I have a simple question regarding R's lubridate package. I've a series of timestamps in seconds since epoch. I want to convert this to YYYY-MM-DD-HH format. In base R, I can do something like this to first convert it to a date format
> x = as.POSIXct(1356129107,origin = "1970-01-01",tz = "GMT")
> x
[1] "2012-12-21 22:31:47 GMT"
Note the above just converts it to a date format, not the YYYY-MM-DD-HH format. How would I do this in lubridate? How would I do it using base R?
Thanks much in advance
lubridate has an as_datetime() that happens to have UNIX epoch time as the default origin time to make this really simple:
> as_datetime(1356129107)
[1] "2012-12-21 22:31:47 UTC"
more details can be found here: https://rdrr.io/cran/lubridate/man/as_date.html
Dirk is correct. However, if you are intent on using lubridate functions:
paste( year(dt), month(dt), mday(dt), hour(dt) sep="-")
If on the other hand you want to handle the POSIXct objects the way they were supposed to be used then this should satisfy:
format(x, format="%Y-%m-%d-%H")
I use the lubridate solution provided by #leerssej
But in case anyone prefers #IRTFM's solution in base R, but also wants minutes and seconds, here's an example of how to do that:
as.POSIXct("2019-03-15 16:17:42" , format="%Y-%m-%d %H:%M:%OS")
Related
So I realized that this isn't a common date type to deal with at least with using as.Date(). When I do the following , the output isn't correct.
> as.Date(Sys.Date(), format = "yyyy.mm.dd")
[1] "2022-06-21"
Is there an easy way to this with lubridate or base R?
We can use format instead of as.Date as Sys.Date() is already in Date class, however as commented, format returns only a character class
format(Sys.Date(), '%Y.%m.%d')
A client sent me an Excel file with dates formatted as e.g 3/15/2012 for March 15. I saved this as a .csv file and then used
camm$Date <- as.Date(camm$Date, "%m/%d/%y")
but this gave me values starting in the year 2020!
I tried to reformat the dates in the original csv file so that they were e.g. 03/14/2013 but was unable to do so.
Any help appreciated
Use capital Y in as.Date call instead. This should do the trick:
> as.Date("3/15/2012", "%m/%d/%Y")
[1] "2012-03-15"
From the help file's examples you can realize when year is full specified you should use %Y otherwise %y for example:
> dates <- c("02/27/92", "02/27/92", "01/14/92", "02/28/92", "02/01/92")
> as.Date(dates, "%m/%d/%y")
[1] "1992-02-27" "1992-02-27" "1992-01-14" "1992-02-28" "1992-02-01"
You can see that in your example the Year format is 2012 then you should use %Y, and in the other example (taken from the as.Date help file) Year format is 92 then using %y is the correct way to go. See as.Date for further details.
You might also give a try to the lubridate package if you do not want to deal with the hieroglyphics :)
> library(lubridate)
> parse_date_time('3/15/2012', 'mdy')
1 parsed with %m/%d/%Y
[1] "2012-03-15 UTC"
PS.: of course I do not encourage anyone to use any extra dependencies, this answer was just posted here as an alternative (and quick to remeber) solution
To complete the picture, you might also try the recently introduced (2016-09) package anytime which takes advantage of the Boost C++ libraries:
anytime::anytime("3/15/2012")
#[1] "2012-03-15 CET"
We can use mdy from lubridate
lubridate::mdy('3/15/2012')
#[1] "2012-03-15"
Or parse_date from readr which uses same format as as.Date
readr::parse_date('3/15/2012', '%m/%d/%Y')
#[1] "2012-03-15"
This should be simple but I can't figure it out. How should I go about formatting dates that are '20150703' into '07-03-2015'? Thanks
You may use format after converting to 'Date' class
format(as.Date(dates, '%Y%m%d'), '%m-%d-%Y')
#[1] "07-03-2015"
data
dates <- '20150703'
Also take a look at lubridate package here which makes it easier to work with dates.
ymd("20150703")
gives
[1] "2015-07-03 UTC"
A client sent me an Excel file with dates formatted as e.g 3/15/2012 for March 15. I saved this as a .csv file and then used
camm$Date <- as.Date(camm$Date, "%m/%d/%y")
but this gave me values starting in the year 2020!
