Timing benefits of mapply() vs. for loop was: Wrap a loop inside a function

like.mat <- function(score, items, theta){

system.time(for(i in 1:100) like.mat(score,items,theta))

like.mat2 <- function(score, items, theta)

system.time(for(i in 1:100) like.mat2(score,items,theta))

all.equal(like.mat(score, items, theta), like.mat2(score, items, theta))

like.mat3 <- function(score, items, theta)

system.time(for(i in 1:100) like.mat3(score,items,theta))

m3 <- like.mat3(score, items, theta)
dimnames(m3) <- NULL
all.equal(like.mat(score, items, theta), m3)

List:

Thank you for the replies to my post yesterday. Gabor and Phil also gave
useful replies on how to improve the function by relying on mapply rather
than the explicit for loop. In general, I try and use the family of apply
functions rather than the looping constructs such as for, while etc as a
matter of practice.

However, it seems the mapply function in this case is slower (in terms of
CPU speed) than the for loop. Here is an example.

# data needed for example

items <- list(item1 = c(0,1,2), item2 = c(0,1), item3 = c(0,1,2,3,4), item4
= c(0,1), item5=c(0,1,2,3,4),
item6=c(0,1,2,3))
score <- c(2,1,3,1,3,2)
theta <- c(-1,-.5,0,.5,1)

# My old function using the for loop

like.mat <- function(score, items, theta){
   like.mat <- matrix(numeric(length(items) * length(theta)), ncol =
length(theta))
   for(i in 1:length(items)) like.mat[i, ] <- pcm(theta, items[[i]],
score[[i]])
   like.mat
   }

system.time(like.mat(score,items,theta))
[1]  0  0  0 NA NA

# Revised using mapply

like.mat <- function(score, items, theta){
matrix(mapply(pcm,rep(theta,length(items)),items,score),ncol=length(theta),byrow=TRUE)
}

system.time(like.mat(score,items,theta))

[1] 0.03 0.00 0.03   NA   NA


It is obviously slower to use mapply, but nominaly. So, let's actually look
at this within the context of the full program I am working on. For context,
I am evaluating an integral using Gaussian quadrature. This is a
psychometric problem where the function 'pcm" is Master's partial credit
model and 'score' is the student's score on test item i. When an item has
two categories (0,1), pcm reduces to the Rasch model for dichotomous data.
The dnorm is set at N(0,1) by default, but the parameters of the population
distribution are estimated from a separate procedure and are normally input
into the function, but this default works for the example.

Here is the full program.

library(statmod)

# Master's partial credit model
pcm <- function(theta,d,score){
     exp(rowSums(outer(theta,d[1:score],'-')))/
     apply(exp(apply(outer(theta,d, '-'), 1, cumsum)), 2, sum)
   }

like.mat <- function(score, items, theta){
   like.mat <- matrix(numeric(length(items) * length(theta)), ncol =
length(theta))
   for(i in 1:length(items)) like.mat[i, ] <- pcm(theta, items[[i]],
score[[i]])
   like.mat
   }

# turn this off for now
#like.mat <- function(score, items, theta){
#matrix(mapply(pcm,rep(theta,length(items)),items,score),ncol=length(theta),byrow=TRUE)
#}

class.numer <- function(score,items, prof_cut, mu=0, sigma=1, aboveQ){
   gauss_numer <- gauss.quad(49,kind="laguerre")
   if(aboveQ==FALSE){
      mat <- rbind(like.mat(score,items, (prof_cut-gauss_numer$nodes)),
dnorm(prof_cut-gauss_numer$nodes, mean=mu, sd=sigma))

      } else { mat <- rbind(like.mat(score,items,
(gauss_numer$nodes+prof_cut)),
dnorm(gauss_numer$nodes+prof_cut, mean=mu,
sd=sigma))

   }
   f_y <- rbind(apply(mat, 2, prod), exp(gauss_numer$nodes),
gauss_numer$weights)
   sum(apply(f_y,2,prod))
   }

class.denom <- function(score,items, mu=0, sigma=1){
   gauss_denom <- gauss.quad.prob(49, dist='normal', mu=mu, sigma=sigma)
   mat <-
rbind(like.mat(score,items,gauss_denom$nodes),gauss_denom$weights)
   sum(apply(mat, 2, prod))
   }

class.acc <-function(score,items,prof_cut, mu=0, sigma=1,
aboveQ=TRUE){
   result <- class.numer(score,items,prof_cut, mu,sigma,
aboveQ)/class.denom(score,items, mu, sigma)
   return(result)
   }

