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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