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"
Related
I imported csv data saved from excel and I am using a Mac if that matters.
To simplify things, I have been working with one particular entry from my data "Jan-12". This is of type character and in the form Month-Year.
This is what I have tried:
as.Date("Jan-12", format = "%b-%y")
I keep getting NA. I have browsed through other answers but haven't been able to figure out whats happening.
as.Date() help page says:
If the date string does not specify the date completely, the returned
answer may be system-specific.
If you really need to use as.Date() you can append some fixed day (say 1st) to date and convert
mydate <- "Jan-12"
as.Date(paste0("01-", mydate), format= "%d-%b-%y")
"2012-01-01"
Or you can use lubridate::fast_strptime():
library(lubridate)
fast_strptime("Jan-12", "%b-%y")
"2012-01-01 UTC"
Here is another option using zoo, then you can use as.Date.
library(zoo)
as.Date(as.yearmon("Jan-12", "%b-%y"))
# [1] "2012-01-01"
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"
I am trying to parse the following datetime with the following format:
library(lubridate)
a <- "2004-05-07 18:24:58.666424"
I tried the following, but returned NAs
b <- lubridate::mdy_hms(a)
c <- lubridate::mdy(a)
Could anyone please explain how to parse this. I am also fine if lubridate is not used.
With lubridate, you can specify that your seconds have a decimal with the special S! or OS formats; see ?parse_date_time for more parsing options.
> parse_date_time("2004-05-07 18:24:58.666424", 'ymd HMS!')
[1] "2004-05-07 18:24:58 UTC"
Alternately, it seems to parse fine with just the usual default "ymd HMS":
parse_date_time("2004-05-07 18:24:58.666424", 'ymd HMS')
or the shorthand
ymd_hms("2004-05-07 18:24:58.666424")
Try
options(digits.secs=6)
as.POSIXct(a,"%Y-%m-%d %H:%M:%S.%OS")
#[1] "2004-05-07 18:24:58.666424"
mdy=Month day year, your data is setup as ymd
Try ymd_hms or ymd
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"
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")