> -----Original Message-----
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of
> rex.dw...@syngenta.com
> Sent: Thursday, March 10, 2011 8:47 AM
> To: lig...@statistik.tu-dortmund.de; arun.kumar.s...@gmail.com
> Cc: r-help@r-project.org
> Subject: Re: [R] using lapply
>
> But no one answered Kushan's question about performance
> implications of for-loop vs lapply.
> With apologies to George Orwell:
> "for-loops BAAAAAAD, no loops GOOOOOOD."
While using no loops is faster, lapply has
a loop in it and isn't much different in
speed from the equvialent for loop. The big
advantage of the *apply functions is that
they can make your code easier to understand.
Here are some times for various ways of computing
log(1:1000000). This example is probably close
to a worst-case scenario for the for loop, since
the time is dominated by the [<- operation.
Using the various *apply functions can get you a
speed-up of c. 4x, which is nice, but the vectorized
log gives a speed-up of c. 15x over the fastest of
the loops. I think the for-loop method is ungainly
because it obscures to flow of the data, but there is
no accounting for taste.
> system.time({ val.for <- numeric(1e6);for(i in
seq_len(1e6))val.for[i]<-log(i)})
user system elapsed
7.03 0.02 7.19
> system.time({ val.sapply <- sapply(seq_len(1e6), log) })
user system elapsed
6.59 0.03 6.80
> system.time({ val.lapply <- unlist(lapply(seq_len(1e6), log)) })
user system elapsed
2.48 0.00 2.52
> system.time({ val.vapply <- vapply(seq_len(1e6), log, FUN.VALUE=0)
})
user system elapsed
1.74 0.00 1.76
> system.time({ val.log <- log(seq_len(1e6)) })
user system elapsed
0.12 0.00 0.12
> identical(val.vapply,val.sapply) && identical(val.vapply,val.for) &&
identical(val.vapply,val.lapply) && identical(val.vapply,val.log)
[1] TRUE
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 Uwe Ligges
> Sent: Thursday, March 10, 2011 4:38 AM
> To: Arun Kumar Saha
> Cc: r-help@r-project.org
> Subject: Re: [R] using lapply
>
>
>
> On 10.03.2011 08:30, Arun Kumar Saha wrote:
> > On reply to the post
> > http://r.789695.n4.nabble.com/using-lapply-td3345268.html
>
> Hmmm, can you please reply to the original post and quote it?
> You mail was not recognized to be in the same thread as the message of
> the original poster (and hence I wasted time to answer it again).
>
> Thanks,
> Uwe Ligges
>
>
>
>
> > Dear Kushan, this may be a good start:
> >
> > ## assuming 'instr.list' is your list object and you are applying
> > my.strat() function on each element of that list, you can use lapply
> > function as
> > lapply(instr.list, function(x) return(my.strat(x)))
> >
> > Here resulting element will again be another list with
> length is same as the
> > length of your original list 'instr.list.'
> >
> > Instead if the returned object for my.strat() function is a
> single number
> > then you might want to create a vector instead list, in
> that case just use
> > 'sapply'
> >
> > HTH
> >
> > Arun,
> >
> > [[alternative HTML version deleted]]
> >
> > ______________________________________________
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> > PLEASE do read the posting guide
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