Converting character string to numeric value in user defined function R

Question

I am trying to create a function that will take an input of character strings "mm/did/yyyy" and return an output vector in numeric form the month, day and year. I essentially want to combine this new function with the weekday() function created below to ultimately return the day of the week the input character string corresponds with.

weekday<-function(q,r,s)
{ (if(q>= 3)
 m<-(q-2)
else
m<-(q+10))
k<-r
 c<-floor(s/100)
(if(q>=3)
y<-s%%100
else
y<-(s%%100)-1)
 f<-(floor((2.6*m)-0.2)+k+y+floor(y/4)+floor(c/4)-(2*c))%%7
if(f==0){return("Sunday")}
   else 
if(f==1){return("Monday")}
 else
 if(f==2){return("Tuesday")}
 else
 if(f==3){return("Wednesday")}
 else
  if(f==4){return("Thursday")}
 else
  if(f==5){return("Friday")}
else
  if(f==6){return("Saturday")}}

I tried using something along the lines of type.convert but this isn't producing the desired output. Any help would be great thanks!

 dateconvert<-function("q/r/s")
  {
    type.convert(dateconvert(), na.strings = )

    weekday(convertedanswer)
    Return (weekday)


 }

R has functions to convert character string to POSIX or datetime format then you can get either months, weekdays, days, year, or time from that datetime format. Why dont you use it? — addicted
– addicted, Commented Oct 6, 2017 at 3:04
Oddly, there is a plural version of your user built function that is included in base R: ?weekdays. — lmo
– lmo, Commented Oct 6, 2017 at 11:08

MingH · Accepted Answer · 2017-10-06 03:19:13Z

1

Have you tried lubridate package?

input <- "12/30/2017"

# change into as.date format
inputdate <- strptime(input, "%m/%d/%Y")

library("lubridate")
day(inputdate)
# [1] 30
month(inputdate)
# [1] 12
year(inputdate)
# [1] 2017

It seems like a roundabout way to get to the day of week though. You should try using wday() that comes with lubridate package.

wday(inputdate, label=T)
# [1] Sat
# Levels: Sun < Mon < Tues < Wed < Thurs < Fri < Sat. 
# as a Ordered factor (Sunday is first)

wday(inputdate)
# [1] 7 
# wday returns the day of the week as a decimal number (01-07, Sunday is 1) or an .

answered Oct 6, 2017 at 3:19

MingH

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Comments

addicted · Accepted Answer · 2017-10-06 03:23:41Z

1

Alternatively without package:

date = "06/10/2017"
POSIXdate = as.POSIXlt(date, format = "%d/%m/%Y")
strftime(POSIXdate, "%A")
#Friday

# Or if you  like one-liner
strftime(as.POSIXlt("dd/mm/yyyy", format= "%d/%m/%Y"), "%A")

answered Oct 6, 2017 at 3:23

addicted

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3 Comments

lmo Over a year ago

Also weekdays in base R.

Lin Over a year ago

Is there any way to do this so that a function created newfunction<-("10/06/17") would produce the vector 10 06 17 ? Ideally I would like for this to happen, so that I can put this result into my weekday() function above.

addicted Over a year ago

Do you want to have input of character d <- "10/06/17" and output a vector of character c("10" , "06", "17")? If so: strsplit("10/06/17", "/")[[1]].

Lucy · Accepted Answer · 2017-10-07 02:19:04Z

1

The lubridate package is great for this! You can use lubridate::mdy() to convert a date in the "mm/dd/yyyy" formate you mentioned, and then lubridate::week() to get the week.

lubridate::week(lubridate::mdy("10/05/2017"))
#> 5

If you want the day itself, rather than the numeric output, you can use

lubridate::wday(lubridate::mdy("10/05/2017"),
                label = TRUE)
#> [1] Thurs

edited Oct 7, 2017 at 2:19

answered Oct 6, 2017 at 3:17

Lucy

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