Below...
On Sun, 23 Sep 2018, Sorkin, John wrote:
At the risk of asking something fundamental . . . .
does log(c1[-1]/c1[-len]
You dropped the closing parenthesis.
log( c1[-1] / c1[-len] )
do the following
(1) use all elements of c and perform the calculation
No. a) "c" is the base "concatenate" function, and b) it is using two
different subsets of the elements in c1.
(2) delete the first element of c and perform the calculation,
It does not change c1. c1[-1] is an expression that creates an entirely
new (but unnamed) vector that contains everything but the first element of
c1.
(2) delete the first two elements of c and perform the calculation,
You are wandering into the weeds here...
. . .
(n) use only the last element of c and perform the calculation.
No, c1[-len] creates a temporary array that contains all elements except
the one(s) in the variable "len". Note that the more conventional syntax
here is c1[ length(c1) ].
c1 <- 1:3
c1[ -1 ]
#> [1] 2 3
c1[ -length(c1) ]
#> [1] 1 2
c1[ -1 ] / c1[ -length( c1 ) ] # c(2,3)/c(1,2)
#> [1] 2.0 1.5
log( c1[ -1 ] / c1[ -length( c1 ) ] ) # log( c(2, 1.5) )
#> [1] 0.6931472 0.4054651
#' Created on 2018-09-23 by the [reprex package](http://reprex.tidyverse.org)
(v0.2.0).
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-boun...@r-project.org> on behalf of Wensui Liu
<liuwen...@gmail.com>
Sent: Sunday, September 23, 2018 2:26 PM
To: Ista Zahn
Cc: r-help@r-project.org
Subject: Re: [R] For Loop
CAUTION: This message originated from a non UMB, UMSOM, FPI, or UMMS email
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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 <istaz...@gmail.com> wrote:
On Sun, Sep 23, 2018 at 1:46 PM Wensui Liu <liuwen...@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 <istaz...@gmail.com> wrote:
On Sun, Sep 23, 2018 at 10:09 AM Wensui Liu <liuwen...@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 <istaz...@gmail.com> wrote:
On Sat, Sep 22, 2018 at 9:06 PM Wensui Liu <liuwen...@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 <rsher...@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
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