I have many strings like "200046". The first four digits are the year, and the last two is the number of the week per year. I'm trying to find the 7 days of the week for that week. I tried something like
date = as.Date(str, "%Y%M")
but it returns "2000-01-29" which is not the 46th week of 2000. How can I do that?
Add day of the week to str.
str <- '200046'
as.Date(paste0(str, 1), "%Y%U%u")
#[1] "2000-11-13"
This is 1st day (Monday) of 46th week of 2000.
Now to get all days of the week you can do :
as.Date(paste0(str, 1), "%Y%U%u") + 0:6
#[1] "2000-11-13" "2000-11-14" "2000-11-15" "2000-11-16" "2000-11-17" "2000-11-18" "2000-11-19"
Related
Given a year and a day (in Julian day), how can I extract the month? For e.g.
Year <- '2000'
Doy <- '159'
I want to extract the month for the above Year and Doy. I thought first I will convert this into date and then extract the month out of it using format(mydate,"%m")
# first convert into date and then extract the month
as.Date(paste0(Year'-',Doy), format = '%Y-%d')
NA
This gives me NA.
%d is for day of month. %j is for day of year where Jan 1 is day of year 1, Jan 2 is day of year 2, ..., Dec 31 is day of year 365 (or 366 on leap years). See ?strptime for the percent codes.
Year <- '2000'
Doy <- '159'
date <- as.Date(paste(Year, Doy), format = "%Y %j"); date
## [1] "2000-06-07"
as.numeric(format(date, "%m")) # month number
## [1] 6
I would like to get the same day last year given any year. How can I best do this in R. For example, given Sunday 2010/01/03, I would like to obtain the Sunday of the same week the year before.
# "Sunday"
weekdays(as.Date("2010/01/03", format="%Y/%m/%d"))
# "Saturday"
weekdays(as.Date("2009/01/03", format="%Y/%m/%d"))
To find the same weekday one year ago, simply subtract 52 weeks or 364 days from the given date:
d <- as.Date("2010-01-03")
weekdays(d)
#[1] "Sunday"
d - 52L * 7L
#[1] "2009-01-04"
weekdays(d - 52L * 7L)
#[1] "Sunday"
Please note that the calendar year has 365 days (or 366 days in a leap year) which is one or two days more than 52 weeks. So, the calendar date of the same weekday one year ago moves on by one or two days. (Or, it explains why New Year's Eve is always on a different weekday.)
Using lubridate the following formula will give you the corresponding weekday in the same week in the previous year:
as.Date(dDate - 364 - ifelse(weekdays( dDate - 363) == weekdays( dDate ), 1, 0))
Where dDate is some date, i.e. dDate <- as.Date("2016-02-29"). The ifelse accounts for leap years.
Here's a simple algorithm. subtract 365 days from the day of interest. Adjust that day to the closest matching day of the week using the Tableau code below (easily translatable into other languages). This is equivalent to the rule in the table below (with 1 = Monday and 7 = Sunday). Basically you adjust day - 365 to be on the correct day of the week either in the same week if that moves <= 3 days otherwise you use the matching weekday from the previous/next week. It'll choose whichever leads to the least difference in terms of # of days.
[day prior year raw] = [day] - 365
[matching day prior year] =
if abs(datepart('weekday',[day]) - datepart('weekday',[day prior year raw]))<= 3
then [day prior year raw]+datepart('weekday',[day]) - datepart('weekday',[day prior year raw])
else [day prior year raw]+(if datepart('weekday',[day]) > datepart('weekday',[day prior year raw])
then -7+(datepart('weekday',[day]) - datepart('weekday',[day prior year raw]))
else 7+(datepart('weekday',[day]) - datepart('weekday',[day prior year raw])) end
)
end)
Look at ?years in package lubridate. This creates a period object which correctly spans a period, across leap years.
> library(lubridate)
> # set the reference date
> d1 = as.Date("2017/01/03", format="%Y/%m/%d")
>
> # verify across years and leap years
> d1 - years(1)
[1] "2016-01-03"
> d1 - years(2)
[1] "2015-01-03"
> d1 - years(3)
[1] "2014-01-03"
> d1 - years(4)
[1] "2013-01-03"
> d1 - years(5)
[1] "2012-01-03"
>
> weekdays(d1 - years(1))
[1] "Sunday"
> weekdays(d1 - years(2))
[1] "Saturday"
>
> # feb 29 on year period in yields NA
> ymd("2016/02/29") - years(1)
[1] NA
>
> # feb 29 in a non-leap year fails to convert
> ymd("2015/02/29") - years(1)
[1] NA
Warning message:
All formats failed to parse. No formats found.
>
> # feb 29, leap year with 4 year period works.
> ymd("2016/02/29") - years(4)
[1] "2012-02-29"
>
I have dates in an R dataframe column formatted as character strings as WK01Q32014.
I want to turn each date into a Date() object.
So I altered the format to make it look like 01-3-2014. I want to try to do something like as.Date("01-3-2014","%W-%Q-%Y") for example, but there is no format code for quarters that I know of.
