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"
Related
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"
If I have a given date, how do I find the first and last days of the next month?
For example,
today <- as.Date("2009-04-04")
I want to find
# first date in next month
"2009-05-01"
# last date in next month
"2009-05-31"
You can do this with base R:
today <- as.Date("2009-04-04")
first <- function(x) {
x <- as.POSIXlt(x)
x$mon[] <- x$mon + 1
x$mday[] <- 1
x$isdst[] <- -1L
as.Date(x)
}
first(today)
#[1] "2009-05-01"
first(first(today)) - 1
#[1] "2009-05-31"
lubridate has some useful tools for this purpose.
library(lubridate)
today <- ymd("2009-04-12")
# First day of next month
first <- ceiling_date(today, unit = "month")
# Last day of next month
last <- ceiling_date(first, unit= "month") -1
first
#"2009-05-01"
last
#"2009-05-31"
Here are some solutions. We use today from the question to test. In both cases the input may be a Date class vector.
1) Base R Define function fom to give the first of the month of its Date
argument. Using that we can get the date of the first and last of the next month as follows. We use the facts that 31 and 62 days after the first of the month is necessarily a date in the next month and month after the next month.
fom <- function(x) as.Date(cut(x, "month"))
fom(fom(today) + 31)
## [1] "2009-05-01"
fom(fom(today) + 62) - 1
## [1] "2009-05-31"
2) yearmon yearmon class objects internally represent a year and month as the year plus 0 for January, 1/12 for Febrary, 2/12 for March and so on. Using as.Date.yearmon the frac argument specifies the fraction of the way through the month to output. The default is frac = 0 and results in the first of the month being output and frac = 1 means the end of the month.
library(zoo)
as.Date(as.yearmon(today) + 1/12)
## [1] "2009-05-01"
as.Date(as.yearmon(today) + 1/12, frac = 1)
## [1] "2009-05-31"
Is there a way in R to get the 1st and the last day for a specified month.
Eg.
Input: September 2018 or any other format to specify month and year
Expected output:
1st day function (Input) -> 01-Sep-2018 or any other valid date format
Last day function (Input) -> 30-Sep-2018 or any other valid date format
We can create a function in base R
get_first_and_last_date <- function(month_year) {
start_date = as.Date(paste0("01 ", month_year), "%d %b %Y")
end_date = (seq(start_date, length.out = 2, by = "month") - 1)[2]
c(start_date, end_date)
}
get_first_and_last_date('Dec 2018')
#[1] "2018-12-01" "2018-12-31"
get_first_and_last_date('Sep 2016')
#[1] "2016-09-01" "2016-09-30"
Whatever format you enter make sure it is consistent throughout. Here I have considered the input would always be a month name and complete year.
Using the lubridate library:
require(lubridate)
d <- as.Date('2018-09-01')
last_day <- d
day(last_day) <- days_in_month(last_day)
For a base R solution, we can define a helper method to add months. Then, the last day of a given month can be computed by adding one month to the first of the month and subtracting one day:
add.months <- function(date, n) seq(date, by=paste (n, "months"), length=2 [2]
d <- as.Date('2018-09-01') # first of the month
last_day <- add.months(d, 1) - 1 # last of the month
Credit for the add.months helper function is given to this SO question.
How can a date/time object in R be transformed on the fraction of a julian day?
For example, how can I turn this date:
date <- as.POSIXct('2006-12-12 12:00:00',tz='GMT')
into a number like this
> fjday
[1] 365.5
where julian day is elapsed day counted from the january 1st. The fraction 0.5 means that it's 12pm, and therefore half of the day.
This is just an example, but my real data covers all the 365 days of year 2006.
Since all your dates are from the same year (2006) this should be pretty easy:
julian(date, origin = as.POSIXct('2006-01-01', tz = 'GMT'))
If you or another reader happen to expand your dataset to other years, then you can set the origin for the beginning of each year as follows:
sapply(date, function(x) julian(x, origin = as.POSIXct(paste0(format(x, "%Y"),'-01-01'), tz = 'GMT')))
Have a look at the difftime function:
> unclass(difftime('2006-12-12 12:00:00', '2006-01-01 00:00:00', tz="GMT", units = "days"))
[1] 345.5
attr(,"units")
[1] "days"
A function to convert POSIX to julian day, an extension of the answer above, source it before using.
julian_conv <- function(x) {
if (is.na(x)) { # Because julian() cannot accept NA values
return(NA)
}
else {
j <-julian(x, origin = as.POSIXlt(paste0(format(x, "%Y"),'-01-01')))
temp <- unclass(j) # To unclass the object julian day to extract julian day
return(temp[1] + 1) # Because Julian day 1 is 1 e.g., 2016-01-01
}
}
Example:
date <- as.POSIXct('2006-12-12 12:00:00')
julian_conv(date)
#[1] 345.5
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"