R help - Dec 2005 - Time data

Dear All, I wonder how to compute the age from the date of birth and the date of
examination. Suppose that have the following data:
   
  df <-
as.data.frame(rbind(c(1,"10/08/1950","15/03/1998"),
c(1,"10/08/1950","20/07/1998"),
c(1,"10/08/1950","23/10/1998")))
   
  names(df) <- c("ID", "date_birth",
"date_exam")
   
  where:
  date_birth: is the date of birth
  date_exam: date of examination
   
  I used the following program to compute the value of the age  :
   
  difftime(strptime(as.character(df$date_exam), '%d/%m/%Y'),
strptime(as.character(df$date_birth), '%d/%m/%Y'))/365.25
   
  which gives me as an output:
  
> Time differences of 47.59491, 47.94251, 48.20260 days
   
  theses values are actually the 3 ages (but 

  My questions are:
   
  1- Why in the output it says "days" instead of "years")
   
  2- How can I obtain the output as a numeric vector, without the statement
"Time difference of ....". This is in order to use it in my
calculations.
   
  3- Is there a way quicker and less redondant  to compute the age form the
date_birth and date_exam?
   
  Thanks a lot,
   
  Bernard,
   

		
---------------------------------

	[[alternative HTML version deleted]]

Hi


On 20 Dec 2005 at 11:12, Marc Bernard wrote:

Date sent:      	Tue, 20 Dec 2005 11:12:28 +0100 (CET)
From:           	Marc Bernard <bernarduse1 at yahoo.fr>
To:             	r-help at stat.math.ethz.ch
Subject:        	[R] Time data
> Dear All, I wonder how to compute the age from the date of birth and
> the date of examination. Suppose that have the following data:
> 
>   df <-
as.data.frame(rbind(c(1,"10/08/1950","15/03/1998"),
>   c(1,"10/08/1950","20/07/1998"),
c(1,"10/08/1950","23/10/1998")))
> 
>   names(df) <- c("ID", "date_birth",
"date_exam")
> 
>   where:
>   date_birth: is the date of birth
>   date_exam: date of examination
> 
>   I used the following program to compute the value of the age  :
> 
>   difftime(strptime(as.character(df$date_exam), '%d/%m/%Y'),
>   strptime(as.character(df$date_birth), '%d/%m/%Y'))/365.25
> 
>   which gives me as an output:
> 
> > Time differences of 47.59491, 47.94251, 48.20260 days
> 
>   theses values are actually the 3 ages (but 
> 
>   My questions are:
> 
>   1- Why in the output it says "days" instead of
"years")
this is in attributes of an output and is not changed by calculations
> 
>   2- How can I obtain the output as a numeric vector, without the
>   statement "Time difference of ....". This is in order to use it
in
>   my calculations.
as.numeric(as.Date(df$date_exam)-as.Date(df$date_birth))/365

However you can use it in calculations without problem as I suppose 
that "Time difference of" results from print method of a difftime 
object.

> 
>   3- Is there a way quicker and less redondant  to compute the age
>   form the date_birth and date_exam?
as.Date(df$date_exam)-as.Date(df$date_birth)

HTH
Petr

> 
>   Thanks a lot,
> 
>   Bernard,
> 
> 
> 
> ---------------------------------
> 
>  [[alternative HTML version deleted]]
> 
> ______________________________________________
> R-help at stat.math.ethz.ch mailing list
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> PLEASE do read the posting guide!
> http://www.R-project.org/posting-guide.html
Petr Pikal
petr.pikal at precheza.cz

On 12/20/05, Marc Bernard <bernarduse1 at yahoo.fr>
wrote:
> Dear All, I wonder how to compute the age from the date of birth and the
date of examination. Suppose that have the following data:
>
>  df <-
as.data.frame(rbind(c(1,"10/08/1950","15/03/1998"),
c(1,"10/08/1950","20/07/1998"),
c(1,"10/08/1950","23/10/1998")))
>
>  names(df) <- c("ID", "date_birth",
"date_exam")
>
>  where:
>  date_birth: is the date of birth
>  date_exam: date of examination
>
>  I used the following program to compute the value of the age  :
>
First ensure that df stores its dates using "Date" class in the first
place.  If your data is stored in the correct representation then
everything becomes easier subsequently:

fmt <- "%d/%m/%Y"
df[,2] <- as.Date(df[,2], fmt)
df[,3] <- as.Date(df[,3], fmt)

# converting them to numeric gives the number of days since
# the Epoch and one can just subtact those:

(as.numeric(df$date_exam) - as..numeric(df$date_birth)) / 365

R News 4/1 Help Desk article has more info on dates.


>  difftime(strptime(as.character(df$date_exam), '%d/%m/%Y'),
strptime(as.character(df$date_birth), '%d/%m/%Y'))/365.25
>
>  which gives me as an output:
>
> > Time differences of 47.59491, 47.94251, 48.20260 days
>
>  theses values are actually the 3 ages (but
>
>  My questions are:
>
>  1- Why in the output it says "days" instead of
"years")
>
>  2- How can I obtain the output as a numeric vector, without the statement
"Time difference of ....". This is in order to use it in my
calculations.
>
>  3- Is there a way quicker and less redondant  to compute the age form the
date_birth and date_exam?
>
>  Thanks a lot,
>
>  Bernard,
>
>
>
> ---------------------------------
>
>        [[alternative HTML version deleted]]
>
> ______________________________________________
> R-help at stat.math.ethz.ch mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide!
http://www.R-project.org/posting-guide.html
>