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-boun...@r-project.org] On Behalf Of Stefano > Sofia > Sent: Friday, June 10, 2016 12:46 PM > To: Duncan Murdoch <murdoch.dun...@gmail.com>; r-help@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.dun...@gmail.com] > Inviato: giovedì 9 giugno 2016 12.36 > A: Stefano Sofia; r-help@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 > > > ________________________________ > > AVVISO IMPORTANTE: Questo messaggio di posta elettronica può contenere > informazioni confidenziali, pertanto è destinato solo a persone autorizzate > alla ricezione. I messaggi di posta elettronica per i client di Regione Marche > possono contenere informazioni confidenziali e con privilegi legali. Se non > si è > il destinatario specificato, non leggere, copiare, inoltrare o archiviare > questo > messaggio. 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