Dear Jim and dear Enrico,
thank you for your replies.
Unfortunately your hints didn't solve my problem, and I am getting mad.
Can I show you my whole process? I will be as quick as possible.
I start from a data frame called Snow of the form
year month day hh mm hs
2007 11 19 0 0 0.00
2007 11 19 0 30 0.00
2007 11 19 1 0 0.00
2007 11 19 1 30 0.00
2007 11 19 2 0 0.00
2007 11 19 2 30 0.00
2007 11 19 3 0 0.00
2007 11 19 3 30 0.00
2007 11 19 4 0 0.00
2007 11 19 4 30 0.00
...
whth semi-hourly data.
I need to deal with date so I used strptime:
Snow$data_factor <- as.factor(paste(Snow$year, Snow$month, Snow$day, Snow$hh,
Snow$mm, sep="-"))
Snow$data_strptime <- strptime(Snow$data_factor, format = "%Y-%m-%d-%H-%M")
It gives me
year month day hh mm hs data_factor data_strptime
1 2007 11 19 0 0 0 2007-11-19-0-0 2007-11-19 00:00:00
2 2007 11 19 0 30 0 2007-11-19-0-30 2007-11-19 00:30:00
3 2007 11 19 1 0 0 2007-11-19-1-0 2007-11-19 01:00:00
4 2007 11 19 1 30 0 2007-11-19-1-30 2007-11-19 01:30:00
5 2007 11 19 2 0 0 2007-11-19-2-0 2007-11-19 02:00:00
6 2007 11 19 2 30 0 2007-11-19-2-30 2007-11-19 02:30:00
7 2007 11 19 3 0 0 2007-11-19-3-0 2007-11-19 03:00:00
8 2007 11 19 3 30 0 2007-11-19-3-30 2007-11-19 03:30:00
9 2007 11 19 4 0 0 2007-11-19-4-0 2007-11-19 04:00:00
10 2007 11 19 4 30 0 2007-11-19-4-30 2007-11-19 04:30:00
...
The type of the column data_strptime is
$data_strptime
[1] "POSIXlt" "POSIXt"
Because of some days (or part of them) might be missing, given a time interval
I want to create a new data frame with all time-steps and then merge the new
data frame with the old one.
In order to create a new data frame with all time-steps, I thought to use
df_new <- data.frame(data_strptime=seq(init_day, fin_day, by="30 mins"))
and then
Snow_all <- merge(df_new, Snow, by=("data_strptime"), all.x=TRUE)
My problem is in dealing with init_day and fin_day, respectively for example
"200711190000" and "200711210000".
I am not able to create a sequence of class "POSIXlt" "POSIXt", in order to
merge the two data frames.
Could you please help me in this?
Thank you again for your attention
Stefano
________________________________________
Da: Jim Lemon [drjimle...@gmail.com]
Inviato: lunedì 11 aprile 2016 9.47
A: Stefano Sofia
Cc: r-help@r-project.org
Oggetto: Re: [R] Query about use of format in strptime
Hi Stefano,
As the help page says:
"The default for the format methods is "%Y-%m-%d %H:%M:%S" if any
element has a time component which is not midnight, and "%Y-%m-%d"
otherwise. This is because when the result is printed, it uses the
default format. If you want a specified output representation:
format(strptime(init_day, format="%Y-%m-%d-%H-%M"),"%Y-%M-%d %H:%M")
[1] "2015-30-24 00:30"
For the "midnight" case:
format(strptime(init_day, format="%Y-%m-%d-%H-%M"),"%Y-%m-%d %H:%M")
[1] "2015-02-24 00:00"
Jim
On Mon, Apr 11, 2016 at 5:22 PM, Stefano Sofia
<stefano.so...@regione.marche.it> wrote:
> Dear R-list users,
> I need to use strptime because I have to deal with date with hours and
> minutes.
> I read the manual for strptime and I also looked at many examples, but when I
> try to apply it to my code, I always encounter some problems.
> I try to change the default format, with no success. Why? How can I change
> the format?
>
> 1.
> init_day <- as.factor("2015-02-24-00-30")
> strptime(init_day, format="%Y-%m-%d-%H-%M")
> [1] "2015-02-24 00:30:00"
> It works, but why also seconds are shown if in format seconds are not
> specified?
>
> 2.
> init_day <- as.factor("2015-02-24-0-00")
> strptime(init_day, format="%Y-%m-%d-%H-%M")
> [1] "2015-02-24"
> Again, the specified format is not applied. Why?
>
> Thank you for your attention and your help
> Stefano
>
>
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visionata da persone estranee al destinatario.
IMPORTANT NOTICE: This e-mail message is intended to be received only by
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