Is there any way to do this using the lubridate, zoo, or any other libraries?
I dont know of any specific function, but here's a basic one:
convert_WQ_to_Date <- function(D) {
weeks <- as.integer(substr(D, 3, 4))
quarter <- as.integer(substr(D, 6, 6))
year <- substr(D, 7, 10)
days <- 7 * ((quarter - 1) * 13 + (weeks-1))
as.Date(sprintf("%s-01-01", year)) + days
}
Example
D <- c("WK01Q32014", "WK01Q12014", "WK05Q42014", "WK01Q22014", "WK02Q32014")
convert_WQ_to_Date(D)
[1] "2014-07-02" "2014-01-01" "2014-10-29" "2014-04-02" "2014-07-09"
The week, quarter and year does not uniquely define a date so we will have to add some assumption. Here we add the assumption that the first week is the first day of the quarter, the second week is 7 days later and so on,
Below, we extract the qtr-year part and use as.yearqtr in the zoo package to convert that to a yearqtr object and then use as.Date to convert that to a date which is the first of the quarter. We then extract the week, subtract 1 and multiply by 7 to get the days offset. Adding the first of the quarter to the offset gives the result:
library(zoo)
xx <- "01-3-2014" # week-quarter-year
qtr.start <- as.Date(as.yearqtr(sub("...", "", xx), "%q-%Y"))
days <- 7 * (as.numeric(sub("-.*", "", xx)) - 1)
qtr.start + days
## [1] "2014-07-01"
Assuming the traditional notion of each quarter starting respectively at the 1st January, 1st April, 1st July and 1st September (in line with the quarters function), just start at these dates and add 7 days for each week:
x <- c("01-3-2014","01-1-2014","05-4-2014","01-2-2014","02-3-2014")
y <- as.numeric(substr(x,6,9))
m <- as.numeric(substr(x,4,4))
d <- as.numeric(substr(x,1,2))
as.Date(paste(y,(m-1)*3+1,"01",sep="-")) + (7*(d-1))
#[1] "2014-07-01" "2014-01-01" "2014-10-29" "2014-04-01" "2014-07-08"
I was wondering if there is a way to get the begin of the week date based on a week number in R? For example, if I enter week number = 10, it should give me 9th March, 2014.
I know how to get the reverse (aka..given a date, get the week number by using as.POSIX functions).
Thanks!
Prakhar
You can try this:
first.day <- as.numeric(format(as.Date("2014-01-01"), "%w"))
week <- 10
as.Date("2014-01-01") + week * 7 - first.day
# [1] "2014-03-09"
This assumes weeks start on Sundays. First, find what day of the week Jan 1 is, then, just add 7 * number of weeks to Jan 1, - the day of week Jan 1 is.
Note this is slightly different to what you get if you use %W when doing the reverse, as from that perspective the first day of the week seems to be Monday:
format(seq(as.Date("2014-03-08"), by="1 day", len=5), "%W %A %m-%d")
# [1] "09 Saturday 03-08" "09 Sunday 03-09" "10 Monday 03-10" "10 Tuesday 03-11"
# [5] "10 Wednesday 03-12"
but you can adjust the above code easily if you prefer the Monday centric view.
You may try the ISOweek2date function in package ISOweek.
Create a function which takes year, week, weekday as arguments and returns date(s):
date_in_week <- function(year, week, weekday){
w <- paste0(year, "-W", sprintf("%02d", week), "-", weekday)
ISOweek2date(w)
}
date_in_week(year = 2014, week = 10, weekday = 1)
# [1] "2014-03-03"
This date is corresponds to an ISO8601 calendar (see %V in ?strptime). I assume you are using the US convention (see %U in ?strptime). Then some tweeking is needed to convert between ISO8601 and US standard:
date_in_week(year = 2014, week = 10 + 1, weekday = 1) - 1
# [1] "2014-03-09"
You can enter several weekdays, e.g.
date_in_week(year = 2014, week = 10 + 1, weekday = 1:3) - 1
# [1] "2014-03-09" "2014-03-10" "2014-03-11"
You can also use strptime to easily get dates from weeks starting on Mondays:
first_date_of_week <- function(year, week){
strptime(paste(year, week, 1), format = "%Y %W %u")
}
You can accomplish this using the package lubridate
library(lubridate)
start = ymd("2014-01-01")
#[1] "2014-01-01 UTC"
end = start+10*weeks()
end = end-(wday(end)-1)*days()
#[1] "2014-03-09 UTC"
How can I obtain the date that is a set number of periods later/earlier than a given date? Eg if x <- as.Date("2001-01-01"), how do I get a date which is 6 months later (2001-07-01) or earlier (2000-07-01) than x? Complicating factors include the day of month of the initial date, or the number of days per month.
library(lubridate)
ymd("2001-01-01") + months(6)
ymd("2001-01-01") - months(6)
seq( as.Date("2001/01/01"), by = "6 months", length = 2)[2]
# [1] "2001-07-01"