get weekday out of a dateset in R [duplicate] - r

Question

Let's say that I have a date in R and it's formatted as follows.
date
2012-02-01
2012-02-01
2012-02-02
Is there any way in R to add another column with the day of the week associated with the date? The dataset is really large, so it would not make sense to go through manually and make the changes.
df = data.frame(date=c("2012-02-01", "2012-02-01", "2012-02-02"))
So after adding the days, it would end up looking like:
date day
2012-02-01 Wednesday
2012-02-01 Wednesday
2012-02-02 Thursday
Is this possible? Can anyone point me to a package that will allow me to do this?
Just trying to automatically generate the day by the date.

Answer 1

df = data.frame(date=c("2012-02-01", "2012-02-01", "2012-02-02"))
df$day <- weekdays(as.Date(df$date))
df
## date day
## 1 2012-02-01 Wednesday
## 2 2012-02-01 Wednesday
## 3 2012-02-02 Thursday
Edit: Just to show another way...
The wday component of a POSIXlt object is the numeric weekday (0-6 starting on Sunday).
as.POSIXlt(df$date)$wday
## [1] 3 3 4
which you could use to subset a character vector of weekday names
c("Sunday", "Monday", "Tuesday", "Wednesday", "Thursday",
"Friday", "Saturday")[as.POSIXlt(df$date)$wday + 1]
## [1] "Wednesday" "Wednesday" "Thursday"

Answer 2

Use the lubridate package and function wday:
library(lubridate)
df$date <- as.Date(df$date)
wday(df$date, label=TRUE)
[1] Wed Wed Thurs
Levels: Sun < Mon < Tues < Wed < Thurs < Fri < Sat

Answer 3

Look up ?strftime:
%A Full weekday name in the current locale
df$day = strftime(df$date,'%A')

Answer 4

Let's say you additionally want the week to begin on Monday (instead of default on Sunday), then the following is helpful:
require(lubridate)
df$day = ifelse(wday(df$time)==1,6,wday(df$time)-2)
The result is the days in the interval [0,..,6].
If you want the interval to be [1,..7], use the following:
df$day = ifelse(wday(df$time)==1,7,wday(df$time)-1)
... or, alternatively:
df$day = df$day + 1

Answer 5

This should do the trick
df = data.frame(date=c("2012-02-01", "2012-02-01", "2012-02-02"))
dow <- function(x) format(as.Date(x), "%A")
df$day <- dow(df$date)
df
#Returns:
date day
1 2012-02-01 Wednesday
2 2012-02-01 Wednesday
3 2012-02-02 Thursday

Answer 6

start = as.POSIXct("2017-09-01")
end = as.POSIXct("2017-09-06")
dat = data.frame(Date = seq.POSIXt(from = start,
to = end,
by = "DSTday"))
# see ?strptime for details of formats you can extract
# day of the week as numeric (Monday is 1)
dat$weekday1 = as.numeric(format(dat$Date, format = "%u"))
# abbreviated weekday name
dat$weekday2 = format(dat$Date, format = "%a")
# full weekday name
dat$weekday3 = format(dat$Date, format = "%A")
dat
# returns
Date weekday1 weekday2 weekday3
1 2017-09-01 5 Fri Friday
2 2017-09-02 6 Sat Saturday
3 2017-09-03 7 Sun Sunday
4 2017-09-04 1 Mon Monday
5 2017-09-05 2 Tue Tuesday
6 2017-09-06 3 Wed Wednesday

Answer 7

form comment of JStrahl format(as.Date(df$date),"%w"), we get number of current day :
as.numeric(format(as.Date("2016-05-09"),"%w"))