I would like to change only the year format on a POSIX date-time value. I would like to change 2013-12-30 XX:XX:XX to 2012-12-30 XX:XX:XX . I would like this to be general as there are hundreds of incidences with different hours. Is this possible to do while keeping the column as a POSIX value
1) Base R. Convert to POSIXlt, subtract one from the year component and convert back to POSIXct. No packages are used.
yearMinus <- function(x, n = 1) {
lt <- as.POSIXlt(x)
lt$year <- lt$year - n
as.POSIXct(lt)
}
# test
datetimes <- as.POSIXct( c("2013-12-30 03:02:01", "2013-12-30 03:02:01") )
yearMinus(datetimes)
## [1] "2012-12-30 03:02:01 EST" "2012-12-30 03:02:01 EST"
2) gsubfn Convert to character, match 4 digits, convert the match to numeric and subtract 1 (done in the second argument which represents the transformation in formula notation) and then convert back to POSIXct. This is done in one gsubfn call.
library(gsubfn)
as.POSIXct(gsubfn("\\d{4}", ~ as.numeric(year) - 1, as.character(datetimes)))
## [1] "2012-12-30 03:02:01 EST" "2012-12-30 03:02:01 EST"
If you want to subtract a year from the current timestamp
df$time - lubridate::years(1)
If you want to change only specific date without changing the time we can use sub
df$time <- as.POSIXct(sub("2013-12-30", "2012-12-30", df$time))
Related
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"
I have a data frame with the amount of time it takes to do a lap and I'm trying to separate that into individual data frames for each driver.
These time values look like this, being in minutes:seconds.milliseconds, except for the first lap which has a Colon in between seconds and milliseconds.
13:14:50 1:28.322 1:24.561 1:23.973 1:23.733 1:24.752
I'd like to have these in a separate data frame in a seconds format like this.
794.500 88.322 84.561 83.973 83.733 84.752
When I convert this to a numeric it gives the following values.
214 201 174 150 133 183
And when I use strptime or POSIXlt it gives me huge values which are also wrong, even when I use the format codes. However, I subtracted 2 values to find that the time difference was correct, and through that I found that were all off by 1609164020. Also, these values ignore the decimal values which I need.
You can use POSIXlt in conjunction with a conversion to seconds.
First, add a date to your first time element:
ds <- c("13:14:50", "1:28.322", "1:24.561", "1:23.973", "1:23.733", "1:24.752")
ds[1] <- paste( Sys.Date(), ds[1] )
#[1] "2020-12-29 13:14:50" "1:28.322" "1:24.561"
#[4] "1:23.973" "1:23.733" "1:24.752"
Create a function to convert the subsequent minutes:seconds.milliseconds to seconds.milliseconds:
to_sec <- function(x){ as.numeric(sub( ":.*","", x )) * 60 +
as.numeric( sub( ".*:","", x ) ) }
Convert the vector to dates that enable calculation of time differences:
ds[2:6] <- to_sec(ds[2:6])
ds[2:6] <- cumsum(ds[2:6])
dv <- c( as.POSIXlt(ds[1]), as.POSIXlt(ds[1]) + as.numeric(ds[2:6]) )
# [1] "2020-12-29 13:14:50 CET" "2020-12-29 13:16:18 CET"
# [3] "2020-12-29 13:17:42 CET" "2020-12-29 13:19:06 CET"
# [5] "2020-12-29 13:20:30 CET" "2020-12-29 13:21:55 CET"
dv[6] - dv[1]
# Time difference of 7.089017 mins
I have a date column in a dataframe. I have read this df into R using openxlsx. The column is 'seen' as a character vector when I use typeof(df$date).
The column contains date information in several formats and I am looking to get this into the one format.