I tried to reformat the dates in the original csv file so that they were e.g. 03/14/2013 but was unable to do so.
Any help appreciated
Use capital Y in as.Date call instead. This should do the trick:
> as.Date("3/15/2012", "%m/%d/%Y")
[1] "2012-03-15"
From the help file's examples you can realize when year is full specified you should use %Y otherwise %y for example:
> dates <- c("02/27/92", "02/27/92", "01/14/92", "02/28/92", "02/01/92")
> as.Date(dates, "%m/%d/%y")
[1] "1992-02-27" "1992-02-27" "1992-01-14" "1992-02-28" "1992-02-01"
You can see that in your example the Year format is 2012 then you should use %Y, and in the other example (taken from the as.Date help file) Year format is 92 then using %y is the correct way to go. See as.Date for further details.
You might also give a try to the lubridate package if you do not want to deal with the hieroglyphics :)
> library(lubridate)
> parse_date_time('3/15/2012', 'mdy')
1 parsed with %m/%d/%Y
[1] "2012-03-15 UTC"
PS.: of course I do not encourage anyone to use any extra dependencies, this answer was just posted here as an alternative (and quick to remeber) solution
To complete the picture, you might also try the recently introduced (2016-09) package anytime which takes advantage of the Boost C++ libraries:
anytime::anytime("3/15/2012")
#[1] "2012-03-15 CET"
We can use mdy from lubridate
lubridate::mdy('3/15/2012')
#[1] "2012-03-15"
Or parse_date from readr which uses same format as as.Date
readr::parse_date('3/15/2012', '%m/%d/%Y')
#[1] "2012-03-15"
My code:
axis.Date(1,sites$date, origin="1970-01-01")
Error:
Error in as.Date.numeric(x) : 'origin' must be supplied
Why is it asking me for the origin when I supplied it in the above code?
I suspect you meant:
axis.Date(1, as.Date(sites$date, origin = "1970-01-01"))
as the 'x' argument to as.Date() has to be of type Date.
As an aside, this would have been appropriate as a follow-up or edit of your previous question.
My R use 1970-01-01:
>as.Date(15103, origin="1970-01-01")
[1] "2011-05-09"
and this matches the calculation from
>as.numeric(as.Date(15103, origin="1970-01-01"))
So generally this has been solved, but you might get this error message because the date you use is not in the correct format.
I know this is an old post, but whenever I run this I get NA all the way down my date column. My dates are in this format 20150521 – NealC Jun 5 '15 at 16:06
If you have dates of this format just check the format of your dates with:
str(sides$date)
If the format is not a character, then convert it:
as.character(sides$date)
For as.Date, you won't need an origin any longer, because this is supplied for numeric values only. Thus you can use (assuming you have the format of NealC):
as.Date(as.character(sides$date),format="%Y%m%d")
I hope this might help some of you.
Another option is the lubridate package:
library(lubridate)
x <- 15103
as_date(x, origin = lubridate::origin)
"2011-05-09"
y <- 1442866615
as_datetime(y, origin = lubridate::origin)
"2015-09-21 20:16:55 UTC"
From the docs:
Origin is the date-time for 1970-01-01 UTC in POSIXct format. This date-time is the origin for the numbering system used by POSIXct, POSIXlt, chron, and Date classes.
If you have both date and time information in the numeric value, then use as.POSIXct. Data.table package IDateTime format is such a case. If you use fwrite to save a file, the package automatically converts date-times to idatetime format which is unix time. To convert back to normal format following can be done.
Example: Let's say you have a unix time stamp with date and time info: 1442866615
> as.POSIXct(1442866615,origin="1970-01-01")
[1] "2015-09-21 16:16:54 EDT"
by the way, the zoo package, if it is loaded, overrides the base as.Date() with its own which, by default, provides origin="1970-01-01".
(i mention this in case you find that sometimes you need to add the origin, and sometimes you don't.)