I still do not get the point for what task this expansion of data may be
useful, by I guess you want (in this case probably very inefficient, but
other can work out how to improve if interested) to insert after
k <- lapply(h, function (x) x*0)
the lines:
for(i in seq_along(k)){
temp <- rep(k[i], ncol(k[[i]]))
names(temp) <- colnames(k[[i]])
k[i] <- list(temp)
for(j in seq_along(colnames(h[[i]]))){
k[[i]][[j]][j,] <- k[[i]][[j]][,j] <- h[[i]][,j]
}
}
Uwe Ligges
On 17.06.2011 13:48, Mathijs de Vaan wrote:
Sorry, forgot to quote:
Hi,
I am trying to use the objects from the list below to create more objects.
For each year in h I am trying to create as many objects as there are B's
keeping only the values of B. Example for 1999:
$`1999`$`8025`
B
B 8025 8026 8027 8028 8029
8025 1 1 1 0 0
8026 1 0 0 0 0
8027 1 0 0 0 0
8028 0 0 0 0 0
8029 0 0 0 0 0
$`1999`$`8026`
B
B 8025 8026 8027 8028 8029
8025 0 1 0 0 0
8026 1 1 1 0 1
8027 0 1 0 0 0
8028 0 0 0 0 0
8029 0 1 0 0 0
$`1999`$`8027`
B
B 8025 8026 8027 8028 8029
8025 0 0 1 0 0
8026 0 0 1 0 0
8027 1 1 1 0 1
8028 0 0 0 0 0
8029 0 0 1 0 0
$`1999`$`8028`
B
B 8025 8026 8027 8028 8029
8025 0 0 0 0 0
8026 0 0 0 0 0
8027 0 0 0 0 0
8028 0 0 0 1 1
8029 0 0 0 1 0
$`1999`$`8029`
B
B 8025 8026 8027 8028 8029
8025 0 0 0 0 0
8026 0 0 0 0 1
8027 0 0 0 0 1
8028 0 0 0 0 1
8029 0 1 1 1 1
Any suggestions? You help is very much appreciated!
DF = data.frame(read.table(textConnection(" A B C
80 8025 1995
80 8026 1995
80 8029 1995
81 8026 1996
82 8025 1997
82 8026 1997
83 8025 1997
83 8027 1997
90 8026 1998
90 8027 1998
90 8029 1998
84 8026 1999
84 8027 1999
85 8028 1999
85 8029 1999"),head=TRUE,stringsAsFactors=FALSE))
e<- function(y) crossprod(table(DF[DF$C %in% y, 1:2]))
years<- sort(unique(DF$C))
f<- as.data.frame(embed(years, 3))
g<-lapply(split(f, f[, 1]), e)
h<-lapply(g, function (x) ifelse(x>0,1,0))
2011/6/17 Uwe Ligges<lig...@statistik.tu-dortmund.de>
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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