CODEWITHSUNDEEP
rdater-faq
ATMathew
ATMathew

Reputation: 12856

Find the day of a week

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.

Upvotes: 276

Views: 308418

Answers (7)

nograpes
nograpes

Reputation: 18323

Look up ?strftime:

%A Full weekday name in the current locale

df$day = strftime(df$date,'%A')

Upvotes: 79

s_scolary
s_scolary

Reputation: 1399

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

Upvotes: 10

Qbik
Qbik

Reputation: 6147

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"))

Upvotes: 4

GSee
GSee

Reputation: 49810

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"

Upvotes: 368

Peter Lustig
Peter Lustig

Reputation: 953

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

Upvotes: 18

Tyler Rinker
Tyler Rinker

Reputation: 109844

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

Upvotes: 14

Andrie
Andrie

Reputation: 179398

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

Upvotes: 99

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