R min and max function does not work for dates












2














I have a dataframe,df, with a date variable.



   locationID organe    
<int> <date>
1 1940-04-01
2 1938-07-01
3 1938-07-01
4 1938-07-01


I want to compare the dates with a fixed time point,like "1938-12-1", to find the earliest date. I used the min function but failed.



df %>% mutate(earliest=min(organe,as.Date("1938-12-1")))


locationID organe earliest
<int> <date> <date>
1 1940-04-01 1937-09-01
2 1938-07-01 1937-09-01
3 1938-07-01 1937-09-01
4 1938-07-01 1937-09-01


I don't know why min function does not work here, though it works well for the following situation



min(as.Date("1938-07-01"),as.Date("1938-12-1"))
[1] "1938-07-01"


Could anybody help?










share|improve this question



























    2














    I have a dataframe,df, with a date variable.



       locationID organe    
    <int> <date>
    1 1940-04-01
    2 1938-07-01
    3 1938-07-01
    4 1938-07-01


    I want to compare the dates with a fixed time point,like "1938-12-1", to find the earliest date. I used the min function but failed.



    df %>% mutate(earliest=min(organe,as.Date("1938-12-1")))


    locationID organe earliest
    <int> <date> <date>
    1 1940-04-01 1937-09-01
    2 1938-07-01 1937-09-01
    3 1938-07-01 1937-09-01
    4 1938-07-01 1937-09-01


    I don't know why min function does not work here, though it works well for the following situation



    min(as.Date("1938-07-01"),as.Date("1938-12-1"))
    [1] "1938-07-01"


    Could anybody help?










    share|improve this question

























      2












      2








      2







      I have a dataframe,df, with a date variable.



         locationID organe    
      <int> <date>
      1 1940-04-01
      2 1938-07-01
      3 1938-07-01
      4 1938-07-01


      I want to compare the dates with a fixed time point,like "1938-12-1", to find the earliest date. I used the min function but failed.



      df %>% mutate(earliest=min(organe,as.Date("1938-12-1")))


      locationID organe earliest
      <int> <date> <date>
      1 1940-04-01 1937-09-01
      2 1938-07-01 1937-09-01
      3 1938-07-01 1937-09-01
      4 1938-07-01 1937-09-01


      I don't know why min function does not work here, though it works well for the following situation



      min(as.Date("1938-07-01"),as.Date("1938-12-1"))
      [1] "1938-07-01"


      Could anybody help?










      share|improve this question













      I have a dataframe,df, with a date variable.



         locationID organe    
      <int> <date>
      1 1940-04-01
      2 1938-07-01
      3 1938-07-01
      4 1938-07-01


      I want to compare the dates with a fixed time point,like "1938-12-1", to find the earliest date. I used the min function but failed.



      df %>% mutate(earliest=min(organe,as.Date("1938-12-1")))


      locationID organe earliest
      <int> <date> <date>
      1 1940-04-01 1937-09-01
      2 1938-07-01 1937-09-01
      3 1938-07-01 1937-09-01
      4 1938-07-01 1937-09-01


      I don't know why min function does not work here, though it works well for the following situation



      min(as.Date("1938-07-01"),as.Date("1938-12-1"))
      [1] "1938-07-01"


      Could anybody help?







      r date min






      share|improve this question













      share|improve this question











      share|improve this question




      share|improve this question










      asked Nov 12 '18 at 22:20









      Li FeiyueLi Feiyue

      154




      154
























          1 Answer
          1






          active

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          2














          We can use pmin/pmax to get the minimum/maximum when one or both of the input argument is a vector of length greater than 1 (if both have length greater than 1, assumes the length to be same)



          df %>% 
          mutate(earliest=pmin(organe,as.Date("1938-12-1")))
          # A tibble: 4 x 3
          # locationID organe earliest
          # <int> <date> <date>
          #1 1 1940-04-01 1938-12-01
          #2 2 1938-07-01 1938-07-01
          #3 3 1938-07-01 1938-07-01
          #4 4 1938-07-01 1938-07-01




          Or apply min after rowwise



          df %>% 
          rowwise %>%
          mutate(earliest=min(organe, as.Date("1938-12-1")))




          Note that min returns a single value as output i.e.



          min(5:1, 3)
          #[1] 1

          min(5:3, 1)
          #[1] 1


          For a vectorized minimum, use pmin. According to ?min




          pmax*() and pmin*() take one or more vectors as arguments, recycle them to common length and return a single vector giving the ‘parallel’ maxima (or minima) of the argument vectors.







