seq() with multiple increments











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I'd like to create a sequence with three different increements. From 6 to 15 for example an increment of 0.7 . The folowing sequence should start with the last number of previous sequence (in this case 14.4).
By this I want to model the diameter increment of a tree dependant on the diameter-class (small 6-14.99; medium 15 - 29.99; big >30).



    dbh <- c(seq(from = 6, to = 15, by = temp$DBH_growth[temp$dbh_class == "sma"]),
seq(from = 15, to = 30, by = temp$DBH_growth[temp$dbh_class == "med"]),
seq(from = 30, to = 300, by = temp$DBH_growth[temp$dbh_class == "big"]))


Like this code, but starting with the last number of the sequence before.










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  • Hi @fritz, welcome to SO! For better reproducibility make sure to provide some sample data for others to be able to help you out (the data frame temp and thus the increments are unknown).
    – nate.edwinton
    Nov 19 at 15:40

















up vote
0
down vote

favorite












I'd like to create a sequence with three different increements. From 6 to 15 for example an increment of 0.7 . The folowing sequence should start with the last number of previous sequence (in this case 14.4).
By this I want to model the diameter increment of a tree dependant on the diameter-class (small 6-14.99; medium 15 - 29.99; big >30).



    dbh <- c(seq(from = 6, to = 15, by = temp$DBH_growth[temp$dbh_class == "sma"]),
seq(from = 15, to = 30, by = temp$DBH_growth[temp$dbh_class == "med"]),
seq(from = 30, to = 300, by = temp$DBH_growth[temp$dbh_class == "big"]))


Like this code, but starting with the last number of the sequence before.










share|improve this question






















  • Hi @fritz, welcome to SO! For better reproducibility make sure to provide some sample data for others to be able to help you out (the data frame temp and thus the increments are unknown).
    – nate.edwinton
    Nov 19 at 15:40















up vote
0
down vote

favorite









up vote
0
down vote

favorite











I'd like to create a sequence with three different increements. From 6 to 15 for example an increment of 0.7 . The folowing sequence should start with the last number of previous sequence (in this case 14.4).
By this I want to model the diameter increment of a tree dependant on the diameter-class (small 6-14.99; medium 15 - 29.99; big >30).



    dbh <- c(seq(from = 6, to = 15, by = temp$DBH_growth[temp$dbh_class == "sma"]),
seq(from = 15, to = 30, by = temp$DBH_growth[temp$dbh_class == "med"]),
seq(from = 30, to = 300, by = temp$DBH_growth[temp$dbh_class == "big"]))


Like this code, but starting with the last number of the sequence before.










share|improve this question













I'd like to create a sequence with three different increements. From 6 to 15 for example an increment of 0.7 . The folowing sequence should start with the last number of previous sequence (in this case 14.4).
By this I want to model the diameter increment of a tree dependant on the diameter-class (small 6-14.99; medium 15 - 29.99; big >30).



    dbh <- c(seq(from = 6, to = 15, by = temp$DBH_growth[temp$dbh_class == "sma"]),
seq(from = 15, to = 30, by = temp$DBH_growth[temp$dbh_class == "med"]),
seq(from = 30, to = 300, by = temp$DBH_growth[temp$dbh_class == "big"]))


Like this code, but starting with the last number of the sequence before.







r seq






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asked Nov 19 at 14:53









fritz

1




1












  • Hi @fritz, welcome to SO! For better reproducibility make sure to provide some sample data for others to be able to help you out (the data frame temp and thus the increments are unknown).
    – nate.edwinton
    Nov 19 at 15:40




















  • Hi @fritz, welcome to SO! For better reproducibility make sure to provide some sample data for others to be able to help you out (the data frame temp and thus the increments are unknown).
    – nate.edwinton
    Nov 19 at 15:40


















Hi @fritz, welcome to SO! For better reproducibility make sure to provide some sample data for others to be able to help you out (the data frame temp and thus the increments are unknown).
– nate.edwinton
Nov 19 at 15:40






Hi @fritz, welcome to SO! For better reproducibility make sure to provide some sample data for others to be able to help you out (the data frame temp and thus the increments are unknown).
– nate.edwinton
Nov 19 at 15:40














1 Answer
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Something like this



sma <- seq(6, 15, .7)
med <- seq(max(sma), 30, 1)
lar <- seq(max(med), 300, 1.4)
dbh <- c(sma, med, lar)


supposing that .7,1,1.4 are the increments for small, medium, large respectively, or in your case



sma <- seq(6, 15, temp$DBH_growth[temp$dbh_class == "sma"])
med <- seq(max(sma), 30, temp$DBH_growth[temp$dbh_class == "med"])
lar <- seq(max(med), 300, temp$DBH_growth[temp$dbh_class == "big"])

dbh <- c(sma, med, lar)





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    1 Answer
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    Something like this



    sma <- seq(6, 15, .7)
    med <- seq(max(sma), 30, 1)
    lar <- seq(max(med), 300, 1.4)
    dbh <- c(sma, med, lar)


    supposing that .7,1,1.4 are the increments for small, medium, large respectively, or in your case



    sma <- seq(6, 15, temp$DBH_growth[temp$dbh_class == "sma"])
    med <- seq(max(sma), 30, temp$DBH_growth[temp$dbh_class == "med"])
    lar <- seq(max(med), 300, temp$DBH_growth[temp$dbh_class == "big"])

    dbh <- c(sma, med, lar)





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      up vote
      0
      down vote













      Something like this



      sma <- seq(6, 15, .7)
      med <- seq(max(sma), 30, 1)
      lar <- seq(max(med), 300, 1.4)
      dbh <- c(sma, med, lar)


      supposing that .7,1,1.4 are the increments for small, medium, large respectively, or in your case



      sma <- seq(6, 15, temp$DBH_growth[temp$dbh_class == "sma"])
      med <- seq(max(sma), 30, temp$DBH_growth[temp$dbh_class == "med"])
      lar <- seq(max(med), 300, temp$DBH_growth[temp$dbh_class == "big"])

      dbh <- c(sma, med, lar)





      share|improve this answer























        up vote
        0
        down vote










        up vote
        0
        down vote









        Something like this



        sma <- seq(6, 15, .7)
        med <- seq(max(sma), 30, 1)
        lar <- seq(max(med), 300, 1.4)
        dbh <- c(sma, med, lar)


        supposing that .7,1,1.4 are the increments for small, medium, large respectively, or in your case



        sma <- seq(6, 15, temp$DBH_growth[temp$dbh_class == "sma"])
        med <- seq(max(sma), 30, temp$DBH_growth[temp$dbh_class == "med"])
        lar <- seq(max(med), 300, temp$DBH_growth[temp$dbh_class == "big"])

        dbh <- c(sma, med, lar)





        share|improve this answer












        Something like this



        sma <- seq(6, 15, .7)
        med <- seq(max(sma), 30, 1)
        lar <- seq(max(med), 300, 1.4)
        dbh <- c(sma, med, lar)


        supposing that .7,1,1.4 are the increments for small, medium, large respectively, or in your case



        sma <- seq(6, 15, temp$DBH_growth[temp$dbh_class == "sma"])
        med <- seq(max(sma), 30, temp$DBH_growth[temp$dbh_class == "med"])
        lar <- seq(max(med), 300, temp$DBH_growth[temp$dbh_class == "big"])

        dbh <- c(sma, med, lar)






        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered Nov 19 at 15:36









        nate.edwinton

        961314




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