将周数转换为日期

Question

I have a data frame in R with the week of the year that I would like to convert to a date. I know I have to pick a year and a day of the week so I am fixing those values at 2014 and 1. Converting this to a date seems simple:

as.Date(paste(2014,df$Week,1,sep=""),"%Y%U%u")

But this code only works if week is greater than 9. Week 1 - 9 returns NA. If I change the week to 01,02,03... it still returns NA.

Anyone see what I am missing?

Answer 1

as.Date is calling the 1 to 9 as NA as it is expects two digits for the week number and can't properly parse it.

To fix it, add in some - to split things up:

as.Date(paste(2014, df$Week, 1, sep="-"), "%Y-%U-%u")

Answer 2

An alternative solution is to use date arithmetic from the lubridate package:

lubridate::ymd( "2014-01-01" ) + lubridate::weeks( df$Week - 1 )

The -1 is necessary because 2014-01-01 is already week 1. In other words, we want:

• df$Week == 1 to map to 2014-01-01 (which is ymd("2014-01-01") + weeks(1-1) )
• df$Week == 2 to map to 2014-01-08 (which is ymd("2014-01-01") + weeks(2-1) )
• and so on.

Answer 3

Another alternative is to make sure that week numbers have two digits, which can be done using stringr::str_pad() , which will add a pad="0" to make sure there are width=2 digits:

year <- 2015
week <- 1
as.Date(paste(year, week, "1", sep=""), "%Y%U%u")
#> [1] NA
as.Date(paste(year, stringr::str_pad(week,width=2, pad="0"), "1", sep=""), "%Y%U%u")
#> [1] "2015-01-05"
as.Date(paste(year, week, "1", sep="-"), "%Y-%U-%u")
#> [1] "2015-01-05"

^{Created on 2021-04-19 by the reprex package (v1.0.0)}

Answer 4

It will be like using 2nd year = (week-52), 3rd year  = (week -104)...so on

for(i in 1:456548)
{
  if (train[i,2] > 0 & train[i,2] <53)
  {
    train["weekdate"] <- as.Date(paste(2016, train$week, 1, sep="-"), "%Y-%U-%u")
  }
  if (train[i,2] > 52 & train[i,2] <105)
  {
    train["weekdate"] <- as.Date(paste(2017, (train$week-52), 1, sep="-"), "%Y-%U-%u")
  }
  if (train[i,2] > 104 & train[i,2] <150)
  {
    train["weekdate"] <- as.Date(paste(2018, (train$week-104), 1, sep="-"), "%Y-%U-%u")
  }

}

Answer 5

Another option with lubridate

lubridate::parse_date_time(paste(2014, df$Week, 1, sep="/"),'Y/W/w')

W - week number, w - weekday number, 0-6 (Sun-Sat)