lubridate errors in R

Question

I am using the following code to give me the day of the week from a date (in the form dd/mm/yyyy).

Edit: I have uploaded a more relvant dataset.

df <- structure(list(Date = c("18/01/2013", "18/01/2013", "18/01/2013", 
                    "18/01/2013", "18/01/2013"), Time = c("07:25:30", "07:25:40", 
                                                          "07:25:50", "07:26:00", "07:26:10"), Axis1 = c(217L, 320L, 821L, 
                                                                                                         18L, 40L), Steps = c(6L, 7L, 5L, 1L, 1L), wday = c(7, 7, 7, 7, 7)), .Names = c("Date", "Time", "Axis1", "Steps", "wday"), row.names = 18154:18158, class = "data.frame")


library(lubridate)
df$wday = wday(df$Date)
df$wday.name = wday(df$Date, label = TRUE, abbr = TRUE)

The 18/1 was however a Friday and not a Saturday as R reports.

Does anyone have any suggestions of how to rectify this?

EDIT: I tried to follow the suggestions given by Dirk...

as.POSIXlt(df[,1])$wday

... but this still implies that the 18/1 is a Saturday.

My timezone is GMT/UTC (+ 1 for British Summer Time), however because I just want R to read from the date column (which is just d/m/y), I presume I don't need to specify this...

How can I get a correct wday column to be added to my existing R dataframe? (as detailed previously in my original script). I am struggling to get the suggested coding working as I gave the dataframe in the wrong format - apologies.

Answer 1

You can use base R functions for this. Using your df object:

 R> as.POSIXlt(df[,1])$wday  
 [1] 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 
 R> weekdays(as.Date(df[,1])) 
  [1] "Friday"   "Friday"   "Friday"   "Friday"   "Friday"
  [6] "Friday"   "Friday"   "Friday"   "Friday"   "Friday" 
 [11] "Friday"   "Friday"   "Friday"   "Friday"   "Saturday"  
 [16] "Saturday" "Saturday" "Saturday" "Saturday" 
 R>     

There is a spillover into Saturday for the end because the TZ was not specified.

If you do

 R> df <- data.frame(Date=seq(as.POSIXct("05:00", format="%H:%M", tz="UTC"),
 +                  as.POSIXct("23:00", format="%H:%M", tz="UTC"), by="hours"))

then

 R> table(weekdays(as.Date(df[,1], TZ="UTC")))

 Friday
    19
 R> 

I presume the Fri/Sat error may go away under lubridate too, but I tend to use base R functions for this.

Edit: Confirmed.

R> lubridate::wday(as.Date(df[,1]), label=TRUE) 
 [1] Fri Fri Fri Fri Fri Fri Fri Fri Fri Fri Fri Fri Fri Fri 
[15] Fri Fri Fri Fri Fri          
Levels: Sun < Mon < Tues < Wed < Thurs < Fri < Sat  
R>