[R] For Loop
Sorkin, John
j@ork|n @end|ng |rom @om@um@ry|@nd@edu
Sun Sep 23 20:36:17 CEST 2018
At the risk of asking something fundamental . . . .
does log(c1[-1]/c1[-len]
do the following
(1) use all elements of c and perform the calculation
(2) delete the first element of c and perform the calculation,
(2) delete the first two elements of c and perform the calculation,
. . .
(n) use only the last element of c and perform the calculation.
Thank you,
John
John David Sorkin M.D., Ph.D.
Professor of Medicine
Chief, Biostatistics and Informatics
University of Maryland School of Medicine Division of Gerontology and Geriatric Medicine
Baltimore VA Medical Center
10 North Greene Street
GRECC (BT/18/GR)
Baltimore, MD 21201-1524
(Phone) 410-605-7119
(Fax) 410-605-7913 (Please call phone number above prior to faxing)
________________________________
From: R-help <r-help-bounces using r-project.org> on behalf of Wensui Liu <liuwensui using gmail.com>
Sent: Sunday, September 23, 2018 2:26 PM
To: Ista Zahn
Cc: r-help using r-project.org
Subject: Re: [R] For Loop
CAUTION: This message originated from a non UMB, UMSOM, FPI, or UMMS email system. Whether the sender is known or not known, hover over any links before clicking and use caution opening attachments.
what you measures is the "elapsed" time in the default setting. you
might need to take a closer look at the beautiful benchmark() function
and see what time I am talking about.
I just provided tentative solution for the person asking for it and
believe he has enough wisdom to decide what's best. why bother to
judge others subjectively?
On Sun, Sep 23, 2018 at 1:18 PM Ista Zahn <istazahn using gmail.com> wrote:
>
> On Sun, Sep 23, 2018 at 1:46 PM Wensui Liu <liuwensui using gmail.com> wrote:
> >
> > actually, by the parallel pvec, the user time is a lot shorter. or did
> > I somewhere miss your invaluable insight?
> >
> > > c1 <- 1:1000000
> > > len <- length(c1)
> > > rbenchmark::benchmark(log(c1[-1]/c1[-len]), replications = 100)
> > test replications elapsed relative user.self sys.self
> > 1 log(c1[-1]/c1[-len]) 100 4.617 1 4.484 0.133
> > user.child sys.child
> > 1 0 0
> > > rbenchmark::benchmark(pvec(1:(len - 1), mc.cores = 4, function(i) log(c1[i + 1] / c1[i])), replications = 100)
> > test
> > 1 pvec(1:(len - 1), mc.cores = 4, function(i) log(c1[i + 1]/c1[i]))
> > replications elapsed relative user.self sys.self user.child sys.child
> > 1 100 9.079 1 2.571 4.138 9.736 8.046
>
> Your output is mangled in my email, but on my system your pvec
> approach takes more than twice as long:
>
> c1 <- 1:1000000
> len <- length(c1)
> library(parallel)
> library(rbenchmark)
>
> regular <- function() log(c1[-1]/c1[-len])
> iterate.parallel <- function() {
> pvec(1:(len - 1), mc.cores = 4,
> function(i) log(c1[i + 1] / c1[i]))
> }
>
> benchmark(regular(), iterate.parallel(),
> replications = 100,
> columns = c("test", "elapsed", "relative"))
> ## test elapsed relative
> ## 2 iterate.parallel() 7.517 2.482
> ## 1 regular() 3.028 1.000
>
> Honestly, just use log(c1[-1]/c1[-len]). The code is simple and easy
> to understand and it runs pretty fast. There is usually no reason to
> make it more complicated.
> --Ista
>
> > On Sun, Sep 23, 2018 at 12:33 PM Ista Zahn <istazahn using gmail.com> wrote:
> > >
> > > On Sun, Sep 23, 2018 at 10:09 AM Wensui Liu <liuwensui using gmail.com> wrote:
> > > >
> > > > Why?
> > >
> > > The operations required for this algorithm are vectorized, as are most
> > > operations in R. There is no need to iterate through each element.
> > > Using Vectorize to achieve the iteration is no better than using
> > > *apply or a for-loop, and betrays the same basic lack of insight into
> > > basic principles of programming in R.
> > >
> > > And/or, if you want a more practical reason:
> > >
> > > > c1 <- 1:1000000
> > > > len <- 1000000
> > > > system.time( s1 <- log(c1[-1]/c1[-len]))
> > > user system elapsed
> > > 0.031 0.004 0.035
> > > > system.time(s2 <- Vectorize(function(i) log(c1[i + 1] / c1[i])) (1:len))
> > > user system elapsed
> > > 1.258 0.022 1.282
> > >
> > > Best,
> > > Ista
> > >
> > > >
> > > > On Sun, Sep 23, 2018 at 7:54 AM Ista Zahn <istazahn using gmail.com> wrote:
> > > >>
> > > >> On Sat, Sep 22, 2018 at 9:06 PM Wensui Liu <liuwensui using gmail.com> wrote:
> > > >> >
> > > >> > or this one:
> > > >> >
> > > >> > (Vectorize(function(i) log(c1[i + 1] / c1[i])) (1:len))
> > > >>
> > > >> Oh dear god no.
> > > >>
> > > >> >
> > > >> > On Sat, Sep 22, 2018 at 4:16 PM rsherry8 <rsherry8 using comcast.net> wrote:
> > > >> > >
> > > >> > >
> > > >> > > It is my impression that good R programmers make very little use of the
> > > >> > > for statement. Please consider the following
> > > >> > > R statement:
> > > >> > > for( i in 1:(len-1) ) s[i] = log(c1[i+1]/c1[i], base = exp(1) )
> > > >> > > One problem I have found with this statement is that s must exist before
> > > >> > > the statement is run. Can it be written without using a for
> > > >> > > loop? Would that be better?
> > > >> > >
> > > >> > > Thanks,
> > > >> > > Bob
> > > >> > >
> > > >> > > ______________________________________________
> > > >> > > R-help using r-project.org mailing list -- To UNSUBSCRIBE and more, see
> > > >> > > 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 using r-project.org mailing list -- To UNSUBSCRIBE and more, see
> > > >> > 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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