To add to William's remarks, another advantage of the
apply family of functions is that they avoid growing an
object inside a loop, which is very inefficient in R.
In other words, without the *apply functions, users
might do something like this:
answer = NULL
for(i in 1:nrows)
answer = rbind(answer,calculateanewrow(i))
or
answer = NULL
for(i in 1:n)
answer = c(answer,newcalculation(i))
So in addition to making your program easier to understand
(which is a huge advantange in and of itself), they
also help you avoid a programming paradigm that's
very inefficient in R:
mat = matrix(abs(rnorm(10000)),1000,10)
system.time({answer=NULL;for(i in 1:nrow(mat))answer =
rbind(answer,log(mat[i,]))})
user system elapsed
0.052 0.020 0.072
system.time({answer1 = t(apply(mat,1,log))})
user system elapsed
0.012 0.000 0.012
all.equal(answer,answer1)
[1] TRUE
That's a speedup of a factor of 6, which gets even bigger as the size
of the object increases:
mat = matrix(abs(rnorm(100000)),10000,10)
system.time({answer=NULL;for(i in 1:nrow(mat))answer =
rbind(answer,log(mat[i,]))})
user system elapsed
5.960 1.524 7.505
system.time({answer1 = t(apply(mat,1,log))})
user system elapsed
0.120 0.004 0.123
all.equal(answer,answer1)
[1] TRUE
Now it's a speedup of 60 -- essentially an O(n^2) algorithm competing with
an O(n) algorithm.
The lack of scalability of this paradigm often leads new users to believe that
R can't handle large problems. Learning to use the apply family of functions
from the start avoids this misconception.
- Phil Spector
Statistical Computing Facility
Department of Statistics
UC Berkeley
spec...@stat.berkeley.edu
On Thu, 10 Mar 2011, William Dunlap wrote:
-----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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