Since this example is not reproducible (and you have not quuoted any
former code) I can only give advice "in principle":
1. Never use 1:length(x) since this will seriously fail if x is a length
0 object. Instead, use seq_along(x)
2. If k is a list, then you probably want to use doubled brackets in
k[[year]] for a scalar valued "year".
Uwe Ligges
On 16.06.2011 23:49, mdvaan wrote:
I am still thinking about this problem. The solution could look something
like this (it's net yet working):
k<-lapply(h, function (x) x*0) # I keep the same format as h, but set all
values to 0
years<-c(1997:1999) # I define the years
for (t in 1:length(years))
{
year = as.character(years[t])
ids = rownames(h[year][[1]])
}
for (m in 1:length(relevant_firms))
{
k[year][[m]]<-lapply(k[year], function (col) k[year][[1]][,m] =
h[year][[1]][,m]& k[year][[1]][m,] = h[year][[1]][m,])
} # I am creating new list objects that should look like this
k$'1999'$'8029' and I replace the values in the 8029 column and row by the
original ones in h
Any takes on this problem? Thank you very much!
Best
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