[R] create an empty data frame and then fill in it (and then evaluate the mean of semi-hourly data for each day)

[PIKAL Petr]

Hi Sofia

df_snow_day  <- aggregate(df_snow$snow, list(df_snow$day), mean, na.rm=TRUE)

should give you automagically required data frame.

Regards
Petr

> -----Original Message-----
> From: R-help [mailto:r-help-bounces at r-project.org] On Behalf Of Stefano
> Sofia
> Sent: Friday, June 10, 2016 12:46 PM
> To: Duncan Murdoch <murdoch.duncan at gmail.com>; r-help at r-project.org
> Subject: Re: [R] create an empty data frame and then fill in it (and then
> evaluate the mean of semi-hourly data for each day)
>
> Thank you for your answer. Very clear.
> (I don't like the second solution either.) Let me then ask a final question.
> From an initial data frame with semi-hourly data (df_snow, with two
> columns, data_POSIX of type "POSIXct" "POSIXt" and snow of type
> "numeric"), I need to evaluate the mean of for each day.
>
> data_POSIX snow
> 2004-11-01 00:00:00 50
> 2004-11-01 00:30:00 55
> 2004-11-01 01:00:00 60
> ...
>
> I first created a new column of type "Date"
> df_snow$day <- as.Date(df_snow$data_POSIX,"%Y-%m-%d")
>
> then I created a new data frame called df_snow_day to store the mean of
> data grouped by day:
> list_days <- unique(df_snow$day)
> df_snow_day <- data.frame(day=list_days)
>
> Finally I applied lapply in this way:
> df_snow_day$snow <- lapply(df_snow_day$day, function(x)
> round(mean(df_snow$snow[df_snow$day == x], na.rm=T)))
>
> This does not work. I do not understand why the class of df_snow_day$snow
> is of type list either:
>
>        day snow
> NA    <NA>       NULL
> NA.1  <NA>       NULL
> NA.2  <NA>       NULL
>
> Where is my mistake?
>
> Thank you for all your help
> Stefano
>
>
> _____________________________________________
>
> Da: Duncan Murdoch [murdoch.duncan at gmail.com]
> Inviato: giovedì 9 giugno 2016 12.36
> A: Stefano Sofia; r-help at r-project.org
> Oggetto: Re: [R] create an empty data frame and then fill in it
>
> On 09/06/2016 6:22 AM, Stefano Sofia wrote:
> > Dear R list users,
> > sorry for this simple question, but I already spent many efforts to solve it.
> >
> > I create an empty data frame called df_year like
> >
> > df_year <- data.frame(day=as.Date(character()), hs_MteBove=integer(),
> > hs_MtePrata=integer(), hs_Pintura=integer(), hs_Pizzo=integer(),
> > hs_Sassotetto=integer(), hs_Sibilla=integer(), stringsAsFactors=FALSE)
> >
> > and then I start to fill in it with
> >
> > df_year$day <- seq(as.Date("2004-11-01-00-00","%Y-%m-%d"),
> > as.Date("2005-05-01-00-00","%Y-%m-%d"), by="day")
> >
> > but I get the following error:
> > "replacement has 182 rows, data has 0"
> >
> > Where is my silly mistake?
>
> Your dataframe has 0 rows, so you can't put a 182 row vector into the first
> column.
>
> Unlike vectors, dataframes won't grow if you make assignments beyond the
> end of the rows.
>
> There are at least a couple of solutions:
>
> 1.  Don't create columns until you have data ready for them.
>
> You can wait to create the dataframe until your "day" column is ready:
>
> df_year <- data.frame(day = seq(...))
>
> As you compute other columns of the same length, you can add them, e.g.
>
> df_year$hs_MteBove <- ...
>
> 2.  Create your columns with the right length from the beginning:
>
> df_year <- data.frame(day = rep(as.Date(NA), 182), ...)
>
> I don't like this solution as much.
>
> Duncan Murdoch
>
>
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