CODEWITHSUNDEEP
rapply
2sb
2sb

Reputation: 649

How to use R apply without changing a function

I have a basic character date to POSIXct function

charDateToPosixct <- function(chrDate, format, timeZone) {
    as.POSIXct(chrDate, format=format, tz=timeZone)
}

I have a data frame with date as character and timezone. 

chrDate <- c("4/25/2012","4/24/2012","4/16/2012","6/30/2012")
timeZone <- c("US/Eastern","US/Central","US/Pacific","US/Eastern")
df <- data.frame(date=chrDate,timezone=timeZone)
str(df)
'data.frame':   4 obs. of  2 variables:
 $ date    : Factor w/ 4 levels "4/16/2012","4/24/2012",..: 3 2 1 4
 $ timezone: Factor w/ 3 levels "US/Central","US/Eastern",..: 2 1 3 2

I want to change the data type of date to POSIXct and so want to apply this function charDateToPosixct to each row with the format "%m/%d/%y".

Upvotes: 0

Views: 218

Answers (1)

GSee
GSee

Reputation: 49810

Use an anonymous function in apply like this

apply(df, 1, function(x) charDateToPosixct(x[1], "%m/%d/%y", x[2]))
#[1] 1587787200 1587704400 1587020400 1593489600

edit

You could convert the numeric vector you got using apply to POSIXct

as.POSIXct(apply(df, 1, function(x) charDateToPosixct(x[1], "%m/%d/%y", x[2])), origin='1970-01-01')
[1] "2020-04-25 05:00:00 CDT" "2020-04-24 06:00:00 CDT" "2020-04-16 08:00:00 CDT" "2020-06-30 05:00:00 CDT"

Or, you can use lapply to get a list of POSIXct objects. Then you can combine with do.call c, or Reduce (You have to use as.character because you implicitly used stringsAsFactors=TRUE when you created the data.frame)

do.call(c, lapply(seq_len(NROW(df)), function(i) {
  charDateToPosixct(as.character(df$date[i]), "%m/%d/%Y", as.character(df$timezone[i]))
}))

Or, if you have 2 vectors, you don't even need the data.frame

do.call(c, lapply(seq_along(chrDate), function(i) {
    charDateToPosixct(chrDate[i], "%m/%d/%Y", timeZone[i])
}))

Or, using Reduce

Reduce(c, lapply(seq_along(chrDate), function(i) {
    charDateToPosixct(chrDate[i], "%m/%d/%Y", timeZone[i])
}))

My timezone is CDT, so any of the above give me

[1] "2012-04-24 23:00:00 CDT" "2012-04-24 00:00:00 CDT" "2012-04-16 02:00:00 CDT" "2012-06-29 23:00:00 CDT"

Upvotes: 3

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