I have an date column with dates like this
Mon Jun 29 13:28:37 BRT 2020
I would like to convert it to something like 29/06/2020.
How can i do this using R?
Maybe this helps:
date<-Sys.time()
format(date, "%d/%m/%Y")
The output of the call from "format" should look like:
"16/04/2021"
With uppercase lettres of the formatstring format the date into the long version of the year.
format(date, "%d/%m/%y") would result into:
"16/04/21"
Related
Say I want to store 1 February, 2003 as a date using a day-month-year without century format and I do the following:
> date <- as.Date("01-02-03", format = "%d-%m-%y")
And then I forget what format that date was in, and do this:
> date
[1] "2003-02-01"
In order to determine if 01 is the month or the day, I can do this:
> format(date, "%d")
[1] "01"
But is there a way to directly retrieve the format this date was stored in?
R only stores dates Date objects in one format. And technically it's a numeric format. The value is only formatted to look like a date when you print() the value. There is no formatting stored with the Date object. For example
dput(as.Date("2003-02-01"))
# structure(12084, class = "Date")
Dates are always printed yyyy-mm-dd ISO style when using the default print method. You can change that by using format() or strftime(). But the internal representation is always the same.
I want to merge two data frames in R programming language using the date as the primary key. While trying to change the date, time format on one of the data frames to date only, im getting NA on the date column. Below is the date time format which i want to change to mm dd yy only.
4/12/2016 0:00
This is the code chunk i used.
sleep_day <- sleep_day %>%
rename(date = sleepday) %>%
mutate(date = as.Date(date,format ="%m/%d/%Y %I:%M:%S %p" , tz=Sys.timezone()))
i am expecting the date column to change from date, time to date alone. ie from mm dd yy 00:00 to mm dd yy. The result i got on the date column is NA in R programming
Your format is not correct:
test <- "4/12/2016 0:00"
as.Date(test,format ="%m/%d/%Y %H" , tz=Sys.timezone())
will work. Look at ?strptime.
As an advice, prefer to work with lubridate library, with has easy-to-use functions, which parse a lot of different formats:
library(lubridate)
mdy_hm(test)
"2016-04-12 UTC"
I'd like to convert the following date format: Wed 1st Jan 2014 to 01/01/2014
The suffix following the 1 is giving me the most trouble and preventing me from using the standard methods.
One option would be dmy from lubridate to convert to Date class and then format it to required format
library(lubridate)
format(dmy(str1), "%d/%m/%Y")
#[1] "01/01/2014"
data
str1 <- "Wed 1st Jan 2014"
6/9/2013 1:15
7/9/2013 1:15
I have a series of data in a column of this format.
I am trying to convert it into a month and store it in a different column.
How can I do it?
Intended output
June
July
I tried using the lubridate library but not able to.
How about
x <- c("6/9/2013 1:15","7/9/2013 1:15")
strftime(as.Date(x, format="%m/%d/%Y %H:%M"), "%B")
# [1] "June" "July"
First convert to a proper date via as.Date() then use the formatting options in strftime to get the month name.
I have a data.frame (CSV originally) in R with dates are in the following 3 formats:
2011-06-02T17:16:05Z
2012-06-02T17:16:05-07:00
6/2/11 17:16:05
which is year-month-day-time. I don't quite know what the -07:00 is, as it seems to be the same for all timestamps (except for some where it is -08:00), but I guess it's some type of time zone offset.
I am not quite sure what format these are (does anyone know?), but I need to convert it to this format:
6/2/11 17:16:05
which is year-month-day-time
I would like to do this in such a way so that all the dates in the CSV (in one and the same row) is converted to the second format. How can I accomplish this in R?
The full dataset can be found here.
Here's another attempt, assuming your data is text to start with:
test <- c("2011-06-02T17:16:05Z","2012-06-02T17:16:05-07:00")
format(as.POSIXct(test,format="%Y-%m-%dT17:%H:%M"),"%m/%d/%y %H:%M")
[1] "06/02/11 16:05" "06/02/12 16:05"
You can try the following, where myDates would be the column of dates
format(strptime(myDates, format="%Y-%m-%dT17:%H:%M"), format= "%m/%d/%Y %H:%M")
[1] "06/02/2011 16:05" "06/02/2012 16:05"
or with 2-digit year
# Note the lower-case %y at the end
format(strptime(myDates, format="%Y-%m-%dT17:%H:%M"), format= "%m/%d/%y %H:%M")
[1] "06/02/11 16:05" "06/02/12 16:05"
As for the Z, that indicates GMT (think: London).
the -7:00 indicates 7 hours back from GMT (think: Colorado / MST etc)
Please see here for more reference