Pat Schmitz
2009-Aug-18 07:42 UTC
[R] Applying Logical statement to DateTime string as factor
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]]
Gabor Grothendieck
2009-Aug-18 10:18 UTC
[R] Applying Logical statement to DateTime string as factor
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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