Find the day of a week

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"))