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 ______________________________________________ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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.
