0
t1 <- Sys.time()
mean((rnorm(10000))^2)
t2 <- Sys.time() 
print(t2-t1)
print(" ")

t1 <- Sys.time()
mean((rnorm(10000))^2)
t2 <- Sys.time() 
print(difftime(t2, t1, units = "secs")[[1]])

I want to compare the time efficiency of a few algorithms for computing the same target, so I tried the two ways above to extract the time difference computed by Sys.time(). However, neither gives a clear numeric.

[1] 0.9998752
Time difference of 0.03889418 secs
[1] " "
[1] 0.9832738
[1] 0.05183697

I also tried proc.time(). It would be great to extract the 3 numeric values into a vector, but none of as.numeric(t), t[0], t['user'], and t[['user']] works. Those are some relevant solutions I found online. How can I get one (or three, either is fine) neat figure from the timing result?

t1 <- proc.time()
mean((rnorm(10000))^2)
t2 <- proc.time() 
t <- t2 - t1
print(" ")
print(t)


[1] " "
   user  system elapsed 
   0.00    0.02    0.17

Is there an equivalent way in R to do what the code below does in Python?

import numpy as np
from time import process_time

t = process_time()
np.mean(np.random.normal(loc=0,scale=1,size=10000))
t = process_time() - t
print(t)
Paw in Data
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    `as.numeric` has a method for `difftime` objects - `as.numeric(t2 - t1, units="secs")`. Specify the units you want and away you go! – thelatemail Sep 22 '20 at 21:23
  • @thelatemail Perhaps I didn't get what you mean. I tried `as.numeric(t2-t1, units='secs')` and `as.numeric(difftime(t2, t1, units='secs'))`, and neither of them yields a neat figure. – Paw in Data Sep 22 '20 at 21:33
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    Works for me with `Sys.time` results - `t1 <- Sys.time(); Sys.sleep(3); t2 <- Sys.time(); as.numeric(t2 - t1, units="secs")` – thelatemail Sep 22 '20 at 21:38
  • @thelatemail Thank you, but I'm afraid it doesn't work for me. It outputs two unnamed figures: [1] 0.9842054 [1] 0.04188514 – Paw in Data Sep 22 '20 at 21:42
  • Those figures are literally impossible given the code I provided, as I forced a 3 second wait between `t1` and `t2` being computed. – thelatemail Sep 22 '20 at 21:46
  • @thelatemail Right... and yet it doesn't apply to what I need to do I'm afraid. Thank you anyway! – Paw in Data Sep 22 '20 at 21:50
  • `x <- difftime(t2, t1, units = "secs")[[1]]` . Isn't `x` the numeric value generated from the time difference? – Ronak Shah Sep 23 '20 at 00:32

1 Answers1

2

You could use system.time():

system.time({mean((rnorm(1e7))^2)})

       user      system       total 
       0.65        0.00        0.67

or package tictoc:

library(tictoc)
tic()
mean((rnorm(1e7))^2)
#> [1] 0.9998728
toc()
#> 0.66 sec elapsed

For better precision, another alternative is microbenchmark which allows to compare different implementations by running them many times :

microbenchmark::microbenchmark( 
  solution_A ={mean((rnorm(1e4))^2)},
  solution_B ={
    mysum <- 0
    for (i in 1:1e4) {
      mysum <- mysum + rnorm(1)^2
    }
    mysum / 1e4
  }
)

Unit: microseconds
       expr     min       lq      mean   median      uq      max neval cld
 solution_A   557.4   570.20   595.785   589.65   597.5   1161.0   100  a 
 solution_B 16177.3 16918.95 22115.115 17347.85 19315.5 247916.7   100   b

for more details, see this link.

Waldi
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