Reputation: 496
My data, DATA, has a variable, TIME, for which the values print out in this format: "11/14/2006 20:10". For TIME, its mode is numeric and its class is a factor.
I need to convert TIME to a proper date/time variable (DTIME) and add the new DTIME to DATA as date.time. I was told I may have to coerce the time values so that they follow the h:m:s format...Think character string manipulation. Below is my code:
library("chron")
VAR=c(as.character(DATA$TIME))
DT<-t(as.data.frame(strsplit(VAR," ")))
DT[1:3,]
row.names(DT)<-NULL
DT[1:3,]
DTIME<-chron(dates=DT[,1],times=DT[,2],
format=c("m/d/y","h:m"))
But once I run the last line of code, I get the following error message:
Error in convert.times(times., format = format[[2]]) :
format h:m may be incorrect
In addition: Warning message:
In is.na(out$s) : is.na() applied to non-(list or vector) of type 'NULL'
I don't understand what this means, much less how to fix it.
Upvotes: 1
Views: 5736
Reputation: 17189
You can use as.POSIXct
to convert string into time.
TIME <- "11/14/2006 20:10"
as.POSIXct(TIME, format="%m/%d/%Y %H:%M", tz='GMT')
## [1] "2006-11-14 20:10:00 GMT"
Upvotes: 1
Reputation: 269441
Its not clear from the question exactly what you have -- in such cases its best to show the output of dput
applied to your variable -- but assuming you can convert it to character format using as.character
or format
then its just a matter of using as.chron
:
> library(chron)
> TIME <- "11/14/2006 20:10"
> as.chron(TIME, "%m/%d/%Y %H:%M")
[1] (11/14/06 20:10:00)
Upvotes: 4
Reputation: 4094
Use lubridate
- it's a fantastic library that will save you much effort.
library(lubridate)
x <- "11/14/2006 20:10"
> mdy_hm(x)
[1] "2006-11-14 20:10:00 UTC"
All lubridate
's time conversion function follow a very similar pattern: e.g. "2013-04-01"
can be parsed with ymd
, etc.
Upvotes: 1