I have a data frame DP with a column variable in numeric format which is a numeric representation of Date.
Example: 43282 corresponds to 7/1/2018 (try in excel).
But in R when I call as.Date() to convert it to date, I get the wrong date
DP$Time <- as.Date(DP$variable)
variable Time
1 43282 2088-07-02
What am I doing wrong here?
If it is based on excel, then change the origin from default 1970-01-01 to 1899-12-30
as.Date(43282, origin = '1899-12-30')
#[1] "2018-07-01"
Excel's origin of date is "1899-12-30" and R's origin of date is "1970-01-01". Since they have different origins for the date, while importing the data from Excel, you are getting a different date in R.
Specify the right origin and it will print the right values:
DP$Time <- as.Date(DP$variable, origin = "1899-12-30")
Related
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"
SAS documentation states the following for data and datetime values:
SAS time value: is a value representing the number of seconds since midnight of the current day. SAS time values are between 0 and 86400.
SAS datetime value: is a value representing the number of seconds between January 1, 1960 and an hour/minute/second within a specified date.
I'm willing to convert the following date and hour values with R, I have a big doubt for the hour (datetime) conversion, which one of the "HH:MM:SS" values within R_hour1 and R_hour2 is correct ?
I have to separate columns, SAS date = 20562 and SAS hour = 143659, in my table
R: R_date <- as.Date(as.integer(20562), origin="1960-01-01"); R_date
[1] "2016-04-18"
R: R_hour1 <- as.POSIXct(143659, origin = R_date); R_hour1
[1] "2016-04-19 17:54:19 CEST"
R: R_hour2 <- as.POSIXct(143659, origin = "1960-01-01"); R_hour2
[1] "1960-01-02 16:54:19 CET"
Similar to R, SAS Date and DateTime values can have whatever origin you wish them to. The default formats have a default (1/1/1960 for both), but you can use the datetime field to mean any origin you wish, and it will generally still work perfectly well with any of the datetime functions (though it will not display properly unless you write a custom format). It is very possible to have a different origin, as you show above with R_hour1.
As such, you would have to ask the person who generated the data what the meaning of the field is and what its origin should be.
I have a factor of a date given in number with the origin 1899-12-30, e.g.
dat1 <- factor(42648)
If I want to change it to a date format, I'd normally do
as.Date(dat1, origin="1899-12-30")
but the conversion requires a certain format. Doing
as.Date(as.numeric(as.character(dat1)), origin="1899-12-30")
works but it seems a little over the top. Is there a shorter way?
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"
I have a data set in R that contains a column of dates in the format yyyy/mm/dd. I am trying to use as.Date to convert these dates to date objects in R. However, I cannot seem to find the correct argument for origin to input into as.Date. The following code is an example of what I have been trying. I am using a CSV file from Excel, so I used origin="1899/12/30 based on other sites I have looked at.
> as.Date(2001/04/26, origin="1899/12/30")
[1] "1900-01-18"
However, this is not working since the input date 2001/04/26 is returned as "1900-01-18". I need to convert the dates into date objects so I can then convert the dates into julian dates.
You can either is as.Date with a numeric value, or with a character value. When you type just 2001/04/26 into R, that's doing division and getting 19.24 (a numeric value). And numeric values require an origin and the number you supply is the offset from that origin. So you're getting 19 days away from your origin, ie "1900-01-18". A date like Apr 26 2001 would be
as.Date(40659, origin="1899-12-30")
# [1] "2011-04-26"
If your dates from Excel "look like" dates chances are they are character values (or factors). To convert a character value to a Date with as.Date() you want so specify a format. Here
as.Date("2001/04/26", format="%Y/%m/%d")
# [1] "2001-04-26"
see ?strptime for details on the special % variables. Now if you're read your data into a data.frame with read.table or something, there's a chance your variable may be a factor. If that's the case, you'll want do convert to character with'
as.Date(as.character(mydf$datecol), format="%Y/%m/%d")