# Re: [R] Time data

From: Petr Pikal <petr.pikal_at_precheza.cz>
Date: Tue 20 Dec 2005 - 21:50:03 EST

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@yahoo.fr>
To:             	r-help@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,
>
>
>
> ---------------------------------
>
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