I have a data frame with year column as financial year
Year
2001-02
2002-03
2003-04
How can I convert this to as.Date keeping either the whole thing or just the second year i.e 2002,2003,2004. On converting with %Y, I inevitably get 2001-08-08, 2002-08-08, 2003-08-08 etc.
Thanks
library(lubridate)
Year <- c('2001-02', '2002-03', '2003-04')
year(as.Date(gsub('[0-9]{2}-', '', Year), format = '%Y'))
1) ISOdate Clarifying the question, since it refers to yearend and Date we assume that the input is the fiscal Year shown in the question (plus we have added the "1999-00" edge case) as well as the month and day of the yearend. We assume that the output desired is the yearend as a Date object. (If that is not the intended question and you just want the fiscal yearend year as a number then see Note at the end.)
Returning to the assumed problem let us suppose, for example, that March 31st is the yearend. Below we extract the first 4 character of Year using substring, convert that to numeric and add 1. Then we pass that along with month and day to ISODate and finally convert that to Date. No regular expressions or packages are used.
# test inputs
month <- 3
day <- 31
Year <- c("1999-00", "2001-02", "2002-03", "2003-04")
# yearends
as.Date(ISOdate( as.numeric(substring(Year, 1, 4))+1, month, day))
## [1] "2000-03-31" "2002-03-31" "2003-03-31" "2004-03-31"
2) string manipulation An alternative solution using the same inputs is the following. It is similar except that we use sub with a regular expression that matches the minus and following two characters subtituting a zero length string for them, converts to numeric and adds 1. Then it formats a string in a format acceptable to as.Date by using sprintf and finally applies as.Date. No packages are used.
as.Date(sprintf("%d-%d-%d", as.numeric(sub("-..", "", Year))+1, month, day))
## [1] "2000-03-31" "2002-03-31" "2003-03-31" "2004-03-31"
Note: If you only wanted the fiscal yearend year as a number then it would be just this:
as.numeric(substring(Year, 1, 4)) + 1
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
I have a date frame df that simply looks like this:
month values
2012M01 99904
2012M02 99616
2012M03 99530
2012M04 99500
2012M05 99380
2012M06 99103
2013M01 98533
2013M02 97600
2013M03 96431
2013M04 95369
2013M05 94527
2013M06 93783
with month that was written in form of "M01", "M02"... and so on.
Now I want to convert this column to date format, is there a way to do it in R with lubridate?
I also want to select columns that contain one certain month from each year, like only March columns from all these years, what is the best way to do it?
The short answer is that dates require a year, month and day, so you cannot convert directly to a date format. You have 2 options.
Option 1: convert to a year-month format using zoo::as.yearmon.
library(zoo)
df$yearmon <- as.yearmon(df$month, "%YM%m")
# you can get e.g. month from that
months(df$yearmon[1])
# [1] "January"
Option 2: convert to a date by assuming that the day is always the first day of the month.
df$date <- as.Date(paste(df$month, "01", sep = "-"), "%YM%m-%d")
For selection (and I think you mean select rows, not columns), you already have everything you need. For example, to select only March 2013:
library(dplyr)
df %>% filter(month == "2013M03")
Something like this will get it:
raw <- "2012M01"
dt <- strptime(raw,format = "%YM%m")
dt will be in a Posix format. The strptime function will assign a '1' as the default day of month to make it a complete date.
I have a R time series data, where I am calculating the means for all values up to a particular date, and storing this means in the date + 4 quarters. The dates are all month ends. To achieve this, I am looking to increment 4 quarters to a date. My question is how can I add 4 quarters to an R date data-type. An illustration:
a <- as.Date("2006-01-01")
b <- as.Date("2011-01-01")
date_range <- quarter(seq.Date(a, b, by = "quarter"), with_year = TRUE)
> date_range[1] + 1
[1] 2007.1
> date_range[1] + quarter(1)
[1] 2007.1
> date_range[1] + 0.25
[1] 2006.35
One possible way I am thinking is to get year-quarter dates, and then adding 4 to it. But wasn't sure what is the best way to do this?
The problem is that quarters have different lengths. Q1 is shortest because it includes February (though it ties with Q2 in leap years). Things like this make "adding a quarter to a date" poorly defined. Even adding months to a date can be tricky at the ends months - what is 1 month after January 31?
