Re: [R] Turn dates into age
What is the frame of reference to determine the age? Check out 'difftime'. On Sun, Nov 8, 2009 at 1:50 PM, frenchcr frenc...@btinternet.com wrote: Ive got a big column of dates (also some fields dont have a date so they have NA instead), that i have converted into date format as so... dates-as.character(data[,date_commissioned]); # converted dates to characters dates[1:10] [1] 19910101 19860101 19910101 19860101 19910101 19910101 19910101 19910101 19910101 19910101 dateObs - as.Date(dates,format=%Y%m%d) dateObs[1:10] [1] 1991-01-01 1986-01-01 1991-01-01 1986-01-01 1991-01-01 1991-01-01 1991-01-01 1991-01-01 1991-01-01 1991-01-01 Now i need to turn the dates into AGE, how do i do it? Im not worried about fractions of years, whole years would do. -- View this message in context: http://old.nabble.com/Turn-dates-into-age-tp26256656p26256656.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Jim Holtman Cincinnati, OH +1 513 646 9390 What is the problem that you are trying to solve? __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Turn dates into age
As Jim has noted, if the dates you have below are an 'end date', you
need to define the time0 or start date for each to calculate the
intervals. On the other hand, are the dates you have below the start
dates and you need to calculate the time to today? In the latter case,
see ?Sys.Date to get the current system date from your computer.
Generically speaking you can use the following:
(EndDate - StartDate) / 365.25
where EndDate and StartDate are of class Date. This will give you the
time interval in years plus any fraction.
You can then use round() which will give you a whole year with typical
rounding up or down to the nearest whole integer. You can use floor(),
which will give you the nearest whole integer less than the result or
basically a round down result. Keep in mind that the above calculation
does not honor a calendar year, but is an approximation.
If you want to calculate age in years as we typically think of it
using the calendar, you can use the following, where DOB is the Date
of Birth (Start Date) and Date2 is the End Date:
# Calculate Age in Years
# DOB: Class Date
# Date2: Class Date
Calc_Age - function(DOB, Date2)
{
if (length(DOB) != length(Date2))
stop(length(DOB) != length(Date2))
if (!inherits(DOB, Date) | !inherits(Date2, Date))
stop(Both DOB and Date2 must be Date class objects)
start - as.POSIXlt(DOB)
end - as.POSIXlt(Date2)
Age - end$year - start$year
ifelse((end$mon start$mon) |
((end$mon == start$mon) (end$mday start$mday)),
Age - 1, Age)
}
HTH,
Marc Schwartz
On Nov 8, 2009, at 1:30 PM, jim holtman wrote:
What is the frame of reference to determine the age? Check out
'difftime'.
On Sun, Nov 8, 2009 at 1:50 PM, frenchcr frenc...@btinternet.com
wrote:
Ive got a big column of dates (also some fields dont have a date so
they have
NA instead),
that i have converted into date format as so...
dates-as.character(data[,date_commissioned]); # converted dates to
characters
dates[1:10]
[1] 19910101 19860101 19910101 19860101 19910101 19910101
19910101 19910101 19910101 19910101
dateObs - as.Date(dates,format=%Y%m%d)
dateObs[1:10]
[1] 1991-01-01 1986-01-01 1991-01-01 1986-01-01 1991-01-01
1991-01-01 1991-01-01 1991-01-01 1991-01-01 1991-01-01
Now i need to turn the dates into AGE, how do i do it? Im not
worried about
fractions of years, whole years would do.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Turn dates into age
it sure does thank you!
will this work for you
x - c('19910101', '19950302', '20010502')
today - Sys.Date()
x.date - as.Date(x, format=%Y%m%d)
round(as.vector(difftime(today , x.date, units='day') / 365.25))
[1] 19 15 9
On Sun, Nov 8, 2009 at 2:44 PM, frenc...@btinternet.com wrote:
Hi Jim,
Thanks for the quick reply...not sure what you mean by frame of
reference(only been using R for 4 days)...to clarify, i need to turn my
dates from 1999-10-01 into 1999 then i subtract 2009 -1999 to get an age
of 10. The column im working on has 312,000 rows and some have NA in them
as we have no dates for that item.
To recap, the column is just a bunch of dates with some field empty, i
want to change the column from date of commision to age of asset
Cheers
Chris.
jholtman wrote:
What is the frame of reference to determine the age? Check out
'difftime'.
