[R] Applying Logical statement to DateTime string as factor

[Gabor Grothendieck]

Try this:


Lines <- '"1","30 Jun 2009 18:14:59"
"2","02 Jul 2009 07:33:37"
"3","06 Jul 2009 08:22:35"
"4","06 Jul 2009 19:25:50"
"5","08 Jul 2009 08:41:48"
"6","10 Jul 2009 07:31:39"
"7","10 Jul 2009 19:59:25"
"8","13 Jul 2009 07:49:18"
"9","13 Jul 2009 18:52:52"
"10","15 Jul 2009 08:06:56"
"11","15 Jul 2009 19:03:01"
"12","17 Jul 2009 08:02:02"'

library(chron)
DF <- read.csv(textConnection(Lines), header = FALSE, as.is = TRUE)
dd <- as.chron(DF$V2, "%d %b %Y %H:%M:%S")

dd - dates(dd) > times("12:00:00")  # c(TRUE, FALSE, ...)


See R News 4/1 and its references for more.


On Tue, Aug 18, 2009 at 3:42 AM, Pat Schmitz<pschmitz at illinois.edu> wrote:
> R-Help
>
> I have a data set which uses a DateTime string as follows : "2009-06-30
> 18:14:59"
> While I have been able to convert to DateTime properly
>
> time <- strptime(as.character(dat$Time),format='%d %b %Y %T') #Convert to
> dateTime string
>
> I would like to use the time of day "hour" as a *factor* level.  I have
> found that I can convert the date time to a factor
>
> time <- as.ordered(cut(time,  "hour"))
>
> But this produces a factor level for every single day.  The issue I have is
> in figuring out how to apply a logical statement to the datetime string or
> to the factor.  I would like to say that every datetime with an hour <12 is
> the morning factor, and every datetime with an hour >12 is at night. Without
> LOTS of superfluous code.  below I am going as far as splitting the string
> to columns, pulling the time and then setting the factor with a for
> statement...yeesh.  I have example CSV dates at the end.
>
> new.dates <- matrix(unlist(lapply(as.character(dat$Time), function(x)
> strsplit(x," "))),ncol=4, byrow=TRUE)  #Break Time String into columns
> colnames(new.dates) <- c("day","month","year", "time") # Set column names
> dat <- cbind(as.data.frame(new.dates),  as.data.frame(dat)) #bind to data
> frame
>
> tfact<-matrix()
>
> for (i in which(as.character(dat$time) <= "12:00:00")){
>    tfact[i] <- "Morning"
>    }
>
> for (i in which(as.character(dat$time) >= "12:00:00")){
>    tfact[i] <- "Night"
>    }
>
> tfact<-factor(tfact,
>    levels=c("Morning",  "Night"),
>    )
>
> Example CSV Data
> "1","30 Jun 2009 18:14:59"
> "2","02 Jul 2009 07:33:37"
> "3","06 Jul 2009 08:22:35"
> "4","06 Jul 2009 19:25:50"
> "5","08 Jul 2009 08:41:48"
> "6","10 Jul 2009 07:31:39"
> "7","10 Jul 2009 19:59:25"
> "8","13 Jul 2009 07:49:18"
> "9","13 Jul 2009 18:52:52"
> "10","15 Jul 2009 08:06:56"
> "11","15 Jul 2009 19:03:01"
> "12","17 Jul 2009 08:02:02"
>
>
> Thanks
> Pat
>
> --
> Patrick Schmitz
> Graduate Student
> Plant Biology
> 1206 West Gregory Drive
> RM 1500
>
>        [[alternative HTML version deleted]]
>
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