I obtained a time string looking like this:
201902041502, containing year, month, day, hour and minute.
Now I want to reformat this string into the german date-time format like this: 04.02.2019 15:02.
I've already tried as.Date and as.POSIXct but it doesnt work and I want to avoid adding seconds to get POSIXct to work.
Thanks in advance! Cheers
You can use strptime to convert the data to a POSIXlt object,
x <- "201902041502"
xd <- strptime(x,"%Y%m%d%H%M")
# [1] "2019-02-04 15:02:00 CET"
and then use strftime to produce your desired format:
strftime(xd, "%d.%m.%Y %H:%M")
# [1] "04.02.2019 15:02"
Related
Current situation:
mydate <- "14:45"
class(mydate)
The current class of this value is a character. I would like to convert it into a POSIXlt format.
I tried the strptime() function but it unfortunately adds the full date to my hours when I actually only need Hours:Minutes
mydate <- strptime(mydate, format = "%H:%M")
What can I do to get a POSIXlt format uniquely containing hours and minutes ?
Thanks in advance for your returns !
POSIXlt and POSIXct always contain date and time. You can use chron times class to represent times less than 24:00:00.
library(chron)
tt <- times(paste(mydate, "00", sep = ":"))
tt
## [1] 14:45:00
times class objects are represented internally as a fraction of a day so, for example, adding 1/24 will add an hour.
tt + 1/24 # add one hour
## [1] 15:45:00
For me it works like this:
test <- "2016-04-10T12:21:25.4278624"
z <- as.POSIXct(test,format="%Y-%m-%dT%H:%M:%OS")
#output:
z
"2016-04-10 12:21:25 CEST"
The code is form here: R: convert date from character to datetime
I have a data frame with time in character format, I need to convert it into time format. I have tried using strptime and POSIXct but it adds the date also. I just need the time.
For e.g.: TRK_DEP_TIME <- c("22:00", "14:30"......) _____ character datatype
doing........ as.POSIXCT(TRK_DEP_TIME, format = %H:%M")
The result will be ("10/11/17,22:00", "10/11/17, 14:30".....)
I am looking for just the time, I don't need the date to be associated with it. Is there any way I can achieve this?
Use chron "times" class:
library(chron)
ch <- c("22:00", "14:30") # test input (character)
times(paste0(ch, ":00"))
## [1] 22:00:00 14:30:00
I have a table (tags) with a column for timestamp (ts), which is formatted as seconds since 1 Jan, 1970 GMT. I'm trying to create a date column that converts the timestamp from seconds to date and time EST.
The suggested code for R was:
tags$date<-strptime(tags$ts, "%Y-%m-%d")
tags$date<-as.POSIXct(tags$date)
But when I do this, tags$date comes up as NA. Any suggestions for what I might be doing wrong? Thanks.
You should us as.POSIXct function instead:
tags$date <- as.POSIXct(tags$ts, origin="1970-01-01", tz="US/New York")
strptime converts between character representations and dates not between timestamp and dates.
Here's a lubridate version. When we use as_datetime we don't need to explicitly specify an origin as it defaults to the desired origin.
lubridate::as_datetime(1507119276, tz='EST')
# [1] "2017-10-04 07:14:36 EST"
The date July, 1, 2016 1:15pm and 43 seconds is given to me as the string 160701131543.
I have an entire column in my data frame of this date time. How should I go about parsing this column into usable data.
You can use the as.POSIXct function and specify the format, in your case the format is year, month, day, hour, minute, second. Read more about formatting date and time data on the ?strptime help page.
as.POSIXct("160701131543", format = "%y%m%d%H%M%S")
[1] "2016-07-01 13:15:43 EDT"
The timezone can be changed with the 'tz' parameter.
Here is another option with lubridate. The default tz is "UTC". It can be changed by specifying tz
library(lubridate)
ymd_hms("160701131543")
#[1] "2016-07-01 13:15:43 UTC"
I am trying to convert the date as factor to date using “as.date” function in R. I have the date in the following format
2008-01-01 02:30
I tried to use the following command :
as.Date(mydata$Date, format="%y-%m-%d %h:%mm")
Can somebody help me with this ? I was able to convert the format with no hour but getting difficulty with hour included.
Thank you.
Your format string is incorrect :
R> strptime("2008-01-01 02:30", format="%Y-%m-%d %H:%M")
# [1] "2008-01-01 02:30:00"
See ?strptime for the detailed values you can use to define a format.
Also note that as your string is in a standard format, you can also use directly as.POSIXlt :
R> as.POSIXlt("2008-01-01 02:30")
# [1] "2008-01-01 02:30:00"