> No need for all this (see solutions including mine already given) --
> but even without those, this is silly. An identity map is a real waste
> if you just want the simplification bit of sapply() -- you'd be much
> better just using simplify2array()
You are right that simplify2array(p) does everything that sapply(p,function(x)x)
does and does it more quickly and that matrix(unlist(p),nrow=2,byrow=TRUE)
is much faster than either:
> p <- lapply(1:1e6, function(i)c(i, log2(i)))
> system.time(z1 <- t(sapply(p,function(x)x)))
user system elapsed
1.2 0.0 1.2
> system.time(z2 <- t(simplify2array(p)))
user system elapsed
0.91 0.00 0.90
> system.time(z3 <- matrix(unlist(p), ncol=2, byrow=TRUE))
user system elapsed
0.04 0.00 0.04
You can also use vapply instead of sapply - it requires that you supply
the expect shape and type of FUN's output so it is doesn't have to figure
this out from looking at all the outputs of FUN:
> system.time(z4 <- t(vapply(p,FUN=function(x)x,FUN.VALUE=numeric(2))))
user system elapsed
0.56 0.00 0.56
An advantage of vapply is that it stops in its tracks if FUN returns
an unexpected value. sapply() and simplify2array() will silently give
you an unexpected result (a single column matrix of mode list
instead of a vector of numbers) and matrix(unlist()...) gives you a
warning if you are lucky.
> pBad <- p ; pBad[[length(pBad)/2]] <- 666
> system.time(zBad1 <- t(sapply(pBad,function(x)x)))
user system elapsed
1.75 0.00 1.75
> zBad1[,499999:500001] # not what we wanted
[[1]]
[1] 499999.00000 18.93157
[[2]]
[1] 666
[[3]]
[1] 500001.00000 18.93157
> system.time(zBad2 <- t(simplify2array(pBad)))
user system elapsed
0.5 0.0 0.5
> system.time(zBad3 <- matrix(unlist(pBad), ncol=2, byrow=TRUE))
user system elapsed
0.03 0.00 0.03
Warning message:
In matrix(unlist(pBad), ncol = 2, byrow = TRUE) :
data length [1999999] is not a sub-multiple or multiple of the number of
rows [1000000]
> # no warning if length(unlist(p)) were even
> system.time(zBad4 <- t(vapply(pBad,function(x)x,numeric(2))))
Error in vapply(pBad, function(x) x, numeric(2)) :
values must be length 2,
but FUN(X[[500000]]) result is length 1
Timing stopped at: 0.29 0 0.28
Which of the latter two methods you choose depends on how likely errors
in the data are.
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 R. Michael Weylandt
> Sent: Tuesday, December 04, 2012 2:59 PM
> To: arun
> Cc: R help; s...@gnu.org
> Subject: Re: [R] list to matrix?
>
> On Tue, Dec 4, 2012 at 8:17 PM, arun <smartpink...@yahoo.com> wrote:
> > Hi,
> > Try this:
> > list1<-list(c(50000, 101), c(1e+05, 46), c(150000, 31), c(2e+05, 17),
> > c(250000, 19), c(3e+05, 11), c(350000, 12), c(4e+05, 25),
> > c(450000, 19), c(5e+05, 16))
> >
> >
> > res<-t(sapply(list1,function(x) x))
>
> Bah Humbug! (In Christmas cheer)
>
> No need for all this (see solutions including mine already given) --
> but even without those, this is silly. An identity map is a real waste
> if you just want the simplification bit of sapply() -- you'd be much
> better just using simplify2array()
>
> Michael
>
> ______________________________________________
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
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