Take the 2-minute tour ×
Mathematica Stack Exchange is a question and answer site for users of Mathematica. It's 100% free, no registration required.

In the process of simplifying code posted for another question, I noticed AbsoluteTiming returned different results for the same code being timed, Range[0, 99], when the surrounding code was different. It was constantly different.

What is going on? Is Map caching and Do-AppendTo not? Is the call to Range[0, 99] close enough to to cost to retrieve the value from the cache so the sum of timing is only off by ~0.01?

I have changed the order, Map then Do-AppendTo, and still the Map is faster by ~0.01.

I also tried substituting Range[0, Mod[i++, 10000]] to negate the caching effect, same results. I tried timing different code and noticed the same difference.


iterations = 10000;
times1 = {};
Do[AppendTo[times1, AbsoluteTiming[
     Range[0, 99]
     ][[1]]]
  , {iterations}];

times2 = Map[
   AbsoluteTiming[
      Range[0, 99]
      ][[1]]
    &, Range[1, iterations]];


PrintStats[times_] :=
 Print["Sum: ", Fold[Plus, 0., times], "  Min: ", Min[times], 
  "  Max: ", Max[times], "  Mean: ", Mean[times], "  StdDev: ", 
  StandardDeviation[times]]

PrintStats[times1];
ListPlot[times1, PlotRange -> All]
Histogram[times1]

PrintStats[times2];
ListPlot[times2, PlotRange -> All]
Histogram[times2]

Summary Results of Five Runs:

Do-AppendTo:

Sum: 0.042083  Min: 3.*10^-6  Max: 0.000029  Mean: 4.*10^-6  StdDev: 0.*10^-7
Sum: 0.039829  Min: 2.*10^-6  Max: 0.000019  Mean: 4.*10^-6  StdDev: 0.*10^-7
Sum: 0.041112  Min: 2.*10^-6  Max: 0.000030  Mean: 4.*10^-6  StdDev: 0.*10^-7
Sum: 0.042261  Min: 0.  Max: 0.000043  Mean: 4.2261*10^-6  StdDev: 1.07875*10^-6
Sum: 0.040704  Min: 3.*10^-6  Max: 0.000025  Mean: 4.*10^-6  StdDev: 0.*10^-7

Map:

Sum: 0.030643  Min: 2.*10^-6  Max: 0.000012  Mean: 3.*10^-6  StdDev: 0.*10^-7
Sum: 0.030974  Min: 2.*10^-6  Max: 0.000014  Mean: 3.*10^-6  StdDev: 0.*10^-7
Sum: 0.030379  Min: 0.  Max: 0.000012  Mean: 3.0379*10^-6  StdDev: 5.08615*10^-7
Sum: 0.030504  Min: 2.*10^-6  Max: 0.000012  Mean: 3.*10^-6  StdDev: 0.*10^-7
Sum: 0.030612  Min: 2.*10^-6  Max: 0.000012  Mean: 3.*10^-6  StdDev: 0.*10^-7

Detailed Results of First Run:

enter image description here

share|improve this question
2  
You are measuring in the microsecond range, just above the resolution of the timer of your system (you get discrete jumps). I wonder how accurate and how practically relevant such short timings are. It would be interesting to know what exactly causes the difference, but I have doubts about whether this difference is reproducible at practically relevant times. (Can you amplify the times and still show the difference?) (Absolute)Timing cannot return such short times on a Windows system, so I can't play with it. The Win7 resolution is 1 ms, the WinXP one is 15 ms. –  Szabolcs Feb 24 '12 at 20:28
    
@Szabolcs Can you amplify the times and still show the difference? Sure, do you have a suggestion on what I should use? I do have a theory, which is that since the difference is so small, increasing the time might drown out the difference. I agree with your thoughts that it might just be running up against the resolution of my machine. –  mmorris Feb 24 '12 at 22:33
    
I mean, can you provide a code sample which will show the difference on some clearly measurable time scale? (E.g. one would run in 1.1 seconds, the other in 0.8) If this is not possible, that's good news: then we don't need to worry about this when optimizing code. –  Szabolcs Feb 26 '12 at 16:00
2  
@mmorris I did not put any effort into this. If you accept your own, that is very fine. Please just do that. –  Szabolcs Mar 9 '12 at 8:03
1  
@mmorris As far as I know, you don't get rep for accepting your own answer, so it wouldn't be a method of gaming the system. If you feel your answer provides a correct answer to your question and is the best of the set (which it is if it is the only one) feel free to accept it. I would allow for a day or two, just to see if any better answers get in. –  Sjoerd C. de Vries Mar 9 '12 at 12:16
show 4 more comments

1 Answer

up vote 4 down vote accepted

As Szabolcs pointed out, I was running into the resolution of the timer in my system in my attempts to micro benchmark. Also, my attempts at micro benchmarking were incorrect I might add, but more on that later.

I was able to see the light only after scratching my head for a number of days on Szabolcs' comment and working on a performance package, I finally came across $TimeUnit here. I got the idea to use Pause[10*$TimeUnit] as the code being tested in each case. Clearly Pause[10*$TimeUnit] is above the resolution of the timer in my system. Oh, in case you are curious, $TimeUnit is 1/100 on my system (2011 MBP 17" running Lion).

For those of you who want to play along at home, the following is the updated micro benchmarking code: (btw it is still incorrect).

