Calculate the week number (0-53) in year

Question

I have a dataset with locations and dates. I would like to calculate week of the year as number (00–53) but using Thursday as the first day of the week. The data looks like this:

  location <- c(a,b,a,b,a,b)
  date <- c("04-01-2013","26-01-2013","03-02-2013","09-02-2013","20-02-2013","03-03-2013")
  mydf <- data.frame(location, date)
  mydf

I know that there is strftime function for calculating week of year but it is only possible to use Monday or Sunday as the first day of the week. Any help would be highly appreciated.

Eipper

Why would you want to do that, if I may ask? – Maffick

So if a year began on a Tuesday, week 1 is Tue/Wed and week 2 begins on Thursday...? – Cloakanddagger

yes. Week should be from Tue till including Wed. – Eipper

@Cloakanddagger ; The strftime convention for week numbering is 0-53. (Unlike the other indexing in R.) – Dorolice

@joran: So if a year began on a Tuesday, week 0 is Tue/Wed and week 1 begins on Thursday – Bonneau

Answer 1

Just add 4 to the Date-formatted values:

> mydf$Dt <- as.Date(mydf$date, format="%d-%m-%Y")
> weeknum <- as.numeric( format(mydf$Dt+3, "%U"))
> weeknum
[1] 1 4 5 6 7 9

This uses a 0 based counting convention since that is what strftime provides and we are just piggybacking off that code base, so the first Friday in a year that begins on Tuesday as was the case in 2013 would be a 1-week result. Add 1 to the value if you want a 1 based convention. (Fundamentally, Date-formated values are in an integer sequence from the "origin" so they don't really recognize years or weeks. Adding 4 just shifts the reference frame of the underlying Date-integer.)

Edit note. Changed to an add three strategy per Gabor's advice. .... which still does not address the question of how to deal with the last week of the prior year.

Dorolice

There is some question on just what the definition of week number is but at any rate there are 1, 4, 5, 6, 7, 9 Thursdays in the year before the 6 dates in the question, respectively, which raises a question of whether the 8 in the answer is correct. Also format(as.Date("2013-12-31")+4, "%U") gives 00. – Leigh

The "add-4" strategy creates an increment on Thursday so the count of Thursdays would be in agreement with an "add-3" strategy. – Dorolice

A fix for the end of year problem might be to subtract 7 days if its in December and then add 1 to the week count: ifelse(months(d) == "December", as.numeric(format(d-4, "%U"))+1, as.numeric(format(d+3, "%U"))) – Leigh

Answer 2

Since the question stated that week goes from 00-53 we assume that the week number is the number of Thursdays in the year on or before the date in question. Thus, the first Thursday in the year begins week 1 and week 0 is assigned to any days prior to that.

(There were comments that if the first day of the year were Tuesday then that would be week 1 but if that were the case there could never be a week 0 as seems to be required in the subject so some clarification on precisely what the definition of week number is may be required. Here we are going to use the definition in the preceding paragraph but it would not be hard to change it if we knew what the definition was. For example, if we always wanted the first week in the year to be 1 even if it were a short week then we could add !is.thu(jan1(d)) to the result.)

Both of the solutions below are short enough that they could be expressed in one statement; however, we have factored them into several short functions each for clarity. The first is particularly straight forward but the second is automatically vectorized without the need for a sapply and would likely be more efficient.

1. sum Thursdays in year This solution assumes the input d is of class "Date" and just sums the number of Thursdays in the year before or on it:

is.thu <- function(x) weekdays(x) == "Thursday"
jan1 <- function(x) as.Date(cut(x, "year"))

week4 <- function(d) {
  sapply(d, function(d) sum(is.thu(seq(jan1(d), d, by = "day"))))
}

We can test it like this:

d <- as.Date(c("2013-01-04", "2013-01-26", "2013-02-03", "2013-02-09", 
    "2013-02-20", "2013-03-03"))
week4(d) # 1 4 5 6 7 9

2. nextthu

Based on the nextfri function in the zoo quickref vignette we see that the number of days since the Epoch (1970-01-01) of the next Thursday (or the day in question if its already a Thursday) is as given by nextthu in the first line below. Applying this to the first day of the year we derive the result where d is as before:

nextthu <- function(d) 7 * ceiling(as.numeric(d) / 7)

week4a <- function(d) (as.numeric(d) - nextthu(jan1(d))) %/% 7 + 1

and here is a test

week4a(d) # 1 4 5 6 7 9

ADDED: fixed bug in second solution.

Leigh