CODEWITHSUNDEEP
rdatedatetime
Naveen Reddy Marthala
Naveen Reddy Marthala

Reputation: 3133

How to get biweek/fornight numbers from dates?

I'm trying to get biweekly/fortnightly numbers for dates.

the weeks can be done like(from documentation):

data <- read.csv("dummy data.csv")
data$Date <- as.Date(data$Date, "%d-%m-%y")
data$weeks <- format(data$Date, "%U-%Y")

this gives:

> data
   Date      weeks 
2017-01-07  01-2017
2017-01-08  02-2017
2017-01-15  03-2017

I want to calculate biweeks/fortnights from data like i did above.
expected output:

   Date      weeks   fortnights
2017-01-07  01-2017   01-2017
2017-01-08  02-2017   01-2017
2017-01-15  03-2017   02-2017

How do I do this for a dataframe?

Upvotes: 1

Views: 1231

Answers (2)

Ronak Shah
Ronak Shah

Reputation: 389275

How about to get the fortnight, we extract the month which are already passed, multiply it with 2 and add 1 or 2 based on date. So here any date after 14 is considered as next fortnight. 

(as.integer(format(df$Date, "%m")) - 1) * 2 + 
   (as.integer(format(df$Date, "%d")) > 14) + 1
#[1] 1 1 2

If you also want the year we can do

paste0((as.integer(format(df$Date, "%m")) - 1) * 2 + 
       (as.integer(format(df$Date, "%d")) > 14) + 1, format(df$Date, "-%Y"))
#[1] "1-2017" "1-2017" "2-2017"

data

df <- structure(list(Date = structure(c(17173, 17174, 17181), class = "Date"), 
weeks = structure(1:3, .Label = c("01-2017", "02-2017", "03-2017"
), class = "factor")), row.names = c(NA, -3L), class = "data.frame")

Upvotes: 1

stratar
stratar

Reputation: 129

You can use the "week" function from the "lubridate" package. e.g.

lubridate::week(as.Date("2017-06-08")) will return 23.
lubridate::week(as.Date("2017-01-07")) will return 1.
lubridate::week(as.Date("2017-01-08")) will return 2.

To return the fortnights it will be just a matter of dividing each of the above with the number 2 and "ceil" the result e.g.

ceiling(lubridate::week(as.Date("2017-06-08")) / 2)

For information on the ceiling() function please see the documentation, i.e. ?ceiling.

P.S. I think in this case it is safer to use the ceiling() function instead of the round() function, although as long as your arguments are positive, the behaviour will be similar for both.

Upvotes: 3

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