#Example
date <- c("43469.494444444441", "12/31/2019 1:41 PM", "12/01/2019 16:00:00")
#What I want -updated
fixed <- c("2019-04-01", "2019-12-31", "2019-12-01")
I have tried many work arounds including openxlsx::ConvertToDate, lubridate::parse_date_time, lubridate::date_decimal
openxlsx::ConvertToDateso far works best but it will only take 1 format and coerce NAs for the others
update
I realized I actually had one of the above output dates wrong.
Value 43469.494444444441 should convert to 2019-04-01.
Here is one way to do this in two-step. Change excel dates separately and all other dates differently. If you have some more formats of dates that can be added in parse_date_time.
temp <- lubridate::parse_date_time(date, c('mdY IMp', 'mdY HMS'))
temp[is.na(temp)] <- as.Date(as.numeric(date[is.na(temp)]), origin = "1899-12-30")
temp
#[1] "2019-01-04 11:51:59 UTC" "2019-12-31 13:41:00 UTC" "2019-12-01 16:00:00 UTC"
as.Date(temp)
#[1] "2019-01-04" "2019-12-31" "2019-12-01"
You could use a helper function to normalize the dates which might be slightly faster than lubridate.
There are weird origins in MS Excel that depend on platform. So if the data are imported from different platforms, you may want to work woth dummy variables.
normDate <- Vectorize(function(x) {
if (!is.na(suppressWarnings(as.numeric(x)))) # Win excel
as.Date(as.numeric(x), origin="1899-12-30")
else if (grepl("A|P", x))
as.Date(x, format="%m/%d/%Y %I:%M %p")
else
as.Date(x, format="%m/%d/%Y %R")
})
For additional date formats just add another else if. Format specifications can be found with ?strptime.
Then just use as.Date() with usual origin.
res <- as.Date(normDate(date), origin="1970-01-01")
# 43469.494444444441 12/31/2019 1:41 PM 12/01/2019 16:00:00
# "2019-01-04" "2019-12-31" "2019-12-01"
class(res)
# [1] "Date"
Edit: To achieve a specific output format, use format, e.g.
format(res, "%Y-%d-%m")
# 43469.494444444441 12/31/2019 1:41 PM 12/01/2019 16:00:00
# "2019-04-01" "2019-31-12" "2019-01-12"
format(res, "%Y/%d/%m")
# 43469.494444444441 12/31/2019 1:41 PM 12/01/2019 16:00:00
# "2019/04/01" "2019/31/12" "2019/01/12"
To lookup the codes type ?strptime.
I have a date that I convert to a numeric value and want to convert back to a date afterwards.
Converting date to numeric:
date1 = as.POSIXct('2017-12-30 15:00:00')
date1_num = as.numeric(date1)
# 1514646000
Reconverting numeric to date:
as.Date(date1_num, origin = '1/1/1970')
# "4146960-12-12"
What am I missing with the reconversion? I'd expect the last command to return my original date1.
As the numeric vector is created from an object with time component, reconversion can also be in the same way i.e. first to POSIXct and then wrap with as.Date
as.Date(as.POSIXct(date1_num, origin = '1970-01-01'))
#[1] "2017-12-30"
You could use anytime() and anydate() from the anytime package:
R> pt <- anytime("2017-12-30 15:00:00")
R> pt
[1] "2017-12-30 15:00:00 CST"
R>
R> anydate(pt)
[1] "2017-12-30"
R>
R> as.numeric(pt)
[1] 1514667600
R>
R> anydate(as.numeric(pt))
[1] "2017-12-30"
R>
POSIXct counts the number of seconds since the Unix Epoch, while Date counts the number of days. So you can recover the date by dividing by (60*60*24) (let's ignore leap seconds), or convert back to POSIXct instead.
as.Date(as.numeric(date1)/(60*60*24), origin="1970-01-01")
[1] "2017-12-30"
as.POSIXct(as.numeric(date1),origin="1970-01-01")
[1] "2017-12-30 15:00:00 GMT"
Using lubridate :
lubridate::as_datetime(1514646000)
[1] "2017-12-30 15:00:00 UTC"
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"