I have this list of dates:
library(lubridate)
my.dates = ymd(c("2013-12-14", "2014-01-18", "2014-01-27", "2013-12-13", "2013-12-29", "2013-12-06"))
The following lubridate::weekfunctions outputs a numeric vector when I convert these dates to week numbers:
week(my.dates)
[1] 50 3 4 50 52 49
Can I get lubridate to output a date ("POSIXct" "POSIXt") object that converts my.dates to a week number and year number. So output should be a date object (not a character or numeric vector) formatted something like this:
[1] "50-2013" "3-2014" "4-2014" "50-2013" "52-2013" "49-2013"
I'm specifically interested in a solution that uses lubridate.
To convert my.dates to a week-year character vector try the following where week and year are lubridate functions:
> paste(week(my.dates), year(my.dates), sep = "-")
[1] "50-2013" "3-2014" "4-2014" "50-2013" "52-2013" "49-2013"
The sample output in the question did not use leading zeros for the week but if leading zeros were desired for the week then:
> sprintf("%02d-%d", week(my.dates), year(my.dates))
[1] "50-2013" "03-2014" "04-2014" "50-2013" "52-2013" "49-2013"
The above are character representations of week-year and do not uniquely identify a date nor can such a format represent a POSIXt object.
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 have a vector of date strings in the form month_name-2_digit_year i.e.
a = rbind("April-21", "March-21", "February-21", "January-21")
I'm trying to convert that vector into a vector of date objects. I'm aware this question is very similar to this: Convert non-standard date format to date in R posted some years ago, but unfortunately, it has not answered my question.
I have tried the following as.Date() calls to do this, but it just returns a vector of NA. I.e.
b = as.Date(a, format = "%B-%y")
b = as.Date(a, format = "%B%y")
b = as.Date(a, "%B-%y")
b = as.Date(a, "%B%y")
I'm also attempted to do it using the convertToDate function from the openxlsx package:
b = convertToDate(a, format = "%B-%y")
I have also tried all the above but using a single character string rather than a vector, but that produced the same issue.
I'm a little lost as to why this isn't working, as this format has worked in reverse earlier in my script (that is, I had a date object already in dd-mm-yyyy format and converted it to month_name-yy using %B-%y). Is there another way to go from string to date when the string is a non-standard (anything other than dd-mm-yyy or mm-dd-yy if you're in the US) date format?
For the record my R locales are all UK and english.
Thanks in advance.
A Date must have all three of day, month and year. Convert to yearmon class which requires only month and year and then to Date as in (1) and (2) below or add the day as in (3).
(1) and (3) give first of month and (2) gives the end of the month.
(3) uses only functions from base R.
Also consider not converting to Date at all but just use yearmon objects instead since they directly represent a year and month which is what the input represents.
library(zoo)
# test input
a <- c("April-21", "March-21", "February-21", "January-21")
# 1
as.Date(as.yearmon(a, "%B-%y"))
## [1] "2021-04-01" "2021-03-01" "2021-02-01" "2021-01-01"
# 2
as.Date(as.yearmon(a, "%B-%y"), frac = 1)
## [1] "2021-04-30" "2021-03-31" "2021-02-28" "2021-01-31"
# 3
as.Date(paste(1, a), "%d %B-%y")
## [1] "2021-04-01" "2021-03-01" "2021-02-01" "2021-01-01"
In addition to zoo, which #G. Grothendieck mentioned, you can also use clock or lubridate.
clock supports a variable precision calendar type called year_month_day. In this case you'd want "month" precision, then you can set the day to whatever you'd like and convert back to Date.
library(clock)
x <- c("April-21", "March-21", "February-21", "January-21")
ymd <- year_month_day_parse(x, format = "%B-%y", precision = "month")
ymd
#> <year_month_day<month>[4]>
#> [1] "2021-04" "2021-03" "2021-02" "2021-01"
# First of month
as.Date(set_day(ymd, 1))
#> [1] "2021-04-01" "2021-03-01" "2021-02-01" "2021-01-01"
# End of month
as.Date(set_day(ymd, "last"))
#> [1] "2021-04-30" "2021-03-31" "2021-02-28" "2021-01-31"
The simplest solution may be to use lubridate::my(), which parses strings in the order of "month then year". That assumes that you want the first day of the month, which may or may not be correct for you.
library(lubridate)
x <- c("April-21", "March-21", "February-21", "January-21")
# Assumes first of month
my(x)
#> [1] "2021-04-01" "2021-03-01" "2021-02-01" "2021-01-01"
How could I convert this numeric vector of dates into the date format that R can recognize?
date <- c(29101958L, 10121957L, 27091953L, 23021960L,
6031967L, 10011968L, 10101958L, 9101992)
I would like an output like:
'1958-10-29', '1957-12-10', '1953-09-27', '1960-02-23', '1967-03-06', '1968-01-10', '1958-10-10', '1992-10-09'
Then I would like to calculate the age by making the difference from 2016-12-31 with the dates of the vector.
I appreciate any help.
