I'm going to assume that Kunden is a data frame, and it has columns
(variables) with names like
Umstatz_2011
and that you want to create new columns with names like
Kunde_real_2011
If that is so, then try this (not tested):
for (year in 2011:2015) {
nmK <- paste0("Kunde_real_", year)
nmU
Hi Georg,
The " around Kunden$* looks unintentional to me.
Second: haveyou considered using a long table? Then you would fill a known
set of columns.
Third if you must have columns based on year I believe df[[a.column.name]]
will work.
Best
Ulrik
schrieb am Sa., 23. Apr. 2016 07:33:
> Hi a
The direct answer to your question is to look at ?get and ? assign.
The R-ish answer to your question is to store the data as elements of a
list rather than separate files and use lapply() instead.
Sarah
On Friday, April 22, 2016, wrote:
> Hi all,
>
> I would like to use a loop for tasks that
Hi all,
I would like to use a loop for tasks that occurs repeatedly:
# Groups
# Umsatz <= 0: 1 (NICHT kaufend)
# Umsatz > 0: 2 (kaufend)
for (year in c("2011", "2012", "2013", "2014", "2015")) {
paste0("Kunden$Kunde_real_", year) <- (paste0("Kunden$Umsatz_", year) <= 0) *
1 +
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