I was going to suggest
> AB <- df[c("A","B")]
> ls2 <- array(split(df$C, AB), dim=sapply(AB, nlevels), dimnames=sapply(AB,
levels))
which produces a matrix very similar to what Duncan's by() call produces
> ls1 <- by(df$C, df[,1:2], identity)
E.g.,
> ls2[["a","X"]]
[1] 1 2
> ls1[["a","X"]]
[1] 1 2
> ls1[["a","Y"]] # by assigns NULL to unoccupied slots
NULL
> ls2[["a","Y"]] # split gives the same type to all slots, copied from input
numeric(0)
They both are quick because they use split() to avoid the repeated
evaluations of
bigVector[ anotherBigVector == scalar ]
that your nested (not imbricated) loops do. If you really need to convert
the matrix to a list of lists that will probably be a quick transformation.
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
> -----Original Message-----
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
> Behalf Of Duncan Murdoch
> Sent: Wednesday, August 10, 2011 9:43 AM
> To: Frederic F
> Cc: r-help@r-project.org
> Subject: Re: [R] How to quickly convert a data.frame into a structure of lists
>
> On 10/08/2011 10:30 AM, Frederic F wrote:
> > Hello Duncan,
> >
> > Here is a small example to illustrate what I am trying to do.
> >
> > # Example data.frame
> > df=data.frame(A=c("a","a","b","b"), B=c("X","X","Y","Z"), C=c(1,2,3,4))
> > # A B C
> > # 1 a X 1
> > # 2 a X 2
> > # 3 b Y 3
> > # 4 b Z 4
> >
> > ### First way of getting the list structure (ls1) using imbricated lapply
> > loops:
> > # Get the structure and populate it:
> > ls1<-lapply(levels(df$A), function(levelA) {
> > lapply(levels(df$B), function(levelB) {df$C[df$A==levelA&
> > df$B==levelB]})
> > })
> > # Apply the names:
> > names(list_structure)<-levels(df$A)
> > for (i in 1:length(list_structure))
> > {names(list_structure[[i]])<-levels(df$B)}
> >
> > # Result:
> > ls1$a$X
> > # [1] 1 2
> > ls1$b$Z
> > # [1] 4
> >
> > The data.frame will always be 'complete', i.e., there will be a value in
> > every row for every column.
> > I want to produce a structure like this one quickly (I aim at something
> > below 10 seconds) for a dataset containing between 1 and 2 millions of rows.
> >
>
> I don't know what the timing would be like for your real data, but this
> does look like by() would work:
>
> ls1 <- by(df$C, df[,1:2], identity)
>
> When I repeat the rows of df a million times each, this finishes in a
> few seconds. It would definitely be slower if there were more levels of
> A or B.
>
> Now ls1 will be a matrix whose entries are the subsets of C that you
> want, so you can see your two results with slightly different syntax:
>
> > ls1[["a", "X"]]
> [1] 1 2
> > ls1[["b","Z"]]
> [1] 4
>
> Duncan Murdoch
>
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