On Sun, Nov 8, 2009 at 1:50 PM, frenchcr frenc...@btinternet.com wrote:
Ive got a big column of dates (also some fields dont have a date so they
have
NA instead),
that i have converted into date format as so...
dates-as.character(data[,date_commissioned]); # converted dates to
characters
dates[1:10]
[1] 19910101 19860101 19910101 19860101 19910101 19910101
19910101 19910101 19910101 19910101
dateObs - as.Date(dates,format=%Y%m%d)
dateObs[1:10]
[1] 1991-01-01 1986-01-01 1991-01-01 1986-01-01 1991-01-01
1991-01-01 1991-01-01 1991-01-01 1991-01-01 1991-01-01
Now i need to turn the dates into AGE, how do i do it? Im not worried
about
fractions of years, whole years would do.
--
View this message in context:
http://old.nabble.com/Turn-dates-into-age-tp26256656p26256656.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Jim Holtman
Cincinnati, OH
+1 513 646 9390
What is the problem that you are trying to solve?
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
View this message in context:
http://old.nabble.com/Turn-dates-into-age-tp26256656p26257419.html
Sent from the R help mailing list archive at Nabble.com.
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Turn dates into age
why do you use 365.25?
dates-as.character(data[,date_commissioned]); # convert dates to
characters
#dates[1:10]
#[1] 19910101 19860101 19910101 19860101 19910101 19910101
19910101 19910101 19910101 19910101
dateObs - as.Date(dates,format=%Y%m%d)
#dateObs[1:10]
#[1] 1991-01-01 1986-01-01 1991-01-01 1986-01-01 1991-01-01
1991-01-01 1991-01-01 1991-01-01 1991-01-01 1991-01-01
today - Sys.Date()
x.date - as.Date(dateObs, format=%Y%m%d)
AGE - round(as.vector(difftime(today , x.date, units='day') / 365.25))
frenchcr wrote:
it sure does thank you!
will this work for you
x - c('19910101', '19950302', '20010502')
today - Sys.Date()
x.date - as.Date(x, format=%Y%m%d)
round(as.vector(difftime(today , x.date, units='day') / 365.25))
[1] 19 15 9
On Sun, Nov 8, 2009 at 2:44 PM, frenc...@btinternet.com wrote:
Hi Jim,
Thanks for the quick reply...not sure what you mean by frame of
reference(only been using R for 4 days)...to clarify, i need to turn my
dates from 1999-10-01 into 1999 then i subtract 2009 -1999 to get an age
of 10. The column im working on has 312,000 rows and some have NA in them
as we have no dates for that item.
To recap, the column is just a bunch of dates with some field empty, i
want to change the column from date of commision to age of asset
Cheers
Chris.
jholtman wrote:
What is the frame of reference to determine the age? Check out
'difftime'.
On Sun, Nov 8, 2009 at 1:50 PM, frenchcr frenc...@btinternet.com wrote:
Ive got a big column of dates (also some fields dont have a date so they
have
NA instead),
that i have converted into date format as so...
dates-as.character(data[,date_commissioned]); # converted dates to
characters
dates[1:10]
[1] 19910101 19860101 19910101 19860101 19910101 19910101
19910101 19910101 19910101 19910101
dateObs - as.Date(dates,format=%Y%m%d)
dateObs[1:10]
[1] 1991-01-01 1986-01-01 1991-01-01 1986-01-01 1991-01-01
1991-01-01 1991-01-01 1991-01-01 1991-01-01 1991-01-01
Now i need to turn the dates into AGE, how do i do it? Im not worried
about
fractions of years, whole years would do.
--
View this message in context:
http://old.nabble.com/Turn-dates-into-age-tp26256656p26256656.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Jim Holtman
Cincinnati, OH
+1 513 646 9390
What is the problem that you are trying to solve?
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
View this message in context:
http://old.nabble.com/Turn-dates-into-age-tp26256656p26257435.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Turn dates into age
On Nov 8, 2009, at 3:11 PM, frenchcr wrote:
why do you use 365.25?
As opposed to what?
--
David
dates-as.character(data[,date_commissioned]); # convert dates to
characters
#dates[1:10]
#[1] 19910101 19860101 19910101 19860101 19910101 19910101
19910101 19910101 19910101 19910101
dateObs - as.Date(dates,format=%Y%m%d)
#dateObs[1:10]
#[1] 1991-01-01 1986-01-01 1991-01-01 1986-01-01 1991-01-01
1991-01-01 1991-01-01 1991-01-01 1991-01-01 1991-01-01
today - Sys.Date()
x.date - as.Date(dateObs, format=%Y%m%d)
AGE - round(as.vector(difftime(today , x.date, units='day') /
365.25))
frenchcr wrote:
it sure does thank you!
will this work for you
x - c('19910101', '19950302', '20010502')
today - Sys.Date()
x.date - as.Date(x, format=%Y%m%d)
round(as.vector(difftime(today , x.date, units='day') / 365.25))
[1] 19 15 9
On Sun, Nov 8, 2009 at 2:44 PM, frenc...@btinternet.com wrote:
Hi Jim,
Thanks for the quick reply...not sure what you mean by frame of
reference(only been using R for 4 days)...to clarify, i need to
turn my
dates from 1999-10-01 into 1999 then i subtract 2009 -1999 to get
an age
of 10. The column im working on has 312,000 rows and some have NA
in them
as we have no dates for that item.
