# Function to get a sequence of months

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## Function to get a sequence of months

 Hi all, I am looking for a function for following calculation. start.month = "July" end.month = "January" months = f(start.month, end.month, by=1) * f is the function that I am looking for. Actually I want to get months = c("July", "August",.............."January") If start.month = 6 and end.month = 1 then I could use (not properly) seq() function and then I would get month as a vector with elements 6,5,4,3,2, and 1 by choosing "by=-1". Is there any function which can subsitute the seq() function in my case? Regards,         [[alternative HTML version deleted]] ______________________________________________ [hidden email] mailing list https://stat.ethz.ch/mailman/listinfo/r-helpPLEASE do read the posting guide http://www.R-project.org/posting-guide.htmland provide commented, minimal, self-contained, reproducible code.
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## Re: Function to get a sequence of months

 On 11/09/2007 6:36 AM, Arun Kumar Saha wrote: > Hi all, > > I am looking for a function for following calculation. > > start.month = "July" > end.month = "January" > > months = f(start.month, end.month, by=1) > > * f is the function that I am looking for. > > Actually I want to get months = c("July", "August",.............."January") > > If start.month = 6 and end.month = 1 then I could use (not properly) seq() > function and then I would get month as a vector with elements 6,5,4,3,2, and > 1 by choosing "by=-1". Is there any function which can subsitute the seq() > function in my case? I don't think one already exists, but it's easy to write one:  > cyclic_seq <- function(from, to, cycle=12) { +     if (to < from) (from - 1):(to + cycle - 1) %% cycle + 1 +     else from:to + }  > cyclic_seq(5, 1) [1]  5  6  7  8  9 10 11 12  1  > cyclic_seq(5, 5) [1] 5  > cyclic_seq(5, 6) [1] 5 6 This makes various assumptions about the inputs (e.g. it would fail on cyclic_seq(20, 1) ); you might want to validate the inputs if you're giving it to other people. Duncan Murdoch ______________________________________________ [hidden email] mailing list https://stat.ethz.ch/mailman/listinfo/r-helpPLEASE do read the posting guide http://www.R-project.org/posting-guide.htmland provide commented, minimal, self-contained, reproducible code.
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## Re: Function to get a sequence of months

 Duncan Murdoch-2 wrote On 11/09/2007 6:36 AM, Arun Kumar Saha wrote: > Hi all, > > I am looking for a function for following calculation. > > start.month = "July" > end.month = "January" > > months = f(start.month, end.month, by=1) > > * f is the function that I am looking for. > > Actually I want to get months = c("July", "August",.............."January") > > If start.month = 6 and end.month = 1 then I could use (not properly) seq() > function and then I would get month as a vector with elements 6,5,4,3,2, and > 1 by choosing "by=-1". Is there any function which can subsitute the seq() > function in my case? I don't think one already exists, but it's easy to write one:  > cyclic_seq <- function(from, to, cycle=12) { +     if (to < from) (from - 1):(to + cycle - 1) %% cycle + 1 +     else from:to + }  > cyclic_seq(5, 1) [1]  5  6  7  8  9 10 11 12  1  > cyclic_seq(5, 5) [1] 5  > cyclic_seq(5, 6) [1] 5 6 This makes various assumptions about the inputs (e.g. it would fail on cyclic_seq(20, 1) ); you might want to validate the inputs if you're giving it to other people. Duncan Murdoch And the rest would be to use month.name or month.abb variables month.name[cyclic_seq(7,1)]
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## Re: Function to get a sequence of months

 In reply to this post by Arun.stat Here is a solution using character manipulation.  Noting that month.name is built into R we paste together the character string: "January Februrary ... December January ... December" Then we use perl style ungreedy matching to get the shortest substring matching the indicated expression then splitting it back apart and taking the first match: library(gsubfn) strapply(paste(rep(month.name, 2), collapse = " "), "July.*?January",   ~ strsplit(x, split = " "), perl = TRUE, simplify = c)[[1]] Here is the result of running it: > library(gsubfn) > strapply(paste(rep(month.name, 2), collapse = " "), "July.*?January", +   ~ strsplit(x, split = " "), perl = TRUE, simplify = c)[[1]] [1] "July"      "August"    "September" "October"   "November" "December"  "January" mn2 <- c(month.name, month.name) strapply(paste(mn2, "July.*January", ~ strsplit(x, split = " "), simplify = unlist) On 9/11/07, Arun Kumar Saha <[hidden email]> wrote: > Hi all, > > I am looking for a function for following calculation. > > start.month = "July" > end.month = "January" > > months = f(start.month, end.month, by=1) > > * f is the function that I am looking for. > > Actually I want to get months = c("July", "August",.............."January") > > If start.month = 6 and end.month = 1 then I could use (not properly) seq() > function and then I would get month as a vector with elements 6,5,4,3,2, and > 1 by choosing "by=-1". Is there any function which can subsitute the seq() > function in my case? > > Regards, > >        [[alternative HTML version deleted]] > > ______________________________________________ > [hidden email] mailing list > https://stat.ethz.ch/mailman/listinfo/r-help> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html> and provide commented, minimal, self-contained, reproducible code. > ______________________________________________ [hidden email] mailing list https://stat.ethz.ch/mailman/listinfo/r-helpPLEASE do read the posting guide http://www.R-project.org/posting-guide.htmland provide commented, minimal, self-contained, reproducible code.