[R] in continuation with the earlier R puzzle

[Huso, Manuela]

Using Ted Harding's example:

news1o <- runif(1000000)
s2o    <- runif(1000000)

pt1 <- proc.time()
s      <- numeric(length(news1o))-1  # Set all of s to -1
s[news1o>s2o] <-1                    # Change to 1 only those values of s
                                     #  for which news1o>s2o
pt2<- proc.time()
pt2-pt1                              # Takes even less time...   
#   user  system elapsed 
#   0.04    0.00    0.05 

>::<>::<>::<>::<>::<>::<>::<>::<>::<>::<>::<>::<
Please note:  I will be out of the office and out
  of email contact from 7/11-7/25/2010
>::<>::<>::<>::<>::<>::<>::<>::<>::<>::<>::<>::<
Manuela Huso
Consulting Statistician
201H Richardson Hall
Department of Forest Ecosystems and Society
Oregon State University
Corvallis, OR   97331
ph: 541-737-6232
fx: 541-737-1393


-----Original Message-----
From: r-help-bounces at r-project.org [mailto:r-help-bounces at r-project.org] On Behalf Of Ted Harding
Sent: Monday, July 12, 2010 9:36 AM
To: r-help at r-project.org
Cc: Raghu
Subject: Re: [R] in continuation with the earlier R puzzle

On 12-Jul-10 14:09:30, Raghu wrote:
> When I just run a for loop it works. But if I am going to
> run a for loop every time for large vectors I might as well
> use C or any other language.
> The reason R is powerful is becasue it can handle large vectors
> without each element being manipulated? Please let me know where
> I am wrong.
> 
> for(i in 1:length(news1o)){
> + if(news1o[i]>s2o[i])
> + s[i]<-1
> + else
> + s[i]<--1
> + }
> 
> -- 
> 'Raghu'

Many operations over the whole length of vectors can be done
in "vectorised" form, in which an entire vector is changed
in one operation based on the values of the separate elemnts
of other vectors, also all take into account in a single
operation. What happens "behind to scenes" is that the single
element by element operations are performed by a function
in a precompiled (usually from C) library. Hence R already
does what you are suggesting as a "might as well" alternative!

Below is an example, using long vectors. The first case is a
copy of your R loop above (with some additional initialisation
of the vectors). The second achieves the same result in the
"vectorised" form.

  news1o <- runif(1000000)
  s2o    <- runif(1000000)
  s      <- numeric(length(news1o))

  proc.time()
  #    user  system elapsed 
  #   1.728   0.680 450.257 
  for(i in 1:length(news1o)){  ### Using a loop
    if(news1o[i]>s2o[i])
    s[i]<-   1
    else
    s[i]<- (-1)
  }
  proc.time()
  #    user  system elapsed
  #  11.184   0.756 460.340 
  s2 <- 2*(news1o > s2o) - 1   ### Vectorised
  proc.time()
  #    user  system elapsed 
  #  11.348   0.852 460.663

  sum(s2 != s)
  # [1] 0                      ### Results identical

Result: The loop took (11.184 -  1.728) = 9.456 seconds,
  Vectorised, it took (11.348 - 11.184) = 0.164 seconds.

Loop/Vector = (11.184 - 1.728)/(11.348 - 11.184) = 57.65854

i.e. nearly 60 times as long.

Ted.

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E-Mail: (Ted Harding) <Ted.Harding at manchester.ac.uk>
Fax-to-email: +44 (0)870 094 0861
Date: 12-Jul-10                                       Time: 17:36:07
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