To recap, the column is just a bunch of dates with some field
empty, i
want to change the column from date of commision to age of asset
Cheers
Chris.
jholtman wrote:
What is the frame of reference to determine the age? Check out
'difftime'.
On Sun, Nov 8, 2009 at 1:50 PM, frenchcr frenc...@btinternet.com
wrote:
Ive got a big column of dates (also some fields dont have a date
so they
have
NA instead),
that i have converted into date format as so...
dates-as.character(data[,date_commissioned]); # converted
dates to
characters
dates[1:10]
[1] 19910101 19860101 19910101 19860101 19910101
19910101
19910101 19910101 19910101 19910101
dateObs - as.Date(dates,format=%Y%m%d)
dateObs[1:10]
[1] 1991-01-01 1986-01-01 1991-01-01 1986-01-01
1991-01-01
1991-01-01 1991-01-01 1991-01-01 1991-01-01 1991-01-01
Now i need to turn the dates into AGE, how do i do it? Im not
worried
about
fractions of years, whole years would do.
--
View this message in context:
http://old.nabble.com/Turn-dates-into-age-tp26256656p26256656.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Jim Holtman
Cincinnati, OH
+1 513 646 9390
What is the problem that you are trying to solve?
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
View this message in context:
http://old.nabble.com/Turn-dates-into-age-tp26256656p26257435.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
David Winsemius, MD
Heritage Laboratories
West Hartford, CT
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Turn dates into age
To clarify.
Lets turn a date into an age. Given 05/29/1971 in mm/dd/
format. What is the year difference between then and today?
This would be the age requested that starts 05/29/1971 as
one.
Thanks,
Jim
David Winsemius wrote:
On Nov 8, 2009, at 3:11 PM, frenchcr wrote:
why do you use 365.25?
As opposed to what?
-- David
dates-as.character(data[,date_commissioned]); # convert dates to
characters
#dates[1:10]
#[1] 19910101 19860101 19910101 19860101 19910101 19910101
19910101 19910101 19910101 19910101
dateObs - as.Date(dates,format=%Y%m%d)
#dateObs[1:10]
#[1] 1991-01-01 1986-01-01 1991-01-01 1986-01-01 1991-01-01
1991-01-01 1991-01-01 1991-01-01 1991-01-01 1991-01-01
today - Sys.Date()
x.date - as.Date(dateObs, format=%Y%m%d)
AGE - round(as.vector(difftime(today , x.date, units='day') / 365.25))
frenchcr wrote:
it sure does thank you!
will this work for you
x - c('19910101', '19950302', '20010502')
today - Sys.Date()
x.date - as.Date(x, format=%Y%m%d)
round(as.vector(difftime(today , x.date, units='day') / 365.25))
[1] 19 15 9
On Sun, Nov 8, 2009 at 2:44 PM, frenc...@btinternet.com wrote:
Hi Jim,
Thanks for the quick reply...not sure what you mean by frame of
reference(only been using R for 4 days)...to clarify, i need to
turn my
dates from 1999-10-01 into 1999 then i subtract 2009 -1999 to get
an age
of 10. The column im working on has 312,000 rows and some have NA
in them
as we have no dates for that item.
To recap, the column is just a bunch of dates with some field empty, i
want to change the column from date of commision to age of asset
Cheers
Chris.
jholtman wrote:
What is the frame of reference to determine the age? Check out
'difftime'.
On Sun, Nov 8, 2009 at 1:50 PM, frenchcr frenc...@btinternet.com
wrote:
Ive got a big column of dates (also some fields dont have a date
so they
have
NA instead),
that i have converted into date format as so...
dates-as.character(data[,date_commissioned]); # converted dates to
characters
dates[1:10]
[1] 19910101 19860101 19910101 19860101 19910101 19910101
19910101 19910101 19910101 19910101
dateObs - as.Date(dates,format=%Y%m%d)
dateObs[1:10]
[1] 1991-01-01 1986-01-01 1991-01-01 1986-01-01 1991-01-01
1991-01-01 1991-01-01 1991-01-01 1991-01-01 1991-01-01
Now i need to turn the dates into AGE, how do i do it? Im not worried
about
fractions of years, whole years would do.
