perform acf plot for each type of group in R












1














Say, here the mydata (little part)



transport<- structure(list(date = structure(c(1L, 2L, 3L, 4L, 5L, 6L, 7L, 
8L, 9L, 10L, 11L, 12L, 1L, 2L, 3L, 4L, 5L, 6L, 7L, 8L, 9L, 10L,
11L, 12L), .Label = c("01.01.2001", "01.02.2001", "01.03.2001",
"01.04.2001", "01.05.2001", "01.06.2001", "01.07.2001", "01.08.2001",
"01.09.2001", "01.10.2001", "01.11.2001", "01.12.2001"), class = "factor"),
Market_82 = c(7000L, 7272L, 7668L, 7869L, 8057L, 8428L, 8587L,
8823L, 8922L, 9178L, 9306L, 9439L, 3725L, 4883L, 8186L, 7525L,
6335L, 4252L, 5642L, 1326L, 8605L, 3501L, 1944L, 7332L),
transport = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L
), .Label = c("plane", "train"), class = "factor")), .Names = c("date",
"Market_82", "transport"), class = "data.frame", row.names = c(NA,
-24L))


group variable - Transport.



For each type of transport i must get acf plot of time series.



something like this
enter image description here



enter image description here



How perform acf plot for each transport?
I have a lot of groups. How to do that plots were in folder
C:/Users/admin/Documents/myplot










share|improve this question





























    1














    Say, here the mydata (little part)



    transport<- structure(list(date = structure(c(1L, 2L, 3L, 4L, 5L, 6L, 7L, 
    8L, 9L, 10L, 11L, 12L, 1L, 2L, 3L, 4L, 5L, 6L, 7L, 8L, 9L, 10L,
    11L, 12L), .Label = c("01.01.2001", "01.02.2001", "01.03.2001",
    "01.04.2001", "01.05.2001", "01.06.2001", "01.07.2001", "01.08.2001",
    "01.09.2001", "01.10.2001", "01.11.2001", "01.12.2001"), class = "factor"),
    Market_82 = c(7000L, 7272L, 7668L, 7869L, 8057L, 8428L, 8587L,
    8823L, 8922L, 9178L, 9306L, 9439L, 3725L, 4883L, 8186L, 7525L,
    6335L, 4252L, 5642L, 1326L, 8605L, 3501L, 1944L, 7332L),
    transport = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
    1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L
    ), .Label = c("plane", "train"), class = "factor")), .Names = c("date",
    "Market_82", "transport"), class = "data.frame", row.names = c(NA,
    -24L))


    group variable - Transport.



    For each type of transport i must get acf plot of time series.



    something like this
    enter image description here



    enter image description here



    How perform acf plot for each transport?
    I have a lot of groups. How to do that plots were in folder
    C:/Users/admin/Documents/myplot










    share|improve this question



























      1












      1








      1







      Say, here the mydata (little part)



      transport<- structure(list(date = structure(c(1L, 2L, 3L, 4L, 5L, 6L, 7L, 
      8L, 9L, 10L, 11L, 12L, 1L, 2L, 3L, 4L, 5L, 6L, 7L, 8L, 9L, 10L,
      11L, 12L), .Label = c("01.01.2001", "01.02.2001", "01.03.2001",
      "01.04.2001", "01.05.2001", "01.06.2001", "01.07.2001", "01.08.2001",
      "01.09.2001", "01.10.2001", "01.11.2001", "01.12.2001"), class = "factor"),
      Market_82 = c(7000L, 7272L, 7668L, 7869L, 8057L, 8428L, 8587L,
      8823L, 8922L, 9178L, 9306L, 9439L, 3725L, 4883L, 8186L, 7525L,
      6335L, 4252L, 5642L, 1326L, 8605L, 3501L, 1944L, 7332L),
      transport = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
      1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L
      ), .Label = c("plane", "train"), class = "factor")), .Names = c("date",
      "Market_82", "transport"), class = "data.frame", row.names = c(NA,
      -24L))


      group variable - Transport.



      For each type of transport i must get acf plot of time series.



      something like this
      enter image description here



      enter image description here



      How perform acf plot for each transport?
      I have a lot of groups. How to do that plots were in folder
      C:/Users/admin/Documents/myplot










      share|improve this question















      Say, here the mydata (little part)



      transport<- structure(list(date = structure(c(1L, 2L, 3L, 4L, 5L, 6L, 7L, 
      8L, 9L, 10L, 11L, 12L, 1L, 2L, 3L, 4L, 5L, 6L, 7L, 8L, 9L, 10L,
      11L, 12L), .Label = c("01.01.2001", "01.02.2001", "01.03.2001",
      "01.04.2001", "01.05.2001", "01.06.2001", "01.07.2001", "01.08.2001",
      "01.09.2001", "01.10.2001", "01.11.2001", "01.12.2001"), class = "factor"),
      Market_82 = c(7000L, 7272L, 7668L, 7869L, 8057L, 8428L, 8587L,
      8823L, 8922L, 9178L, 9306L, 9439L, 3725L, 4883L, 8186L, 7525L,
      6335L, 4252L, 5642L, 1326L, 8605L, 3501L, 1944L, 7332L),
      transport = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
      1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L
      ), .Label = c("plane", "train"), class = "factor")), .Names = c("date",
      "Market_82", "transport"), class = "data.frame", row.names = c(NA,
      -24L))


      group variable - Transport.



