I would like to convert a YYYYMM-integer to a Date without converting it to character (=is there a mdy-function like in SAS?). I would like to replace this code:
dateint<-201511
datestr<-paste(toString(dateint,length=8),'01')
date<-as.Date(datestr,'%Y%m%d')
print(date)
class(date)
with a working version of this. If possible the resulting class should be a date too:
year<-dateint %% 100
month<-floor(dateint/100)
date2<-ISOdate(year,month,1) # I can't make this work ..
print(date2)
class(date2)
Thanks & kind regards
The package lubridate has a function ymd, which accepts numeric input:
> library(lubridate)
> ymd(20151101)
[1] "2015-11-01 UTC"
You need however to add the day at the end.
Related
I am working with an external package that's converting columns of a dataframe with the lubridate date type Date into numeric type. (Confirmed by running as.numeric() on the columns).
I'm wondering if there's a way to convert it back?
For example, if I have the date "O1-01-2021" then running as.numeric on it returns -719143. How can I turn that back into "O1-01-2021" ?
Note that Date class is part of base R, not lubridate.
You probably assumed that the data was year/month/day by mistake. Using base R to eliminate lubridate as a problem we can replicate the question's result like this:
as.numeric(as.Date("01-01-2021", "%Y-%m-%d"))
## [1] -719143
Had we used day/month/year we would have gotten:
as.numeric(as.Date("01-01-2021", "%d-%m-%Y"))
## [1] 18628
or using lubridate
library(lubridate)
as.numeric(dmy("01-01-2021"))
## [1] 18628
It would be best if you fix the mistake that resulted in -719143 but if you don't control that and are faced with an input of
-719143 and want to get as.Date("2021-01-01") as the output then:
# input x is numeric; result is Date class
fixup <- function(x) as.Date(format(.Date(x), "%y-%m-%d"), "%d-%m-%y")
fixup(-719143)
## [1] "2020-01-01"
Note that we can't tell from the question whether 01-01-2020 is supposed to represent day-month-year or month-day-year so we assumed the first but if it is to represent the second then it should be obvious at this point how to proceed.
EDIT #2: It looks like the original data is being parsed as Jan 20, year 1, which might happen if the year-month-day columns were jumbled while being parsed:
as.numeric(as.Date("01-01-2021", format = "%Y-%m-%d", origin = "1970-01-01"))
[1] -719143
as.numeric(as.Date("0001-01-20", origin = "1970-01-01"))
[1] -719143
Is there a way to share an example of the raw data as you have it? e.g. dput(MY_DATA[1:10, DATE_COL])
EDIT: -719143 is about 1970 years of days, which can't be a coincidence, given that many date/time formats use 1970 as a baseline. I wonder if 01-01-2021 is being interpreted as the numeric formula equal to -2021 and so we're looking at perhaps -2021 seconds/days/[?] before year zero, which would be about -1970 years before the epoch...
-719143/(365)
[1] -1970.255
For instance, we can get something close with:
as.numeric(as.Date("0000-01-01", origin = "1970-01-01"))
[1] -719528
Original answer:
R treats a string describing a date as text:
x <- "01-01-2021"
class(x)
[1] "character"
We can convert it to a Date data type using these two equivalent commands:
base_dt <- as.Date(x, "%m-%d-%Y") # base R version
lubridt <- lubridate::mdy(x) # convenience lubridate function
identical(base_dt, lubridt)
[1] TRUE
Under the hood, a Date object in R is a numeric value with a flag telling R it's a date:
> typeof(lubridt) # What general type of data is it?
[1] "double" # --> numeric, stored as a double
> as.numeric(lubridt)
[1] 18628
> class(lubridt) # Does it have any special class attributes?
[1] "Date" # --> yes, it's a Date
> dput(lubridt) # How would we construct it from scratch?
structure(18628, class = "Date") # --> by giving 18628 a Date attribute
In R, a Date is encoded as the number of days since 1970 began:
> as.Date("1970-01-1") + as.numeric(lubridt)
[1] "2021-01-01"
We could convert it back to the original text using:
format(base_dt, "%m-%d-%Y")
[1] "01-01-2021"
identical(x, format(base_dt, "%m-%d-%Y"))
[1] TRUE
I am trying to separate a date from a column in a database, but the result date format is not proper.
column data = "24-01-2021 19:15"
Code used:
database_1$date <- format(as.Date(database_1$start_time), "%d-%m-%Y")
Result: 20-01-0024
Expected result: 24-01-2021
Update:
I just realized the expected output:
just add format("%d.%m.%Y") to the code:
as.Date(dmy_hm(column_date)) %>%
format("%d.%m.%Y")
[1] "24.01.2021"
With lubridate package you can:
With dmy_hm you read in the character column_date to date format
Then you can add as.Date
library(lubridate)
column_date <- "24-01-2021 19:15"
as.Date(dmy_hm(column_date))
[1] "2021-01-24"
I have a numeric vector as follows
aa <- c(1022011, 2022011, 13022011, 23022011) (this vector is just a sample, it is very long)
Values are written in such a way that first value is day then month and then year.
What I am doing right now is
as.Date(as.character(aa), %d%m%Y")
but,
it is causing problems (returning NA) in case of single digits day numbers. (i.e. 1022011, 2022011).
so basically
as.Date("1022011", "%d%m%Y") does not work
but
as.Date("01022011", "%d%m%Y") (pasting '0' ahead of the number) works.
