Hi,
On 10/09/2014 11:12 PM, PO SU wrote:
Is that mean while may be more effient than for in R? as i know, while and for
are all just functions in R.
Tks for your suggestion to not use apply that way, but i want to know, if
possible, is there any way to break it ?
As Jeff said, you cannot break the loop that happens inside
the sapply() call. Also it is *not* true that for or while are
less efficient than sapply() or lapply():
> a <- numeric(100000)
> system.time(for (i in 1:100000) {a[i] <- i * (i - 1) / 2})
user system elapsed
0.148 0.000 0.147
> system.time(b <- sapply(1:100000, function(i) {i * (i - 1) / 2}))
user system elapsed
0.194 0.007 0.201
> identical(a, b)
[1] TRUE
> system.time(c <- unlist(lapply(1:100000, function(i) {i * (i - 1) / 2})))
user system elapsed
0.116 0.000 0.119
> identical(a, c)
[1] TRUE
OK lapply() is maybe slightly faster but not significantly. And the
more work you need to do inside the loop, the less significant this
difference will be.
Actually, there is a additional question:
x<- c(3,4,5,6,9)
sapply(x ,function(i){
foo(i) #do something to each value in x,how can i know the i's index in x?
)}
You can't. Inside the anonymous function, you only have access to 'i'
which is an element of 'x', not its index in 'x'.
In my way , i always
sapply(seq(x),function(i){
foo(x[i])
})
Yes, if you want to loop on the index instead of the elements, you
need to do something like that. Using seq_along(x) is probably
cleaner than seq(x) for this.
Cheers,
H.
or
Map( function(i,index){
foo(i) # through index to know the i's index in x
},x ,seq(x))
How you solve the problem? I mean just use apply functions.
--
PO SU
mail: desolato...@163.com
Majored in Statistics from SJTU
At 2014-10-10 13:58:29, "Jeff Newmiller" <jdnew...@dcn.davis.ca.us> wrote:
Don't use apply functions if you want to do what you describe. They don't work
that way. Use a while control structure.
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On October 9, 2014 10:24:49 PM PDT, PO SU <rhelpmaill...@163.com> wrote:
Dear expeRts,
i use sapply for loop, and i want to break it when i needed, how to
do that? e.g.
sapply( 1:10, function(i) {
if(i==5) break and jump out of the function sapply
} )
I want to do it because i have to loop 1000000 times, but i don't know
when it will break, that means, it may need break at i=5 or at i=50000,
for the possible of the last case, i don't use for loop, because it
slow(is it right?).
So,if you happen to know it ,may you help me?
--
PO SU
mail: desolato...@163.com
Majored in Statistics from SJTU
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--
Hervé Pagès
Program in Computational Biology
Division of Public Health Sciences
Fred Hutchinson Cancer Research Center
1100 Fairview Ave. N, M1-B514
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E-mail: hpa...@fhcrc.org
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