try this:
> x <- data.frame(time = seq(as.POSIXct("2011-1-1 00:00"), by = '1 hour',
> length = 300))
> # default to night
> x$day <- FALSE
> # now set days to TRUE
> # get hour/minutes
> temp <- format(x$time, "%H%M")
> x$day[temp >= "0600" & temp < "1900"] <- TRUE
>
> head(x, 30)
time day
1 2011-01-01 00:00:00 FALSE
2 2011-01-01 01:00:00 FALSE
3 2011-01-01 02:00:00 FALSE
4 2011-01-01 03:00:00 FALSE
5 2011-01-01 04:00:00 FALSE
6 2011-01-01 05:00:00 FALSE
7 2011-01-01 06:00:00 TRUE
8 2011-01-01 07:00:00 TRUE
9 2011-01-01 08:00:00 TRUE
10 2011-01-01 09:00:00 TRUE
11 2011-01-01 10:00:00 TRUE
12 2011-01-01 11:00:00 TRUE
13 2011-01-01 12:00:00 TRUE
14 2011-01-01 13:00:00 TRUE
15 2011-01-01 14:00:00 TRUE
16 2011-01-01 15:00:00 TRUE
17 2011-01-01 16:00:00 TRUE
18 2011-01-01 17:00:00 TRUE
19 2011-01-01 18:00:00 TRUE
20 2011-01-01 19:00:00 FALSE
21 2011-01-01 20:00:00 FALSE
22 2011-01-01 21:00:00 FALSE
23 2011-01-01 22:00:00 FALSE
On Sun, Jul 10, 2011 at 5:49 PM, Amy Ruiz Goyco
<amy_ruiz_go...@hotmail.com> wrote:
>
>
> Hi Professor:
>
>
> When I read the file, the time in military time and date are stored in the
> same vector as follows:
>
> 2011-01-29 16:15:11.823547
> I want to take all the hours saved in that file and categorize them as day
> and night.
> For
> example, the day would be between 06:00:00-19:00:00 and the night
> would be between 19:00:01-05:00:59. Given that condition, I intend to
> allocate time to their respective categories, and then calculate how
> many hours to read from the file fall into the category of the night and
> many more day. Then, each category be averaged, ie, the hours during
> the day and the hours during the night.
> If one day he drove 10 hours and 3 were day and 7 night, calculate the
> average.
>
> From: amy_ruiz_go...@hotmail.com
> To: rip...@stats.ox.ac.uk
> Subject: RE: [R] Converting the time in a numeric value
> Date: Sun, 10 Jul 2011 17:39:38 -0400
>
>
>
>
>
>
>
>
> When I read the file, the time in military time and date are stored in the
> same vector as follows:
>
> 2011-01-29 16:15:11.823547
> I want to take all the hours saved in that file and categorize them as day
> and night.
> For example, the day would be between 06:00:00-19:00:00 and the night would
> be between 19:00:01-05:00:59. Given that condition, I intend to allocate time
> to their respective categories, and then calculate how many hours to read
> from the file fall into the category of the night and many more day. Then,
> each category be averaged, ie, the hours during the day and the hours during
> the night.
> If one day he drove 10 hours and 3 were day and 7 night, calculate the
> average.
>
>> Date: Sun, 10 Jul 2011 08:15:00 +0100
>> From: rip...@stats.ox.ac.uk
>> To: amy_ruiz_go...@hotmail.com
>> CC: r-help@r-project.org
>> Subject: Re: [R] Converting the time in a numeric value
>>
>> On Sun, 10 Jul 2011, Amy Ruiz Goyco wrote:
>>
>> > Hello:
>>
>> > I am new using R. I have a file that contain in the same columns
>> > date and time like for example 2011/10/03 12:34:45.123423 p.m., but
>> > when I read the file and display the vector, I see of this way
>> > "2011-10-03 12:34:45.123423". I need to convert the time in a
>> > numeric and the date if is possible, but I don't need this to
>> > compute. Thus, I used this tiempo=substr(time,12,26) to selected
>> > the data that I need, but I don't know how I can change this to a
>> > numeric values.
>>
>> You need to clarify what you mean by 'date' and 'numeric' here. At a
>> guess, 'numeric' might mean 'number of seconds past midnight'. We
>> can't even do that, since we don't know the timezone (and it differs
>> on DST transition days). So you need to read a well-informed article
>> on date-times (not the un-refereed one in R-News 4/1) to gain more
>> understanding.
>>
>> Also, your AM/PM indicator is very non-standard.
>>
>> But here are some pieces for you to work with
>>
>> dt <- "2011/10/03 12:34:45.123423 p.m."
>> t0 <- substr(dt, 12, 26)
>> PM <- substr(dt, 28, 31)
>>
>> date <- strptime(dt, "%Y/%m/%d")
>> t1 <- unclass(as.POSIXct(paste("1970-01-01", t0)))
>> time <- ifelse(PM == "p.m.", t1+12*3600, t1)
>>
>> > date
>> [1] "2011-10-03"
>> > print(time, digits=12)
>> [1] 84885.123423
>>
>>
>> > [[alternative HTML version deleted]]
>> >
>> > ______________________________________________
>> > 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.
>> >
>>
>> --
>> Brian D. Ripley, rip...@stats.ox.ac.uk
>> Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/
>> University of Oxford, Tel: +44 1865 272861 (self)
>> 1 South Parks Road, +44 1865 272866 (PA)
>> Oxford OX1 3TG, UK Fax: +44 1865 272595
>
> [[alternative HTML version deleted]]
>
> ______________________________________________
> 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
Data Munger Guru
What is the problem that you are trying to solve?
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