So I built an SE service connection and as a piece of pure curiosity I wanted to see if I could determine what a given user's time-zone was by the times which they answer.

I managed to pull in all the answers using my service connection.

And then I grouped the users and the hours of the day (in Unix time) when they answered and then I got stuck.

Because I can tell you when Mr. Wizard answers questions on average:

In[82]:= Mean@userTimes["Mr.Wizard"] // N

Out[82]= 11.6578

But this alone isn't enough to tell me what his(?) time-zone is (although per his/her profile it's PST: mr.wizard time).

And I can give you an EstimatedDistribution (using NormalDistribution) of Kuba's answer times:

Plot[PDF[userDistributions["Kuba"], x], {x, -5, 30}]

kuba dist

But I don't know how to connect that to his time zone.

I know Mathematica should be able to give me this, maybe by comparing both the peak and the height in the distribution:

 PDF[#, x] & /@ Take[userDistributions, 3] // Values // Evaluate,
 {x, -5, 30},
 PlotLabels -> Keys@Take[userDistributions, 3]

top-three dist

But this is simply not something I know enough about.

So can someone crack the code? As I have it set-up I suppose this breaks down to a statistical argument about how likely it is that a given user has a given time-zone, but is there a way to do this better than just that (if I even knew how to do that)?

  • $\begingroup$ I stumbled onto this today. Some ideas: (1) I am male. (2) I think many people often post at a couple of times each day (e.g. before and after work), so I would try a bimodal distribution for each profile to see if it fits better. (3) You should consider my data a pathological case (unfortunately, it is) – stackoverflow.com/a/5845444/618728 $\endgroup$
    – Mr.Wizard
    Jan 11, 2019 at 13:08

1 Answer 1


So I ended up trying to use Classify on this. First I pulled in all of the users.

Then I found the ones where I could get a property city or admin. div. Entity:

$stateMap =
     ] &@DeleteDuplicates@Normal@users[All, "location"];

$cityMap =
     ] &@Keys@Select[$stateMap, FailureQ];

$locMap = 
  Join[Select[$cityMap, Not@*FailureQ], 
   Select[$stateMap, Not@*FailureQ]];

userLocs = 
    First@# -> (Last@# /. $locMap) & /@ 
     Normal@users[All, {"display_name", "location"}]] // Dataset;

testableUsers = Select[userLocs, MatchQ[_Entity]];

Then I calculated the shifts from UTC for these users:

calcedShifts = <||>;

calcShit[ent_] :=
 Lookup[calcedShifts, ent,
  calcedShifts[ent] =
      QuantityMagnitude[#["OffsetFromUTC"], "Hours"]
      ] &@
    If[EntityTypeName@ent === "AdministrativeDivision",

userShifts = calcShit /@ Normal@testableUsers;

And then I built a classifier mapping the distribution parameters to the time-zone shift:

trainingSet =
      Normal@Keys[userShifts]] ->
   _Missing -> _

classifier = Classify[trainingSet];

timeZoneGuess[user_] :=

 classifier[userDistParams[user], "Probabilities"] // ReverseSort // 
timeZoneGuess[users : {__String}] :=

  Dataset@AssociationMap[Normal@timeZoneGuess[#] &, users];

And then we test:

  timeZoneGuess@Keys@Take[userDistributions, 15] // Normal

Out[349]= <|"Mr.Wizard" -> -5., "Michael E2" -> -5., 
 "m_goldberg" -> -5., "corey979" -> -5., "Szabolcs" -> 2., 
 "kglr" -> -5., "Bob Hanlon" -> -5., "ubpdqn" -> 2., "Kuba" -> 2., 
 "J. M." -> -5., "Carl Woll" -> -5., "george2079" -> -5., "zhk" -> 2.,
  "bill s" -> -5., "David G. Stork" -> -5.|>

And we find that it isn't so great... (obviously this is just the most probable time-zone guess from a fit to a NormalDistribution. Using something more sophisticated might help.)

But it is certainly a start. And it does seem to classify America vs. Europe fine.

  • $\begingroup$ You have "circular" data. Perhaps you should consider using the Von Mises distribution, which is appropriate for such data. If it were me, I might go farther and compute non-parametric density estimates for each user, using the Von Mises distribution as the kernel. $\endgroup$
    – mef
    Aug 14, 2017 at 17:39
  • $\begingroup$ @mef interesting suggestion. I'll give it a try when I have time. $\endgroup$
    – b3m2a1
    Aug 18, 2017 at 21:03

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