# Test the function "class.acc"
items <- list(item1 = c(0,1,2), item2 = c(0,1), item3 = c(0,1,2,3,4), item4
= c(0,1), item5=c(0,1,2,3,4),
item6=c(0,1,2,3))
score <- c(2,1,3,1,3,2)

# This is the system time when using the for loop for the like.mat function
system.time(class.acc(score,items,1,aboveQ=T))
[1] 0.04 0.00 0.04   NA   NA

# This is the system time using the mapply for the like.mat function
system.time(class.acc(score,items,1,aboveQ=T))
[1] 0.70 0.00 0.73   NA   NA


There is a substantial improvement in CPU seconds when the for loop is
applied rather than using the mapply function. I experimented with adding
more items and varying the quadrature points and it always turned out the
for loop was faster.

Given this result, I wonder what advice might be offered. Is there an
inherent reason one might opt for the mapply function (such as reliability,
etc) even when it compromises computational speed? Or, should the issue of
computational speed be considered ahead of common advice to rely on the
family of apply functions instead of the explicit loops.

Thanks for your consideration of my question,
Harold

orm       i386-pc-mingw32
arch           i386
os             mingw32
system         i386, mingw32
status
major          2
minor          3.0
year           2006
month          04
day            24
svn rev        37909
language       R
version.string Version 2.3.0 (2006-04-24)

Note that if you use mapply in the way I suggested, which is not
the same as in your post, then its just as fast.  (Also the version
of mapply in your post gives different numerical results than
the for loop whereas mine gives the same.)   like.mat is the for
loop version, like.mat2 is your mapply version and like.mat3
is my mapply version.

like.mat <- function(score, items, theta){

+   like.mat <- matrix(numeric(length(items) * length(theta)), ncol =
+     length(theta))
+   for(i in 1:length(items)) like.mat[i, ] <- pcm(theta, items[[i]],
score[[i]])
+   like.mat
+ }

system.time(for(i in 1:100) like.mat(score,items,theta))

[1] 1.30 0.00 1.34   NA   NA

like.mat2 <- function(score, items, theta)

+   matrix(mapply(pcm, rep(theta,length(items)), items, score),
+     ncol = length(theta), byrow = TRUE)

system.time(for(i in 1:100) like.mat2(score,items,theta))

[1] 5.70 0.00 5.91   NA   NA

all.equal(like.mat(score, items, theta), like.mat2(score, items, theta))

[1] "Mean relative  difference: 1.268095"

like.mat3 <- function(score, items, theta)

+ t(mapply(pcm, items, score, MoreArgs = list(theta = theta)))

system.time(for(i in 1:100) like.mat3(score,items,theta))

[1] 1.32 0.01 1.39   NA   NA

m3 <- like.mat3(score, items, theta)
dimnames(m3) <- NULL
all.equal(like.mat(score, items, theta), m3)

[1] TRUE

On 7/20/06, Doran, Harold <HDoran at air.org> wrote:

List:

Thank you for the replies to my post yesterday. Gabor and Phil also gave
useful replies on how to improve the function by relying on mapply rather
than the explicit for loop. In general, I try and use the family of apply
functions rather than the looping constructs such as for, while etc as a
matter of practice.

However, it seems the mapply function in this case is slower (in terms of
CPU speed) than the for loop. Here is an example.