I want to avoid pasting '0' in such cases. Is there any other (direct) alternative to convert numeric values to dates at once?
It could be rearranged using sub in which case a plain as.Date with no format works:
x <- c(1022011, 11022011) # test data
pat <- "^(..?)(..)(....)$"
as.Date(sub(pat, "\\3-\\2-\\1", x))
giving:
[1] "2011-02-01" "2011-02-11"
Depending on your platform, you could use sprintf in order to add a zero at the beginning. It seems that Mac is OK with this, but not windows 7 given the discussion with the OP.
aa <- c(1022011, 2022011, 13022011, 23022011)
as.Date(sprintf("%08s", aa), format = "%d%m%Y")
[1] "2011-02-01" "2011-02-02" "2011-02-13" "2011-02-23"
UPDATE
#CathyG kindly mentioned that sprintf("%08i",aa) works on Windows 7.
You can use dmy in lubridate:
library(lubridate)
aa <- c(1022011, 2022011, 13022011, 23022011)
> dmy(aa)
[1] "2011-02-01 UTC" "2011-02-02 UTC" "2011-02-13 UTC" "2011-02-23 UTC"
and if you don't want the timezone just wrap it in as.Date:
> as.Date(dmy(aa))
[1] "2011-02-01" "2011-02-02" "2011-02-13" "2011-02-23"
Thank you #Ben Bolker,
> as.Date(mdy(aa))
[1] "2011-01-02" "2011-02-02" "2012-01-02" "2011-01-02"
I know you don't want to add a "0" but still, in base R, this works :
as.Date(sapply(aa,function(x){ifelse(nchar(x)==8,x,paste("0",x,sep=""))}),format = "%d%m%Y")
Is there an easy way in R for me to itemize all valid days that occurred between two specified dates? For instance, I'd like the following inputs:
itemizeDates(startDate="12-30-11", endDate="1-4-12")
To produce the following dates:
"12-30-11" "12-31-11", "1-1-12", "1-2-12", "1-3-12", "1-4-12"
I'm flexible on classes and formatting of the dates, I just need an implementation of the concept.
You're looking for seq
> seq(as.Date("2011-12-30"), as.Date("2012-01-04"), by="days")
[1] "2011-12-30" "2011-12-31" "2012-01-01" "2012-01-02" "2012-01-03"
[6] "2012-01-04"
Or, you can use :
> as.Date(as.Date("2011-12-30"):as.Date("2012-01-04"), origin="1970-01-01")
[1] "2011-12-30" "2011-12-31" "2012-01-01" "2012-01-02" "2012-01-03"
[6] "2012-01-04"
Note that with : "Non-numeric arguments are coerced internally". Thus, we convert back to class Date, using as.Date method for class 'numeric' and provide origin.
Here's a function to meet your specific request
itemizeDates <- function(startDate="12-30-11", endDate="1-4-12",
format="%m-%d-%y") {
out <- seq(as.Date(startDate, format=format),
as.Date(endDate, format=format), by="days")
format(out, format)
}
> itemizeDates(startDate="12-30-11", endDate="1-4-12")
[1] "12-30-11" "12-31-11" "01-01-12" "01-02-12" "01-03-12" "01-04-12"
I prefer using the lubridate package to solve datetime problems. It is more intuitive and easier to understand and use once you know it.
library(lubridate)
#mdy() in lubridate package means "month-day-year", which is used to convert
#the string to date object
>start_date <- mdy("12-30-11")
>end_date <- mdy("1-4-12")
#calculate how many days in this time interval
>n_days <- interval(start_date,end_date)/days(1)
>start_date + days(0:n_days)
[1]"2011-12-30" "2011-12-31" "2012-01-01" "2012-01-02" "2012-01-03" "2012-01-04"
#convert to original format
format(start_date + days(0:n_days), format="%m-%d-%y")
[1] "12-30-11" "12-31-11" "01-01-12" "01-02-12" "01-03-12" "01-04-12"
Reference:
Dates and Times Made Easy with lubridate
2 similar implementations in lubridate:
library(lubridate)
as_date(mdy("12-30-11"):mdy("1-4-12"))
# OR
seq(mdy("12-30-11"), mdy("1-4-12"), by = "days")
These don't format your dates in month-day-year but you can fix the formatting if you want. But year-month-day is a bit easy to work with when analyzing.
I have an integer which I want to convert to class Date. I assume I first need to convert it to a string, but how?
My attempt:
v <- 20081101
date <- as.Date(v, format("%Y%m%d"))
Error in charToDate(x) : character string is not in a standard
unambiguous format
Using paste() works, but is that really the correct way to do the conversion?
date <- as.Date(paste(v), format("%Y%m%d"))
date
[1] "2008-11-01"
class(date)
# [1] "Date"
as.character() would be the general way rather than use paste() for its side effect
> v <- 20081101
> date <- as.Date(as.character(v), format = "%Y%m%d")
> date
[1] "2008-11-01"
(I presume this is a simple example and something like this:
v <- "20081101"
isn't possible?)
Another way to get the same result:
date <- strptime(v,format="%Y%m%d")
You can use ymd from lubridate
lubridate::ymd(v)
#[1] "2008-11-01"
Or anytime::anydate
anytime::anydate(v)
#[1] "2008-11-01"