On 09/07/10 21:19, Duncan Murdoch wrote:
On 09/07/2010 4:11 PM, Gene Leynes wrote:
I thought the "apply" functions are faster than for loops, but my most
recent test shows that apply actually takes a significantly longer
than a
for loop. Am I missing something?
Probably not. apply() needs to figure out the shape of the results it
gets from each row in order to put them into the final result matrix.
You know that in advance, and set up the result to hold them, so your
calculation would be more efficient.
Plus, in the way it is set up, apply has to make a much larger temporary
matrix. If you do
ds1<- 1+ds
library("rbenchmark")
benchmark(apply=apply(1+ds, 1, cumprod),
forloop=for (i in 1:nrow(ds)){ y2[i,]<-cumprod(1+ds[i,]) },
apply2=apply(ds1, 1, cumprod),
replications=1, columns=c("test","elapsed","relative","user.self","sys.self"),
order="elapsed")
# test elapsed relative user.self sys.self
# 2 forloop 1.863 1.000000 1.861 0.000
# 3 apply2 2.175 1.167472 1.934 0.239
# 1 apply 2.443 1.311326 2.108 0.334
you can sense some of the impact of that.
But if it is speed you are after, sapply may be even faster (you'll need
to t() the result again):
benchmark(forloop=for (i in 1:nrow(ds)){ y2[i,]<-cumprod(1+ds[i,]) },
sapply={dsone<-1+ds;sapply(1:NROW(dsone), function(i)
cumprod(dsone[i,]))},
replications=1, columns=c("test","elapsed","relative","user.self","sys.self"),
order="elapsed")
# test elapsed relative user.self sys.self
# 2 sapply 1.539 1.000000 1.300 0.239
# 1 forloop 1.878 1.220273 1.878 0.000
zz<- sapply(1:NROW(ds1), function(i) cumprod(dsone[i,]))
identical(t(zz), y2)
# [1] TRUE
Hope this helps a little.
Allan
The *apply functions are designed to be convenient and clear to read,
not necessarily fast.
Duncan Murdoch
It doesn't matter much if I do column wise calculations rather than
row wise
## Example of how apply is SLOWER than for loop:
#rm(list=ls())
## DEFINE VARIABLES
mu=0.05 ; sigma=0.20 ; dt=.25 ; T=50 ; sims=1e5
timesteps = T/dt
## MAKE PHI AND DS
phi = matrix(rnorm(timesteps*sims), nrow=sims, ncol=timesteps)
ds = mu*dt + sigma * sqrt(dt) * phi
## USE APPLY TO CALCULATE ROWWISE CUMULATIVE PRODUCT
system.time(y1 <- apply(1+ds, 1, cumprod))
## UNTRANSFORM Y1, BECAUSE ROW APPLY FLIPS THE MATRIX
y1=t(y1)
## USE FOR LOOP TO CALCULATE ROWWISE CUMULATIVE PRODUCT
y2=matrix(NA,nrow(ds),ncol(ds))
system.time(
for (i in 1:nrow(ds)){
y2[i,]<-cumprod(1+ds[i,])
}
)
## COMPARE RESULTS TO MAKE SURE THEY DID THE SAME THING
str(y1)
str(y2)
all(y1==y2)
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