--
View this message in context:
http://old.nabble.com/Turn-dates-into-age-tp26256656p26256656.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Jim Holtman
Cincinnati, OH
+1 513 646 9390
What is the problem that you are trying to solve?
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
View this message in context:
http://old.nabble.com/Turn-dates-into-age-tp26256656p26257435.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
David Winsemius, MD
Heritage Laboratories
West Hartford, CT
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Turn dates into age
Sys.Date()
[1] 2009-11-08
as.Date(05/29/1971, format = %m/%d/%Y)
[1] 1971-05-29
as.numeric((Sys.Date() - as.Date(05/29/1971, format = %m/%d/
%Y)) / 365.25)
[1] 38.44764
or perhaps more clearly:
EndDate - Sys.Date()
StartDate - as.Date(05/29/1971, format = %m/%d/%Y)
as.numeric((EndDate - StartDate) / 365.25)
[1] 38.44764
We coerce to numeric here, to return a standard numeric value, rather
than the result being of class difftime with an attribute of 'days'
for units:
str((Sys.Date() - as.Date(05/29/1971, format = %m/%d/%Y)) /
365.25)
Class 'difftime' atomic [1:1] 38.4
..- attr(*, units)= chr days
HTH,
Marc Schwartz
On Nov 8, 2009, at 5:22 PM, Jim Burke wrote:
To clarify.
Lets turn a date into an age. Given 05/29/1971 in mm/dd/
format. What is the year difference between then and today?
This would be the age requested that starts 05/29/1971 as
one.
Thanks,
Jim
David Winsemius wrote:
On Nov 8, 2009, at 3:11 PM, frenchcr wrote:
why do you use 365.25?
As opposed to what?
-- David
dates-as.character(data[,date_commissioned]); # convert dates to
characters
#dates[1:10]
#[1] 19910101 19860101 19910101 19860101 19910101
19910101
19910101 19910101 19910101 19910101
dateObs - as.Date(dates,format=%Y%m%d)
#dateObs[1:10]
#[1] 1991-01-01 1986-01-01 1991-01-01 1986-01-01
1991-01-01
1991-01-01 1991-01-01 1991-01-01 1991-01-01 1991-01-01
today - Sys.Date()
x.date - as.Date(dateObs, format=%Y%m%d)
AGE - round(as.vector(difftime(today , x.date, units='day') /
365.25))
frenchcr wrote:
it sure does thank you!
will this work for you
x - c('19910101', '19950302', '20010502')
today - Sys.Date()
x.date - as.Date(x, format=%Y%m%d)
round(as.vector(difftime(today , x.date, units='day') / 365.25))
[1] 19 15 9
On Sun, Nov 8, 2009 at 2:44 PM, frenc...@btinternet.com wrote:
Hi Jim,
Thanks for the quick reply...not sure what you mean by frame of
reference(only been using R for 4 days)...to clarify, i need to
turn my
dates from 1999-10-01 into 1999 then i subtract 2009 -1999 to get
an age
of 10. The column im working on has 312,000 rows and some have NA
in them
as we have no dates for that item.
To recap, the column is just a bunch of dates with some field
empty, i
want to change the column from date of commision to age of
asset
Cheers
Chris.
jholtman wrote:
What is the frame of reference to determine the age? Check out
'difftime'.
On Sun, Nov 8, 2009 at 1:50 PM, frenchcr frenc...@btinternet.com
wrote:
Ive got a big column of dates (also some fields dont have a date
so they
have
NA instead),
that i have converted into date format as so...
dates-as.character(data[,date_commissioned]); # converted
dates to
characters
dates[1:10]
[1] 19910101 19860101 19910101 19860101 19910101
19910101
19910101 19910101 19910101 19910101
dateObs - as.Date(dates,format=%Y%m%d)
dateObs[1:10]
[1] 1991-01-01 1986-01-01 1991-01-01 1986-01-01
1991-01-01
1991-01-01 1991-01-01 1991-01-01 1991-01-01 1991-01-01
Now i need to turn the dates into AGE, how do i do it? Im not
worried
about
fractions of years, whole years would do.
--
View this message in context:
http://old.nabble.com/Turn-dates-into-age-
tp26256656p26256656.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible
code.
--
Jim Holtman
Cincinnati, OH
+1 513 646 9390
What is the problem that you are trying to solve?
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
View this message in context:
http://old.nabble.com/Turn-dates-into-age-tp26256656p26257435.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
David Winsemius, MD
Heritage Laboratories
West Hartford, CT
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[[alternative HTML version deleted]]
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