does white space slow down processing

Question

I have huge amounts of data to analyze, I tend to leave space between words or variable names as I write my code, So the question is, incases where efficiency is the number 1 priority, does the white space have cost?

Is c<-a+b more efficient than c <- a + b

Answer 1

To a first, second, third, ..., approximation, no, it won't cost you any time at all.

The extra time you spend pressing the space bar is orders of magnitude more costly than the cost at run time (and neither matter at all).

The much more significant cost will come from any any decreased readability that results from leaving out spaces, which can make code harder (for humans) to parse.

Answer 2

In a word, no!

library(microbenchmark)

f1 <- function(x){
    j   <- rnorm( x , mean = 0 , sd = 1 )         ;
    k   <-      j    *      2         ;
    return(    k     )
}

f2 <- function(x){j<-rnorm(x,mean=0,sd=1);k<-j*2;return(k)}


microbenchmark( f1(1e3) , f2(1e3) , times= 1e3 )
    Unit: microseconds
     expr     min       lq  median      uq      max neval
 f1(1000) 110.763 112.8430 113.554 114.319  677.996  1000
 f2(1000) 110.386 112.6755 113.416 114.151 5717.811  1000

#Even more runs and longer sampling
microbenchmark( f1(1e4) , f2(1e4) , times= 1e4 )
  Unit: milliseconds
      expr      min       lq   median       uq       max neval
 f1(10000) 1.060010 1.074880 1.079174 1.083414 66.791782 10000
 f2(10000) 1.058773 1.074186 1.078485 1.082866  7.491616 10000

EDIT

It seems like using microbenchmark would be unfair because the expressions are parsed before ever they are run in the loop. However using source should mean that with each iteration the sourced code must be parsed and whitespace removed. So I saved the functions to two seperate files, with the last line of the file being a call of the function, e.g.so my file f2.R looks like this:

f2 <- function(x){j<-rnorm(x,mean=0,sd=1);k<-j*2;return(k)};f2(1e3)

And I test them like so:

microbenchmark( eval(source("~/Desktop/f2.R")) ,  eval(source("~/Desktop/f1.R")) , times = 1e3)
  Unit: microseconds
                           expr     min       lq   median      uq       max neval
 eval(source("~/Desktop/f2.R")) 649.786 658.6225 663.6485 671.772  7025.662  1000
 eval(source("~/Desktop/f1.R")) 687.023 697.2890 702.2315 710.111 19014.116  1000

And a visual representation of the difference with 1e4 replications....

Maybe it does make a minuscule difference in the situation where functions are repeatedly parsed but this wouldn't happen in normal use cases.