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 <ligges@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
>>
>>
>> --
>> View this message in context: http://r.789695.n4.nabble.com/**
>> Multiply-list-objects-**tp3595719p3603871.html<http://r.789695.n4.nabble.com/Multiply-list-objects-tp3595719p3603871.html>
>> Sent from the R help mailing list archive at Nabble.com.
>>
>> ______________________________**________________
>> R-help@r-project.org mailing list
>> https://stat.ethz.ch/mailman/**listinfo/r-help<https://stat.ethz.ch/mailman/listinfo/r-help>
>> PLEASE do read the posting guide http://www.R-project.org/**
>> posting-guide.html <http://www.R-project.org/posting-guide.html>
>> and provide commented, minimal, self-contained, reproducible code.
>>
>

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