PrintHeader[] :=
  Print["        Total    Mean     Min      Max      StdDev "];

PrintStats[doTimes_List , mapTimes_List] :=
 (
  Print["Do    ", "  ", doTotal = Total[doTimes], "  ", 
   doMean = Mean[doTimes], "  ", Min[doTimes], "  ", Max[doTimes], 
   "  ", StandardDeviation[doTimes]];
  Print["Map   ", "  ", mapTotal = Total[mapTimes], "  ", 
   mapMean = Mean[mapTimes] "  ", Min[mapTimes], "  ", Max[mapTimes], 
   "  ", StandardDeviation[mapTimes]];
  Print["Do-Map", "  ", doTotal - mapTotal, "  ", doMean - mapMean, 
   "\n"];
  )

PrintHeader[];
iterations = 1000;
Do[(
   doTimes = {};
   Do[AppendTo[doTimes,
     AbsoluteTiming[Pause[10*$TimeUnit]][[1]]]
    , {iterations}];

   mapTimes = Map[
     AbsoluteTiming[Pause[10*$TimeUnit]][[1]]
      &, Range[1, iterations]];

   PrintStats[doTimes, mapTimes];
   ), {5}];

Results

        Total    Mean     Min      Max      StdDev 
Do      100.339  0.100339  0.100027  0.101277  0.000330
Map     100.343  0.100343  0.100035  0.101294  0.000330
Do-Map  -0.004  -4.*10^-6

Do      100.351  0.100351  0.100036  0.101276  0.000327
Map     100.338  0.100338  0.100035  0.101247  0.000330
Do-Map  0.013  0.000013

Do      100.328  0.100328  0.100025  0.101328  0.000319
Map     100.346  0.100346  0.100027  0.101280  0.000332
Do-Map  -0.018  -0.000018

Do      100.347  0.100347  0.100029  0.101238  0.000330
Map     100.332  0.100332  0.100021  0.101235  0.000338
Do-Map  0.015  0.000015

Do      100.330  0.100330  0.100023  0.101258  0.000318
Map     100.327  0.100327  0.100028  0.101238  0.000324
Do-Map  0.003  3.*10^-6

The results are as expected, there is no longer a consistent difference between the Do an Map implementations.

Incorrect Micro Benchmarking Method

So why are were my two implementations of micro benchmarking above incorrect? I wanted to capture timing values for each call of the code being tested over a large number of iterations so that I could perform various analytics on them (sum, mean, min, max, std dev, plotting, histogram, ...).

I thought AbsoluteTiming was incredibly precise. When I evaluated the simple expression 1, AbsoluteTiming[1] returned {0.000014, 1}. 0.000014 is way more precise than $TimeUnit on my system which was 0.01 right? Wrong!

Lets look at two different ways of micro benching marking the expression 1 for 1000000 iterations. But first, lets establish a base case to help us understand the results.


Base Case

iterations = 1000000;
oneTime = AbsoluteTiming[1][[1]];
expectation = oneTime*iterations;

Print["oneTime: ", oneTime, "   expectation for ", iterations, ": ", expectation];

Results:

oneTime: 0.000014   expectation for 1000000: 14.

Collection of Individual AbsoluteTiming Values

individualTimes = Map[AbsoluteTiming[1][[1]] &, Range[1, iterations]];

Print["Total: ", Total[individualTimes], "  Min: ", 
 Min[individualTimes], "  Max: ", Max[individualTimes], "  Mean: ", 
 Mean[individualTimes], "  StdDev: ", 
 StandardDeviation[individualTimes]];

Results:

Total: 2.26236  Min: 0.  Max: 0.000167  Mean: 2.26236*10^-6  StdDev: 6.79631*10^-7

First thing we notice is that the Total(2.26236) is less than our expectation(14.) . Why the difference? Well first look at the Min(0.) and Max(0.000167), they are out side our oneTime(0.000014), which is good. If there are a significant number of individual times less than our oneTime value, that could explain the difference. It just so happened that Length[Select[individualTimes, # < oneTime & ]] returned 999939 .

Anyone know how Min can be 0. ? Is it because the expression was cached? If it was cached, it still would have taken some time. Maybe the time it took is less than what AbsoluteTiming can observe, hmmmm.


One AbsoluteTiming Value for All Iterations

totalTime = AbsoluteTiming[Do [1, {iterations}]][[1]];

Print["totalTime: ", totalTime];

Results:

totalTime: 0.015782

Woe wait a minute, the totalTime(0.015782) is no where close to the Total of the individualTimes(2.26236) not to mention our expectation(14.)! I must have messed something up. Okay let me think about this, could it be AbsoluteTime is not as precise as I originally thought? Ding ding, ding, give that man a cigar!

As the chaps at Wolfram said it here. AbsoluteTiming does not have the precision past $TimeUnit. So the oneTime and all of the individually collected timing values were prone to error as a result of the resolution of the timer of the system. Thanks again Szabolcs!

The the one timing value for all iterations also has the same error, but only since it is only called one time for 1000000 iterations and is less than $TimeUnit (0.01) it has a much smaller effect.

Okay, yeah I know, I should rerun the One AbsoluteTiming Value for All Iterations with iterations = 1000000000 to get a more accurate value. totalTime: 12.440098 which makes making one call to 1 take 1.2440098*10^-8 seconds, which is a much much much smaller number than $TimeUnit. Poor AbsoluteTiming, it never stood a chance. ;)

share|improve this answer
1  
Also note that AbsoluteTiming gives the total time between the beginning and end of the calculation, including the time when your code wasn't running at all because the processor was working on something else (like checking whether something on your screen needed updating, deciding what to do with some packet which happened to arrive on your network, your virus checker running in the background etc.). So even if your system would allow reliable microsecond timing, you'd still get quite varying times, depending on whether there happened to be an interrupt or even a task switch in between. –  celtschk Mar 9 '12 at 8:21
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.