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]]

>

> ______________________________________________

>

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>

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.

>

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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.