Beginnings of months are more straightforward, and I would recommend you use the 1st day of quarters rather than the last (if you must use a specific date). lubridate provides functions like floor_date() and ceiling_date() to which you can pass unit = "quarter" and they will return the first day of the current or subsequent quarter, respectively. You can also always add months(3) to a day at the beginning of a month, though of course if your intention is to add 4 quarters you may as well just add 1 year.
Just add 12 months or a year instead?
Or if it must be quarters, define yourself a function, like so:
quarters <- function(x) {
months(3*x)
}
and then use it to add to the date sequence:
date_range <- seq.Date(a, b, by = "quarter")
date_range + quarters(4)
Lubridate has a function for quarters already included. This is a much better solution than creating your own function.
https://www.rdocumentation.org/packages/lubridate/versions/1.7.4/topics/quarter
Old answer but to those arriving here, lubridate has a function %m+%that adds months and preserves monthends.
a <- as.Date("2006-01-01")
Add future months worth of dates:
The original poster wanted 4 quarters in future so that will be 12 months.
future_date <- a %m+% months(12)
future_date
[1] "2007-01-01"
You could also do years as the period:
future_date <- a %m+% years(1)
Remove months from date:
Subtract dates with %m-%
If you wanted a date 3 months ago from 1/1/2006:
past_date <- a %m-% months(3)
past_date
[1] "2005-10-01"
Example with dates not at end of months:
mplus will preserve days in month:
as.Date("2022-10-10") %m-% months(3)
[1] "2022-07-10"
For more, see documentation on "Add and subtract months to a date without exceeding the last day of the new month"
Note that other answers that use Date class will give irregularly spaced series and so are unsuitable for time series analysis.
To do this in such a way that time series analyses can be performed and noting the zoo tag on the question, the yearmon class represents year/month as year + fraction where fraction is 0 for Jan, 1/12 for Feb, 2/12 for Mar, ..., 11/12 for Dec. Thus adding 4 quarters is just a matter of adding 1. (Adding x quarters is done by adding x/4.)
library(zoo)
ym <- yearmon(2006) + 0:11/12 # months in 2006
ym + 1 # one year later
Also this converts yearmon objects to end-of-month Date and in the second line Date to yearmon. Using frac = 0 or omitting frac in the first line would convert to beginning of month dates.
d <- as.Date(ym, frac = 1) # d is Date vector of end-of-months
as.yearmon(d) # convert Date vector to yearmon
If your input dates represent quarters then there is also the yearqtr class which represents a year/quarter as year + fraction where fraction is 0, 1/4, 2/4, 3/4 for the 4 quarters of a year. Adding 4 quarters is done by adding 1 (or to add x quarters add x/4).
yq <- as.yearqtr(2006) + 0:3/4 # all quarters in 2006
yq + 1 # one year later
Conversions work similarly to yearmon:
d <- as.Date(ym, frac = 1) # d is Date vector of end-of-quarters
as.yearqtr(d) # convert Date vector to yearqtr
I have date recorded as: Month/Day/Year or MM/DD/YYYY
I would like to write code that creates two new variables from that information.
I would like a year variable alone
I would like to create a quarter variable
The Quarter Variables would not be influenced by year. I would want this variable to apply to all years.
Quarter 1 would be January 1 - March 31
Quarter 2 would be April 1 - June 30
Quarter 3 would be July 1 - September 30
Quarter 4 would be October 1 - December 31
Any assistance would be greatly appreciated. I cannot seem to get the nuance of how to do these functions in R.
