Doing as much as possible with vectors instead of loops of a good thing.
Fooling yourself that apply functions are vectorized is, well, not a good thing.
If you want to write a function to use instead of sapply, fine, but don't call
it *apply because those functions always give you one result for each input
item you start with, and anyone who reads your code will be confused if you
rename a standard function. I think you need to use ?while.
---------------------------------------------------------------------------
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Sent from my phone. Please excuse my brevity.
On October 10, 2014 1:21:23 AM PDT, PO SU <rhelpmaill...@163.com> wrote:
>
>OK, it seems that i misunderstand something, i forget how and when i
>pick up the monition in my mind that " as possible as avoid using for
>loop".
>TKS for all your suggestions!
>But i still want the way to break sapply, if not exsits now, create it
>.....
>such as:
> sapply<-function(...){
>out<-FALSE
>.....
>if(out==TRUE) return
>}
>sapply(1:10, function(i){
>if(i=5) out<-TRUE
>}
>)
>That means to rewrite sapply, and create a virable in it , let called
>OUT, then mayebe in sapply(1:10,FUN)
>FUN can use OUT. because i think in sapply, FUN is an inner
>function,so it can use OUT.
>when it let OUT=TRUE. sapply should be break..........
>
>
>
>
>
>
>
>--
>
>PO SU
>mail: desolato...@163.com
>Majored in Statistics from SJTU
>
>
>
>
>At 2014-10-10 14:44:47, "Hervé Pagès" <hpa...@fhcrc.org> wrote:
>>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.
>>>>
>>>>
>---------------------------------------------------------------------------
>>>> Jeff Newmiller The ..... ..... Go
>Live...
>>>> DCN:<jdnew...@dcn.davis.ca.us> Basics: ##.#. ##.#.
>Live Go...
>>>> Live: OO#.. Dead: OO#..
>Playing
>>>> Research Engineer (Solar/Batteries O.O#. #.O#.
>with
>>>> /Software/Embedded Controllers) .OO#. .OO#.
>rocks...1k
>>>>
>---------------------------------------------------------------------------
>>>> Sent from my phone. Please excuse my brevity.
>>>>
>>>> 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
>>>>> ______________________________________________
>>>>> 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.
>>>>
>>> ______________________________________________
>>> 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.
>>>
>>
>>--
>>Hervé Pagès
>>
>>Program in Computational Biology
>>Division of Public Health Sciences
>>Fred Hutchinson Cancer Research Center
>>1100 Fairview Ave. N, M1-B514
>>P.O. Box 19024
>>Seattle, WA 98109-1024
>>
>>E-mail: hpa...@fhcrc.org
>>Phone: (206) 667-5791
>>Fax: (206) 667-1319
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