We can use dmy from lubridate
library(lubridate)
newdate <- dmy(date)
newdate
#[1] "1958-10-29" "1957-12-10" "1953-09-27" "1960-02-23" "1967-03-06" "1968-01-10" "1958-10-10" "1992-10-09"
and get the difference between the new date in years
as.integer(difftime(as.Date('2016-12-31'), newdate, units = 'days')/365)
#[1] 58 59 63 56 49 49 58 24
Base R option using as.Date :
as.Date(sprintf('%08d', date), '%d%m%Y')
#[1] "1958-10-29" "1957-12-10" "1953-09-27" "1960-02-23" "1967-03-06"
#[6] "1968-01-10" "1958-10-10" "1992-10-09"
Using sprintf we add leading zeroes for single digit dates.
I am given a data frame where one of the column parameters is a year and month value (ex "2019-05"). I need to only display rows where the date value is later than a certain value. For instance, if I only wanted to show data later than a given year-month "2018-11".
You can convert them to dates, but in R < and > work on characters too, so you could just do something like this (assuming there's a leading 0 in the months with only 1 digit)
examp <- c('2011-01', '2013-08', '2018-04', '2018-12', '2019-05')
examp[examp > '2018-11']
#[1] "2018-12" "2019-05"
If you want to convert to dates, add a day to the end and use as.Date
examp <- as.Date(paste0(examp, '-01'))
examp
# [1] "2011-01-01" "2013-08-01" "2018-04-01" "2018-12-01" "2019-05-01"
examp[examp > as.Date('2018-11-01')]
# [1] "2018-12-01" "2019-05-01"
I imported Excel data into R and I have a problem to convert dates.
In R, my data are character and look like :
date<-c('1971-02-00 00:00:00', '1979-06-00 00:00:00')
I would like to convert character into date (MM/YYYY) but the '00' value used for days poses a problem and 'NA' are returned systematically.
It works when I manually replace '00' with '01' and then use as.yearmon, ymd and format. But I have lots of dates to change and I don't know how to change all my '00' into '01' in R.
# data exemple
date1<-c('1971-02-00 00:00:00', '1979-06-00 00:00:00')
# removing time -> doesn't work because of the '00' day
date1c<-format(strptime(date1, format = "%Y-%m-%d"), "%Y/%m/%d")
date1c<-format(strptime(date1, format = '%Y-%m'), '%Y/%m')
# trying to convert character into date -> doesn't work either
date1c<-ymd(date1)
date1c<-strptime(date1, format = "%Y-%m-%d %H:%M:%S")
date1c<-as.Date(date1, format="%Y-%m-%d %H:%M:%S")
date1c<as.yearmon(date1, format='%Y%m')
# everything works if days are '01'
date2<-c('1971-02-01 00:00:00', '1979-06-01 00:00:00')
date2c<-as.yearmon(ymd(format(strptime(date2, format = "%Y-%m-%d"), "%Y/%m/%d")))
date2c
If you have an idea to do it or an another idea to solve my problem, I would be thankful!
Use gsub to replace -00 with -01.
date1<-c('1971-02-01 00:00:00', '1979-06-01 00:00:00')
date1 <- gsub("-00", "-01", date1)
date1c <-format(strptime(date1, format = "%Y-%m-%d"), "%Y/%m/%d")
> date1c
[1] "1971/02/01" "1979/06/01"
Another possibility could be:
as.Date(paste0(substr(date1, 1, 9), "1"), format = "%Y-%m-%d")
[1] "1971-02-01" "1979-06-01"
Here it extracts the first nine characters, pastes it together with 1 and then converts it into a date object.
These alternatives each accept a vector input and produce a vector as output.
Date output
These all will accept a vector as input and produce a Date vector as the output.
# 1. replace first occurrence of '00 ' with '01 ' and then convert to Date
as.Date(sub("00 ", "01 ", date1))
## [1] "1971-02-01" "1979-06-01"
# 2. convert to yearmon class and then to Date
library(zoo)
as.Date(as.yearmon(date1, "%Y-%m"))
## [1] "1971-02-01" "1979-06-01"
# 3. insert a 1 and then convert to Date
as.Date(paste(1, date1), "%d %Y-%m")
## [1] "1971-02-01" "1979-06-01"
yearmon output
Note that if you really are trying to represent just months and years then yearmon class directly represents such objects without the kludge of using an unused day of the month. Such objects are internally represented as a year plus a fraction of a year, i.e. year + 0 for January, year + 1/12 for February, etc. They display in a meaningful way, they sort in the expected manner and can be manipulated, e.g. take the difference between two such objects or add 1/12 to get the next month, etc. As with the others it takes a vector in and produces a vector out.
library(zoo)
as.yearmon(date1, "%Y-%m")
## [1] "Feb 1971" "Jun 1979"
character output
If you want character output rather than Date or yearmon output then these variations work and again accept a vector as input and produce a vector as output:
# 1. replace -00 and everything after that with a string having 0 characters
sub("-00.*", "", date1)
## [1] "1971-02" "1979-06"
# 2. convert to yearmon and then format that
library(zoo)
format(as.yearmon(date1, "%Y-%m"), "%Y-%m")
## [1] "1971-02" "1979-06"
# 3. convert to Date class and then format that
format(as.Date(paste(1, date1), "%d %Y-%m"), "%Y-%m")
## [1] "1971-02" "1979-06"
# 4. pick off the first 7 characters
substring(date1, 1, 7)
## [1] "1971-02" "1979-06"