I have log files where the date is mentioned in the ordinal date format.
wikipedia page for ordinal date
i.e 14273 implies 273'rd day of 2014 so 14273 is 30-Sep-2014.
is there a function in R to convert ordinal date (14273) to (30-Sep-2014).
Tried the date package but didn come across a function that would do this.
Try as.Date with the indicated format:
as.Date(sprintf("%05d", 14273), format = "%y%j")
## [1] "2014-09-30"
Notes
For more information see ?strptime [link]
The 273 part is sometimes referred to as the day of the year (as opposed to the day of the month) or the day number or the julian day relative to the beginning of the year.
If the input were a character string of the form yyjjj (rather than numeric) then as.Date(x, format = "%y%j") will do.
Update Have updated to also handle years with one digit as per comments.
Data example
x<-as.character(c("14273", "09001", "07031", "01033"))
Data conversion
x1<-substr(x, start=0, stop=2)
x2<-substr(x, start=3, stop=5)
x3<-format(strptime(x2, format="%j"), format="%m-%d")
date<-as.Date(paste(x3, x1, sep="-"), format="%m-%d-%y")
You can use lubridate package as follows:
>library(lubridate)
# Create a template date object
>date <- as.POSIXlt("2009-02-10")
# Update the date using
> update(date, year=2014, yday=273)
[1] "2014-09-30 JST"
Related
I want to convert date of birth to age using the following code
df$age <- round(as.numeric(Sys.Date()-as.Date(df$DOB),format="%d/%m/%y")/365)
The format of DOB is f.e. 11-10-1969.
In the dataframe I see an age of 2012 (instead of 52).
I really dont know what I've done wrong. Can someone help me?
Thank you in advance!
"11-10-1969" (month day year or day month year) is not an unambiguous date format. To get it properly converted you will need to specify the format argument to as.Date()
Note also that a 4-digit year needs a capital Y in the format string: "%d-%m-%Y" (or "%d/%m/%Y" for /). Sys.Date() is already a Date object, so you don't need the format argument with the /s in it.
> as.numeric(Sys.Date() - as.Date("11-10-1969", format="%d-%m-%Y")) / 365.25
#> [1] 52.56674
EDIT: use 365.25 to approximate leap years per Henry's suggestion in comment
how do you work with a column of mixed date types, for example 8/2/2020,2/7/2020, and all are reflecting February,
I have tried zoo::as.Date(mixeddatescolumn,"%d/%m/%Y").The first one is right but the second is wrong.
i have tried solutions here too
Fixing mixed date formats in data frame? but the questions seems different from what i am handling.
It is really tricky to know even for a human if dates like '8/2/2020' is 8th February or 2nd August. However, we can leverage the fact that you know all these dates are in February and remove the "2" part of the date which represents the month and arrange the date in one standard format and then convert the date to an actual Date object.
x <- c('8/2/2020','2/7/2020')
lubridate::mdy(paste0('2/', sub('2/', '', x, fixed = TRUE)))
#[1] "2020-02-08" "2020-02-07"
Or same in base R :
as.Date(paste0('2/', sub('2/', '', x, fixed = TRUE)), "%m/%d/%Y")
Since we know that every month is in February search for /2/ or /02/ and if found the middle number is the month; otherwise, the first number is the month. In either case set the format appropriately and use as.Date. No packages are used.
dates <- c("8/2/2020", "2/7/2020", "2/28/2000", "28/2/2000") # test data
as.Date(dates, ifelse(grepl("/0?2/", dates), "%d/%m/%Y", "%m/%d/%Y"))
## [1] "2020-02-08" "2020-02-07" "2000-02-28" "2000-02-28"
How can set R to count months instead of dates when converting integers to dates?
After reading several threads on how to convert dates in R, it seems like nobody has asked how it is possible to convert numeric dates if the numerics is given in monthly timeseries. E.g. 552 represents January 2006.
I have tried several things, such as using as.Date(dates,origin="1899-12-01"), but I reckognize that R counts days instead of months. Thus, the code on year-month number 552 above yields "1901-06-06" instead of the correct 2006-01-01.
Sidenote: I also want the format to be YEARmonth, but does R allow displaying dates without days?
I think your starting date should be '1960-01-01'.
anyway you can solve this problem using the package lubridate.
in this case you can start from a date and add months.
library(lubridate)
as.Date('1960-01-01') %m+% months(552)
it gives you
[1] "2006-01-01"
you can display only the year and month of a date, but in that case R coerces the date into a character.
format(as.Date('2006-01-01'), "%Y-%m")
I have a column of my dataframe as
date
17-Feb
17-Mar
16-Dec
16-Nov
16-Sep
17-Feb
I am trying to convert it into a date column from string. I am using the following pieces of code:
as.Date(df$Date, format="%y-%b")
and
as.POSIXct(df$Date, format="%y-%b")
Both of them give NAs
I am getting the format from this link
The starting number is year. Sorry for the confusion.
I assume from your approach that the 17 and 16 refer to the year 2017 and 2016 respectively. You need to also specify the day of month. If you don't care about it, then set it to the 1st.
A slight modification to your code will work, by appending '-01' to the date then updating your format argument to reflect this:
df = data.frame(Date = c("17-Feb", "17-Mar", "16-Dec"))
as.Date(paste0(df$Date, "-01"), format="%y-%b-%d")
I am running into some date issues when working with Dates in R.
Here's my situation-
I have a data set based on dates and finally got the Date field converted from character to Date in R using the following code
o1$Date <- as.Date(o1$Date , "%m/%d/%y")
(My dataset is o1 and Date is the name of my Date column)
My Date column has the following values
"1/1/2013" "1/1/2014" "1/10/2013" "1/10/2014" "1/11/2013" "1/11/2014"
However when I convert the Char to Date I get the following Dates
"2020-01-01" "2020-01-01" "2020-01-10" "2020-01-10" "2020-01-11"
Any suggestions on what the problem could be and how to work around it?
look at ?strptime to see the formatting options for times and dates. You need to use %Y rather than %y which is for a 2 digit year.