Hi Peter,
Thanks for these information.
I used a column concatenating the listBy data to do this aggregation : (I
don't know if it's the best solution, but it seems to work).
aggregateMultiBy <- function(x, by, FUN){
tableBy = data.frame(by)
tableBy$byKey = ""
for(colBy in
On Aug 2, 2011, at 19:09 , Guillaume wrote:
> Hi Peter,
>
> Yes I have a large number of factors in the listBy table.
>
> Do you mean that aggregate() creates a complete cartesian product of the
> "by" columns ? (and creates combinations of values that do not exist in the
> orignial "by" table,
Hi Peter,
Yes I have a large number of factors in the listBy table.
Do you mean that aggregate() creates a complete cartesian product of the
"by" columns ? (and creates combinations of values that do not exist in the
orignial "by" table, before removing them when returning the aggregated
table?)
On Aug 2, 2011, at 17:10 , Guillaume wrote:
> Hi Peter,
> Thanks for your answer.
> I made a mistake in the script I copied sorry !
>
> The description of the object : listX has 3 column, listBy has 4 column, and
So what is the contents of listBy? If they are all factors with 100 levels,
t
Hi Peter,
Thanks for your answer.
I made a mistake in the script I copied sorry !
The description of the object : listX has 3 column, listBy has 4 column, and
they have 9000 rows :
print(paste("ncol x ", length((listX
print(paste("ncol By ", length((listBy
print(paste("nrow ", length(
On Aug 2, 2011, at 11:45 , Guillaume wrote:
> Dear all,
> I am trying to aggregate a table (divided in two lists here), but get a
> memory error.
> Here is the code I'm running :
>
> sessionInfo()
>
> print(paste("memory.limit() ", memory.limit()))
> print(paste("memory.size() ",
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