Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
408 views
in Technique[技术] by (71.8m points)

r - Why is mean() so slow?

Everything is in the question! I just tried to do a bit of optimization, and nailing down the bottle necks, out of curiosity, I tried that:

t1 <- rnorm(10)
microbenchmark(
  mean(t1),
  sum(t1)/length(t1),
  times = 10000)

and the result is that mean() is 6+ times slower than the computation "by hand"!

Does it stem from the overhead in the code of mean() before the call to the Internal(mean) or is it the C code itself which is slower? Why? Is there a good reason and thus a good use case?

See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)

It is due to the s3 look up for the method, and then the necessary parsing of arguments in mean.default. (and also the other code in mean)

sum and length are both Primitive functions. so will be fast (but how are you handling NA values?)

t1 <- rnorm(10)
microbenchmark(
  mean(t1),
  sum(t1)/length(t1),
  mean.default(t1),
  .Internal(mean(t1)),
  times = 10000)

Unit: nanoseconds
                expr   min    lq median    uq     max neval
            mean(t1) 10266 10951  11293 11635 1470714 10000
  sum(t1)/length(t1)   684  1027   1369  1711  104367 10000
    mean.default(t1)  2053  2396   2738  2739 1167195 10000
 .Internal(mean(t1))   342   343    685   685   86574 10000

The internal bit of mean is faster even than sum/length.

See http://rwiki.sciviews.org/doku.php?id=packages:cran:data.table#method_dispatch_takes_time (mirror) for more details (and a data.table solution that avoids .Internal).

Note that if we increase the length of the vector, then the primitive approach is fastest

t1 <- rnorm(1e7)
microbenchmark(
     mean(t1),
     sum(t1)/length(t1),
     mean.default(t1),
     .Internal(mean(t1)),
+     times = 100)

Unit: milliseconds
                expr      min       lq   median       uq      max neval
            mean(t1) 25.79873 26.39242 26.56608 26.85523 33.36137   100
  sum(t1)/length(t1) 15.02399 15.22948 15.31383 15.43239 19.20824   100
    mean.default(t1) 25.69402 26.21466 26.44683 26.84257 33.62896   100
 .Internal(mean(t1)) 25.70497 26.16247 26.39396 26.63982 35.21054   100

Now method dispatch is only a fraction of the overall "time" required.


与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...