      For each type of transport i must get acf plot of time series.



      something like this
      enter image description here



      enter image description here



      How perform acf plot for each transport?
      I have a lot of groups. How to do that plots were in folder
      C:/Users/admin/Documents/myplot







      r ggplot2 time-series






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited 21 hours ago









      markus

      10.7k1029




      10.7k1029










      asked yesterday









      D.Joe

      6191513




      6191513
























          2 Answers
          2






          active

          oldest

          votes


















          2














          We can do this with Acf from forecast



          library(forecast)
          par(mfrow = c(2, 1))
          lapply(split(transport['Market_82'], transport$transport), Acf)




          If we also want the title, then



          lst <- lapply(split(transport['Market_82'], transport$transport), acf, plot = FALSE)
          par(mfrow = c(2, 1))
          lapply(names(lst), function(x) plot(lst[[x]], main = x))


          enter image description here






          share|improve this answer































            4














            akrun's answer is spot on. Since you tagged the question with ggplot2 you could also use ggAcf from the forcast package.



            The first step is to split your data.



            transport_split <- split(transport, transport$transport)


            If you want to include the respective element of column transport in the title, subtitle etc. try with Map



            out <- Map(
            f = function(x, y)
            forecast::ggAcf(x$Market_82) + labs(title = y),
            x = transport_split,
            y = names(transport_split)
            )

            out$train


            enter image description here






            share|improve this answer























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              2 Answers
              2






              active

              oldest

              votes








              2 Answers
              2






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              2














              We can do this with Acf from forecast



              library(forecast)
              par(mfrow = c(2, 1))
              lapply(split(transport['Market_82'], transport$transport), Acf)




              If we also want the title, then



              lst <- lapply(split(transport['Market_82'], transport$transport), acf, plot = FALSE)
              par(mfrow = c(2, 1))
              lapply(names(lst), function(x) plot(lst[[x]], main = x))


              enter image description here






              share|improve this answer




























                2














                We can do this with Acf from forecast



                library(forecast)
                par(mfrow = c(2, 1))
                lapply(split(transport['Market_82'], transport$transport), Acf)




                If we also want the title, then



                lst <- lapply(split(transport['Market_82'], transport$transport), acf, plot = FALSE)
                par(mfrow = c(2, 1))
                lapply(names(lst), function(x) plot(lst[[x]], main = x))


                enter image description here






                share|improve this answer


























                  2












                  2








                  2






                  We can do this with Acf from forecast



                  library(forecast)
                  par(mfrow = c(2, 1))
                  lapply(split(transport['Market_82'], transport$transport), Acf)




                  If we also want the title, then



                  lst <- lapply(split(transport['Market_82'], transport$transport), acf, plot = FALSE)
                  par(mfrow = c(2, 1))
                  lapply(names(lst), function(x) plot(lst[[x]], main = x))


                  enter image description here






                  share|improve this answer














                  We can do this with Acf from forecast



                  library(forecast)
                  par(mfrow = c(2, 1))
                  lapply(split(transport['Market_82'], transport$transport), Acf)




                  If we also want the title, then



                  lst <- lapply(split(transport['Market_82'], transport$transport), acf, plot = FALSE)
                  par(mfrow = c(2, 1))
                  lapply(names(lst), function(x) plot(lst[[x]], main = x))


                  enter image description here







                  share|improve this answer














                  share|improve this answer



                  share|improve this answer








                  edited yesterday

























                  answered yesterday









                  akrun

                  397k13187260




                  397k13187260

























                      4














                      akrun's answer is spot on. Since you tagged the question with ggplot2 you could also use ggAcf from the forcast package.



                      The first step is to split your data.



                      transport_split <- split(transport, transport$transport)


                      If you want to include the respective element of column transport in the title, subtitle etc. try with Map



                      out <- Map(
                      f = function(x, y)
                      forecast::ggAcf(x$Market_82) + labs(title = y),
                      x = transport_split,
                      y = names(transport_split)
                      )

                      out$train


                      enter image description here






                      share|improve this answer




























                        4














                        akrun's answer is spot on. Since you tagged the question with ggplot2 you could also use ggAcf from the forcast package.



                        The first step is to split your data.



                        transport_split <- split(transport, transport$transport)


                        If you want to include the respective element of column transport in the title, subtitle etc. try with Map



                        out <- Map(
                        f = function(x, y)
                        forecast::ggAcf(x$Market_82) + labs(title = y),
                        x = transport_split,
                        y = names(transport_split)
                        )

                        out$train


                        enter image description here






                        share|improve this answer


























                          4












                          4








                          4






                          akrun's answer is spot on. Since you tagged the question with ggplot2 you could also use ggAcf from the forcast package.



                          The first step is to split your data.



                          transport_split <- split(transport, transport$transport)


                          If you want to include the respective element of column transport in the title, subtitle etc. try with Map



                          out <- Map(
                          f = function(x, y)
                          forecast::ggAcf(x$Market_82) + labs(title = y),
                          x = transport_split,
                          y = names(transport_split)
                          )

                          out$train


                          enter image description here






                          share|improve this answer














                          akrun's answer is spot on. Since you tagged the question with ggplot2 you could also use ggAcf from the forcast package.



                          The first step is to split your data.



                          transport_split <- split(transport, transport$transport)


                          If you want to include the respective element of column transport in the title, subtitle etc. try with Map



                          out <- Map(
                          f = function(x, y)
                          forecast::ggAcf(x$Market_82) + labs(title = y),
                          x = transport_split,
                          y = names(transport_split)
                          )

                          out$train


                          enter image description here







                          share|improve this answer














                          share|improve this answer



                          share|improve this answer








                          edited 8 hours ago

























                          answered yesterday









                          markus

                          10.7k1029




                          10.7k1029






























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