Selections in lists
On Aug 25, 2011, at 11:24 AM, mdvaan wrote:
Hi, I have produced a list g and I would like to reduce the amount of information contained in each object in g. For each matrix I would like to keep the values where the column name equals g[year][[1]][[x]] and the row names equals g[year][[1]][[-x]]. So in g$`1999`$`8029`, year = 1999 and x = 8029. I have been experimenting with the subset function, but have been unsuccesful. Thanks for your help!
I suspect this is because you g object does not have the structure that you imagine for it: > str(g) List of 3 $ 1997: num [1:4, 1:4] 3 2 1 1 2 3 0 1 1 0 ... ..- attr(*, "dimnames")=List of 2 .. ..$ B: chr [1:4] "8025" "8026" "8027" "8029" .. ..$ B: chr [1:4] "8025" "8026" "8027" "8029" $ 1998: num [1:4, 1:4] 2 1 1 0 1 3 1 1 1 1 ... ..- attr(*, "dimnames")=List of 2 .. ..$ B: chr [1:4] "8025" "8026" "8027" "8029" .. ..$ B: chr [1:4] "8025" "8026" "8027" "8029" $ 1999: num [1:5, 1:5] 2 1 1 0 0 1 3 2 0 1 ... ..- attr(*, "dimnames")=List of 2 .. ..$ B: chr [1:5] "8025" "8026" "8027" "8028" ... .. ..$ B: chr [1:5] "8025" "8026" "8027" "8028" ... The things you think are list indices are really in the 'dimnames' attributes of the objects.
David
> The result for g$`1999`$`8029` should be:
>
> $`1999`$`8029`
> B
> B 8029
> 8026 1
> 8027 1
> 8028 1
>
> The result for g$`1999`$`8028` should be:
>
> $`1999`$`8028`
> B
> B 8028
> 8029 1
>
>
> The result for g$`1999`$`8027` should be:
>
> $`1999`$`8027`
> B
> B 8027
> 8025 1
> 8026 2
> 8029 1
>
>
> Example:
>
> 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)
> years<-names(g)
> for (t in seq(years))
> {
> year=as.character(years[t])
> g[[year]]<-sapply(colnames(g[[year]]), function(var)
> g[[year]][g[[year]][,var]>0, g[[year]][var,]>0])
> }
>
> --
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> Sent from the R help mailing list archive at Nabble.com.
>
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David Winsemius, MD
West Hartford, CT