这是由于s3查找该方法，然后需要对mean.default中的参数进行解析. (以及其他代码)

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和length都是基本函数.这样会很快(但是您如何处理NA值?)

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

mean的内部位甚至比sum/length还要快.

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

请参见 http://rwiki.sciviews. org/doku.php?id = packages:cran:data.table#method_dispatch_takes_time (

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).

请注意，如果我们增加向量的长度，那么原始方法是最快的

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

现在方法分派只是所需的总时间"的一小部分.

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