Why is apply() method slower than a for loop in R?

Question

As a matter of best practices, I\'m trying to determine if it\'s better to create a function and apply() it across a matrix, or if it\'s better to simply loop a

Answer 1

As Chase said: Use the power of vectorization. You're comparing two bad solutions here.

To clarify why your apply solution is slower:

Within the for loop, you actually use the vectorized indices of the matrix, meaning there is no conversion of type going on. I'm going a bit rough over it here, but basically the internal calculation kind of ignores the dimensions. They're just kept as an attribute and returned with the vector representing the matrix. To illustrate :

> x <- 1:10
> attr(x,"dim") <- c(5,2)
> y <- matrix(1:10,ncol=2)
> all.equal(x,y)
[1] TRUE

Now, when you use the apply, the matrix is split up internally in 100,000 row vectors, every row vector (i.e. a single number) is put through the function, and in the end the result is combined into an appropriate form. The apply function reckons a vector is best in this case, and thus has to concatenate the results of all rows. This takes time.

Also the sapply function first uses as.vector(unlist(...)) to convert anything to a vector, and in the end tries to simplify the answer into a suitable form. Also this takes time, hence also the sapply might be slower here. Yet, it's not on my machine.

IF apply would be a solution here (and it isn't), you could compare :

> system.time(loop_million <- mash(million))
   user  system elapsed 
   0.75    0.00    0.75    
> system.time(sapply_million <- matrix(unlist(sapply(million,squish,simplify=F))))
   user  system elapsed 
   0.25    0.00    0.25 
> system.time(sapply2_million <- matrix(sapply(million,squish)))
   user  system elapsed 
   0.34    0.00    0.34 
> all.equal(loop_million,sapply_million)
[1] TRUE
> all.equal(loop_million,sapply2_million)
[1] TRUE