create an empty data frame and then fill in it (and then evaluate the mean of semi-hourly data for each day)
On 10/06/2016 6:45 AM, Stefano Sofia wrote:
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:
lapply() returns a list. Petr's solution is probably better, but you could likely get what you want using vapply() instead: df_snow_day$snow <- vapply(df_snow_day$day, function(x) round(mean(df_snow$snow[df_snow$day == x], na.rm=T)), 0) The 0 at the end is an example of the numeric function result you want, so that vapply() knows to create a numeric vector. Duncan Murdoch
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: gioved? 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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