R Programming 30 day Months - r

Question

I'm currently writing a script in the R Programming Language and I've hit a snag.
I have time series data organized in a way where there are 30 days in each month for 12 months in 1 year. However, I need the data organized in a proper 365 days in a year calendar, as in 30 days in a month, 31 days in a month, etc.
Is there a simple way for R to recognize there are 30 days in a month and to operate within that parameter? At the moment I have my script converting the number of days from the source in UNIX time and it counts up.
For example:
startingdate <- "20060101"
endingdate <- "20121230"
date <- seq(from = as.Date(startingdate, "%Y%m%d"), to = as.Date(endingdate, "%Y%m%d"), by = "days")
This would generate an array of dates with each month having 29 days/30 days/31 days etc. However, my data is currently organized as 30 days per month, regardless of 29 days or 31 days present.
Thanks.

Answer 1

The first 4 solutions are basically variations of the same theme using expand.grid. (3) uses magrittr and the others use no packages. The last two work by creating long sequence of numbers and then picking out the ones that have month and day in range.
1) apply This gives a series of yyyymmdd numbers such that there are 30 days in each month. Note that the line defining yrs in this case is the same as yrs <- 2006:2012 so if the years are handy we could shorten that line. Omit as.numeric in the line defining s if you want character string output instead. Also, s and d are the same because we have whole years so we could omit the line defining d and use s as the answer in this case and also in general if we are always dealing with whole years.
startingdate <- "20060101"
endingdate <- "20121230"
yrs <- seq(as.numeric(substr(startingdate, 1, 4)), as.numeric(substr(endingdate, 1, 4)))
g <- expand.grid(yrs, sprintf("%02d", 1:12), sprintf("%02d", 1:30))
s <- sort(as.numeric(apply(g, 1, paste, collapse = "")))
d <- s[ s >= startingdate & s <= endingdate ] # optional if whole years
Run some checks.
head(d)
## [1] 20060101 20060102 20060103 20060104 20060105 20060106
tail(d)
## 20121225 20121226 20121227 20121228 20121229 20121230
length(d) == length(2006:2012) * 12 * 30
## [1] TRUE
2) no apply An alternative variation would be this. In this and the following solutions we are using yrs as calculated in (1) so we omit it to avoid redundancy. Also, in this and the following solutions, the corresponding line to the one setting d is omitted, again, to avoid redundancy -- if you don't have whole years then add the line defining d in (1) replacing s in that line with s2.
g2 <- expand.grid(yr = yrs, mon = sprintf("%02d", 1:12), day = sprintf("%02d", 1:30))
s2 <- with(g2, sort(as.numeric(paste0(yr, mon, day))))
3) magrittr This could also be written using magrittr like this:
library(magrittr)
expand.grid(yr = yrs, mon = sprintf("%02d", 1:12), day = sprintf("%02d", 1:30)) %>%
with(paste0(yr, mon, day)) %>%
as.numeric %>%
sort -> s3
4) do.call Another variation.
g4 <- expand.grid(yrs, 1:12, 1:30)
s4 <- sort(as.numeric(do.call("sprintf", c("%d%02d%02d", g4))))
5) subset sequence Create a sequence of numbers from the starting date to the ending date and if each number is of the form yyyymmdd pick out those for which mm and dd are in range.
seq5 <- seq(as.numeric(startingdate), as.numeric(endingdate))
d5 <- seq5[ seq5 %/% 100 %% 100 %in% 1:12 & seq5 %% 100 %in% 1:30]
6) grep Using seq5 from (5)
d6 <- as.numeric(grep("(0[1-9]|1[0-2])(0[1-9]|[12][0-9]|30)$", seq5, value = TRUE))

Answer 2

Here's an alternative:
date <- unclass(startingdate):unclass(endingdate) %% 30L
month <- rep(1:12, each = 30, length.out = NN <- length(date))
year <- rep(1:(NN %/% 360 + 1), each = 360, length.out = NN)
(of course, we can easily adjust by adding constants to taste if you want a specific day to be 0, or a specific month, etc.)