup question would be "How can I apply to this my existing variable?". I tried to use dplyr::mutate and apply the as.Date function you provided above but I get an error message.
            – SteveM
            Nov 19 at 14:07










          • You want to change DateVar so that the origin dates become class Date?
            – iod
            Nov 19 at 14:10










          • Yes - or alternatively change the other date format be be YYYY-MM-DD since that is the format the excel origin dates are in. I just want to have a consistent date format for my DateVar variable.
            – SteveM
            Nov 19 at 14:18










          • Added code above.
            – iod
            Nov 19 at 14:19










          • This worked beautifully. Thanks so much!
            – SteveM
            Nov 19 at 14:40















          up vote
          1
          down vote



          accepted










          Couple of things: The for loop is unnecessary - just subset DateVar with [indicator]. Second, it's as.Date, not as.date (note the "D"). Third, since it's a character vector, you need to pass the origin numbers through as.integer for as.Date to be able to work with them:



          as.Date(as.integer(DateVar[indicator]), origin = "1899-12-30")


          (or, without the intervening indicator assignment:



          as.Date(as.integer(DateVar[!grepl("-",DateVar)]), origin = "1899-12-30")

          [1] "2018-05-09" "2018-02-06"


          If you wish to input these dates back into DateVar, you again use the subset function:



          DateVar[indicator]<-format(as.Date(as.integer(DateVar[indicator]), origin = "1899-12-30"), "%m-%d-%Y")





          share|improve this answer























          • Thank you for your detailed reply as I am fairly new to "R". This appears to work for my case, however my followup question would be "How can I apply to this my existing variable?". I tried to use dplyr::mutate and apply the as.Date function you provided above but I get an error message.
            – SteveM
            Nov 19 at 14:07










          • You want to change DateVar so that the origin dates become class Date?
            – iod
            Nov 19 at 14:10










          • Yes - or alternatively change the other date format be be YYYY-MM-DD since that is the format the excel origin dates are in. I just want to have a consistent date format for my DateVar variable.
            – SteveM
            Nov 19 at 14:18










          • Added code above.
            – iod
            Nov 19 at 14:19










          • This worked beautifully. Thanks so much!
            – SteveM
            Nov 19 at 14:40













          up vote
          1
          down vote



          accepted







          up vote
          1
          down vote



          accepted






          Couple of things: The for loop is unnecessary - just subset DateVar with [indicator]. Second, it's as.Date, not as.date (note the "D"). Third, since it's a character vector, you need to pass the origin numbers through as.integer for as.Date to be able to work with them:



          as.Date(as.integer(DateVar[indicator]), origin = "1899-12-30")


          (or, without the intervening indicator assignment:



          as.Date(as.integer(DateVar[!grepl("-",DateVar)]), origin = "1899-12-30")

          [1] "2018-05-09" "2018-02-06"


          If you wish to input these dates back into DateVar, you again use the subset function:



          DateVar[indicator]<-format(as.Date(as.integer(DateVar[indicator]), origin = "1899-12-30"), "%m-%d-%Y")





          share|improve this answer














          Couple of things: The for loop is unnecessary - just subset DateVar with [indicator]. Second, it's as.Date, not as.date (note the "D"). Third, since it's a character vector, you need to pass the origin numbers through as.integer for as.Date to be able to work with them:



          as.Date(as.integer(DateVar[indicator]), origin = "1899-12-30")


          (or, without the intervening indicator assignment:



          as.Date(as.integer(DateVar[!grepl("-",DateVar)]), origin = "1899-12-30")

          [1] "2018-05-09" "2018-02-06"


          If you wish to input these dates back into DateVar, you again use the subset function:



          DateVar[indicator]<-format(as.Date(as.integer(DateVar[indicator]), origin = "1899-12-30"), "%m-%d-%Y")






          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited Nov 19 at 20:45

























          answered Nov 19 at 13:23









          iod

          3,3741620




          3,3741620












          • Thank you for your detailed reply as I am fairly new to "R". This appears to work for my case, however my followup question would be "How can I apply to this my existing variable?". I tried to use dplyr::mutate and apply the as.Date function you provided above but I get an error message.
            – SteveM
            Nov 19 at 14:07










          • You want to change DateVar so that the origin dates become class Date?
            – iod
            Nov 19 at 14:10










          • Yes - or alternatively change the other date format be be YYYY-MM-DD since that is the format the excel origin dates are in. I just want to have a consistent date format for my DateVar variable.
            – SteveM
            Nov 19 at 14:18










          • Added code above.
            – iod
            Nov 19 at 14:19










          • This worked beautifully. Thanks so much!
            – SteveM
            Nov 19 at 14:40


















          • Thank you for your detailed reply as I am fairly new to "R". This appears to work for my case, however my followup question would be "How can I apply to this my existing variable?". I tried to use dplyr::mutate and apply the as.Date function you provided above but I get an error message.
            – SteveM
            Nov 19 at 14:07










          • You want to change DateVar so that the origin dates become class Date?
            – iod
            Nov 19 at 14:10










          • Yes - or alternatively change the other date format be be YYYY-MM-DD since that is the format the excel origin dates are in. I just want to have a consistent date format for my DateVar variable.
            – SteveM
            Nov 19 at 14:18










          • Added code above.
            – iod
            Nov 19 at 14:19










          • This worked beautifully. Thanks so much!
            – SteveM
            Nov 19 at 14:40
















          Thank you for your detailed reply as I am fairly new to "R". This appears to work for my case, however my followup question would be "How can I apply to this my existing variable?". I tried to use dplyr::mutate and apply the as.Date function you provided above but I get an error message.
          – SteveM
          Nov 19 at 14:07




          Thank you for your detailed reply as I am fairly new to "R". This appears to work for my case, however my followup question would be "How can I apply to this my existing variable?". I tried to use dplyr::mutate and apply the as.Date function you provided above but I get an error message.
          – SteveM
          Nov 19 at 14:07












          You want to change DateVar so that the origin dates become class Date?
          – iod
          Nov 19 at 14:10




          You want to change DateVar so that the origin dates become class Date?
          – iod
          Nov 19 at 14:10












          Yes - or alternatively change the other date format be be YYYY-MM-DD since that is the format the excel origin dates are in. I just want to have a consistent date format for my DateVar variable.
          – SteveM
          Nov 19 at 14:18




          Yes - or alternatively change the other date format be be YYYY-MM-DD since that is the format the excel origin dates are in. I just want to have a consistent date format for my DateVar variable.
          – SteveM
          Nov 19 at 14:18












          Added code above.
          – iod
          Nov 19 at 14:19




          Added code above.
          – iod
          Nov 19 at 14:19












          This worked beautifully. Thanks so much!
          – SteveM
          Nov 19 at 14:40




          This worked beautifully. Thanks so much!
          – SteveM
          Nov 19 at 14:40


















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