Thanks,
Jared
Assuming that the date variable is of class POSIX** you could do:
#example date
date <- as.POSIXlt( "05/12/2015", format='%m/%d/%Y')
In order to return the year from a date data.table has already a function to do it and that is year:
library(data.table)
> year(date)
[1] 2015
As for the quarter it can easily be created from the function below (uses data.table::month that returns the number of a month):
quarter <- function(x) {
rep(c('quarter 1','quarter 2','quarter 3','quarter 4'), each=3)[month(x)]
}
> quarter(date)
[1] "quarter 2"
Using only the base packages:
Try formatting your dates with the strptime fxn, so that all dates are now in the Year-Month-Day format. This format constrains the each element of the date to be the same character length and in the same position. Look at the strptime documentation for the appropriate formatting argument.
date.vec<-c(1/1/1999,2/2/1999)
fmt.date.vec<-strptime(date.vec, "%m/%d/%Y")
With the dates in this format it is easy to extract the year, month, and day using the substring function
Year<-substring(fmt.date.vec,1,4)
Month<-substring(fmt.date.vec,6,7)
Day<-substring(fmt.date.vec,9,10)
With this information you can now generate your Quarter vector any number of ways. For example if a data.frame "df" has a Month column:
df$Quarter<-"Quarter_1"
df[df$Month %in% c("04","05","06"),]$Quarter<-"Quarter_2"
df[df$Month %in% c("07","08","09"),]$Quarter<-"Quarter_3"
df[df$Month %in% c("10","11","12"),]$Quarter<-"Quarter_4"
I have this .txt file:
http://pastebin.com/raw.php?i=0fdswDxF
First column (Date) shows date in month/day
So 0601 is the 1st of June
When I load this into R and I show the data, it removes the first 0 in the data.
So when loaded it looks like:
601
602
etc
For 1st of June, 2nd of June
For the months 10,11,12, it remains unchanged.
How do I change it back to 0601 etc.?
What I am trying to do is to change these days into the day of the year, for instance,
1st of January (0101) would be 1, and 31st of December would be 365.
There is no leap year to be considered.
I have the code to change this, if my data was shown as 0601 etc, but not as 601 etc.
copperNew$Date = as.numeric(as.POSIXct(strptime(paste0("2013",copperNew$Date), format="%Y%m%d")) -
as.POSIXct("2012-12-31"), units = "days")
Where Date of course is from the file linked above.
Please ask if you do not consider the description to be good enough.
You can use colClasses in the read.table function, then convert to POSIXlt and extract the year date. You are over complicating the process.
copperNew <- read.table("http://pastebin.com/raw.php?i=0fdswDxF", header=TRUE,
colClasses=c("character", "integer", rep("numeric", 3)))
tmp <- as.POSIXlt( copperNew$Date, format='%m%d' )
copperNew$Yday <- tmp$yday
The as.POSIXct function is able to parse a string without a year (assumes the current year) and computes the day of the year for you.
d<-as.Date("0201", format = "%m%d")
strftime(d, format="%j")
#[1] "032"
First you parse your string and obtain Date object which represents your date (notice that it will add current year, so if you want to count days for some specific year add it to your string: as.Date("1988-0201", format = "%Y-%m%d")).
Function strftime will convert your Date to POSIXlt object and return day of year. If you want the result to be a numeric value, you can do it like this: as.numeric(strftime(d, format = "%j"))(Thanks Gavin Simpson)
Convert it to POSIXlt using a year that is not a leap-year, then access the yday element and add 1 (because yday is 0 on January 1st).
strptime(paste0("2011","0201"),"%Y%m%d")$yday+1
# [1] 32
From start-to-finish:
x <- read.table("http://pastebin.com/raw.php?i=0fdswDxF",
colClasses=c("character",rep("numeric",5)), header=TRUE)
x$Date <- strptime(paste0("2011",x$Date),"%Y%m%d")$yday+1
In which language?
If it's something like C#, Java or Javascript, I'd follow these steps:
1-) parse a pair of integers from that column;
2-) create a datetime variable whose day and month are taken from the integers from step one. Set the year to some fixed value, or to the current year.
3-) create another datetime variable, whose date is the 1st of February of the same year as the one in step 2.
The number of the day is the difference in days between the datetime variables, + 1 day.
This one worked for me:
copperNew <- read.table("http://pastebin.com/raw.php?i=0fdswDxF",
header=TRUE, sep=" ", colClasses=c("character",
"integer",
rep("numeric", 3)))
copperNew$diff = difftime(as.POSIXct(strptime(paste0("2013",dat$Date),
format="%Y%m%d", tz="GMT")),
as.POSIXct("2012-12-31", tz="GMT"), units="days")
I had to specify the timezone (tz argument in as.POSIXct), otherwise I got two different timezones for the vectors I am subtracting and therefore non-integer days.