          share|improve this answer























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            1 Answer
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            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            2














            We can use pmin/pmax to get the minimum/maximum when one or both of the input argument is a vector of length greater than 1 (if both have length greater than 1, assumes the length to be same)



            df %>% 
            mutate(earliest=pmin(organe,as.Date("1938-12-1")))
            # A tibble: 4 x 3
            # locationID organe earliest
            # <int> <date> <date>
            #1 1 1940-04-01 1938-12-01
            #2 2 1938-07-01 1938-07-01
            #3 3 1938-07-01 1938-07-01
            #4 4 1938-07-01 1938-07-01




            Or apply min after rowwise



            df %>% 
            rowwise %>%
            mutate(earliest=min(organe, as.Date("1938-12-1")))




            Note that min returns a single value as output i.e.



            min(5:1, 3)
            #[1] 1

            min(5:3, 1)
            #[1] 1


            For a vectorized minimum, use pmin. According to ?min




            pmax*() and pmin*() take one or more vectors as arguments, recycle them to common length and return a single vector giving the ‘parallel’ maxima (or minima) of the argument vectors.







            share|improve this answer




























              2














              We can use pmin/pmax to get the minimum/maximum when one or both of the input argument is a vector of length greater than 1 (if both have length greater than 1, assumes the length to be same)



              df %>% 
              mutate(earliest=pmin(organe,as.Date("1938-12-1")))
              # A tibble: 4 x 3
              # locationID organe earliest
              # <int> <date> <date>
              #1 1 1940-04-01 1938-12-01
              #2 2 1938-07-01 1938-07-01
              #3 3 1938-07-01 1938-07-01
              #4 4 1938-07-01 1938-07-01




              Or apply min after rowwise



              df %>% 
              rowwise %>%
              mutate(earliest=min(organe, as.Date("1938-12-1")))




              Note that min returns a single value as output i.e.



              min(5:1, 3)
              #[1] 1

              min(5:3, 1)
              #[1] 1


              For a vectorized minimum, use pmin. According to ?min




              pmax*() and pmin*() take one or more vectors as arguments, recycle them to common length and return a single vector giving the ‘parallel’ maxima (or minima) of the argument vectors.







              share|improve this answer


























                2












                2








                2






                We can use pmin/pmax to get the minimum/maximum when one or both of the input argument is a vector of length greater than 1 (if both have length greater than 1, assumes the length to be same)



                df %>% 
                mutate(earliest=pmin(organe,as.Date("1938-12-1")))
                # A tibble: 4 x 3
                # locationID organe earliest
                # <int> <date> <date>
                #1 1 1940-04-01 1938-12-01
                #2 2 1938-07-01 1938-07-01
                #3 3 1938-07-01 1938-07-01
                #4 4 1938-07-01 1938-07-01




                Or apply min after rowwise



                df %>% 
                rowwise %>%
                mutate(earliest=min(organe, as.Date("1938-12-1")))




                Note that min returns a single value as output i.e.



                min(5:1, 3)
                #[1] 1

                min(5:3, 1)
                #[1] 1


                For a vectorized minimum, use pmin. According to ?min




                pmax*() and pmin*() take one or more vectors as arguments, recycle them to common length and return a single vector giving the ‘parallel’ maxima (or minima) of the argument vectors.







                share|improve this answer














                We can use pmin/pmax to get the minimum/maximum when one or both of the input argument is a vector of length greater than 1 (if both have length greater than 1, assumes the length to be same)



                df %>% 
                mutate(earliest=pmin(organe,as.Date("1938-12-1")))
                # A tibble: 4 x 3
                # locationID organe earliest
                # <int> <date> <date>
                #1 1 1940-04-01 1938-12-01
                #2 2 1938-07-01 1938-07-01
                #3 3 1938-07-01 1938-07-01
                #4 4 1938-07-01 1938-07-01




                Or apply min after rowwise



                df %>% 
                rowwise %>%
                mutate(earliest=min(organe, as.Date("1938-12-1")))




                Note that min returns a single value as output i.e.



                min(5:1, 3)
                #[1] 1

                min(5:3, 1)
                #[1] 1


                For a vectorized minimum, use pmin. According to ?min




                pmax*() and pmin*() take one or more vectors as arguments, recycle them to common length and return a single vector giving the ‘parallel’ maxima (or minima) of the argument vectors.








                share|improve this answer














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                edited Nov 12 '18 at 22:35

























                answered Nov 12 '18 at 22:21









                akrunakrun

                400k13189264




                400k13189264






























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