# data needed for example

items <- list(item1 = c(0,1,2), item2 = c(0,1), item3 = c(0,1,2,3,4), item4
= c(0,1), item5=c(0,1,2,3,4),
item6=c(0,1,2,3))
score <- c(2,1,3,1,3,2)
theta <- c(-1,-.5,0,.5,1)

# My old function using the for loop

like.mat <- function(score, items, theta){
   like.mat <- matrix(numeric(length(items) * length(theta)), ncol =
length(theta))
   for(i in 1:length(items)) like.mat[i, ] <- pcm(theta, items[[i]],
score[[i]])
   like.mat
   }

system.time(like.mat(score,items,theta))
[1]  0  0  0 NA NA

# Revised using mapply

like.mat <- function(score, items, theta){
matrix(mapply(pcm,rep(theta,length(items)),items,score),ncol=length(theta),byrow=TRUE)
}

system.time(like.mat(score,items,theta))

[1] 0.03 0.00 0.03   NA   NA


It is obviously slower to use mapply, but nominaly. So, let's actually look
at this within the context of the full program I am working on. For context,
I am evaluating an integral using Gaussian quadrature. This is a
psychometric problem where the function 'pcm" is Master's partial credit
model and 'score' is the student's score on test item i. When an item has
two categories (0,1), pcm reduces to the Rasch model for dichotomous data.
The dnorm is set at N(0,1) by default, but the parameters of the population
distribution are estimated from a separate procedure and are normally input
into the function, but this default works for the example.

Here is the full program.

library(statmod)

# Master's partial credit model
pcm <- function(theta,d,score){
     exp(rowSums(outer(theta,d[1:score],'-')))/
     apply(exp(apply(outer(theta,d, '-'), 1, cumsum)), 2, sum)
   }

like.mat <- function(score, items, theta){
   like.mat <- matrix(numeric(length(items) * length(theta)), ncol =
length(theta))
   for(i in 1:length(items)) like.mat[i, ] <- pcm(theta, items[[i]],
score[[i]])
   like.mat
   }

# turn this off for now
#like.mat <- function(score, items, theta){
#matrix(mapply(pcm,rep(theta,length(items)),items,score),ncol=length(theta),byrow=TRUE)
#}

class.numer <- function(score,items, prof_cut, mu=0, sigma=1, aboveQ){
   gauss_numer <- gauss.quad(49,kind="laguerre")
   if(aboveQ==FALSE){
      mat <- rbind(like.mat(score,items, (prof_cut-gauss_numer$nodes)),
dnorm(prof_cut-gauss_numer$nodes, mean=mu, sd=sigma))

      } else { mat <- rbind(like.mat(score,items,
(gauss_numer$nodes+prof_cut)),
dnorm(gauss_numer$nodes+prof_cut, mean=mu,
sd=sigma))

   }
   f_y <- rbind(apply(mat, 2, prod), exp(gauss_numer$nodes),
gauss_numer$weights)
   sum(apply(f_y,2,prod))
   }

class.denom <- function(score,items, mu=0, sigma=1){
   gauss_denom <- gauss.quad.prob(49, dist='normal', mu=mu, sigma=sigma)
   mat <-
rbind(like.mat(score,items,gauss_denom$nodes),gauss_denom$weights)
   sum(apply(mat, 2, prod))
   }

class.acc <-function(score,items,prof_cut, mu=0, sigma=1,
aboveQ=TRUE){
   result <- class.numer(score,items,prof_cut, mu,sigma,
aboveQ)/class.denom(score,items, mu, sigma)
   return(result)
   }

# Test the function "class.acc"
items <- list(item1 = c(0,1,2), item2 = c(0,1), item3 = c(0,1,2,3,4), item4
= c(0,1), item5=c(0,1,2,3,4),
item6=c(0,1,2,3))
score <- c(2,1,3,1,3,2)

# This is the system time when using the for loop for the like.mat function
system.time(class.acc(score,items,1,aboveQ=T))
[1] 0.04 0.00 0.04   NA   NA

# This is the system time using the mapply for the like.mat function
system.time(class.acc(score,items,1,aboveQ=T))
[1] 0.70 0.00 0.73   NA   NA


There is a substantial improvement in CPU seconds when the for loop is
applied rather than using the mapply function. I experimented with adding
more items and varying the quadrature points and it always turned out the
for loop was faster.

Given this result, I wonder what advice might be offered. Is there an
inherent reason one might opt for the mapply function (such as reliability,
etc) even when it compromises computational speed? Or, should the issue of
computational speed be considered ahead of common advice to rely on the
family of apply functions instead of the explicit loops.

Thanks for your consideration of my question,
Harold