On 17/02/2011 11:02 AM, Alex F. Bokov wrote:
Motivation: during each iteration, my code needs to collect tabular data (and
use it only during that iteration), but the rows of data may vary. I thought I
would speed it up by preinitializing the matrix that collects the data with
zeros to what I know to be the maximum number of rows. I was surprised by what
I found...
# set up (not the puzzling part)
x<-matrix(runif(20),nrow=4); y<-matrix(0,nrow=12,ncol=5); foo<-c();
# this is what surprises me... what the?
> system.time(for(i in 1:100000){n<-sample(1:4,1);y[1:n,]<-x[1:n,];});
user system elapsed
1.510 0.000 1.514
> system.time(for(i in 1:100000){n<-sample(1:4,1);foo<-x[1:n,];});
user system elapsed
1.090 0.000 1.085
These results are very repeatable. So, if I'm interpreting them correctly,
dynamically allocating 'foo' each time to whatever the current output size is
runs faster than writing to a subset of a preallocated 'y'? How is that
possible?
The expression
y[1:n,]<-x[1:n,]
creates a new temporary variable to hold the result of the expression
x[1:n,], then copies the elements of it to y[1:n,].
The expression
foo <- x[1:n,]
creates the same temporary, and then binds foo to it without doing any
copying. Much less work.
And, more generally, I'm sure other people have encountered this type of
situation. Am I reinventing the wheel? Is there a best practice for storing
temporary loop-specific data?
Storing the value of an expression in a new variable will always be
faster than copying it into part of an existing variable.
Duncan Murdoch
P.S. You might be aware of this, but there's one other thing that might
be a surprise to you: x[1:1,] will be a vector, while x[1:n,] will be
a matrix for n>1. Use the "drop=FALSE" argument if you always want a
matrix result.
Thanks.
PS: By the way, though I cannot write to foo[,] because the size is different
each time, I tried writing to foo[] and the runtime was worse than either of
the above examples.
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