Just for stats I want to know which nation has its dominance or maximum user on physics.stackexhange.

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    $\begingroup$ How do you think that is possible to tell, given that users are not required to enter their country at all? $\endgroup$
    – ACuriousMind Mod
    Commented Feb 5, 2016 at 18:55
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    $\begingroup$ may be those who tell can be counted. Also there are IP addresses of people which can be tracked down using some softwares. $\endgroup$ Commented Feb 5, 2016 at 19:02
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    $\begingroup$ This really isn't a question about the site, but you might be able to scrub some data over at data.stackexchange.com $\endgroup$
    – Kyle Kanos
    Commented Feb 5, 2016 at 19:09
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    $\begingroup$ @Kyle Kanos It shows invalid URL. $\endgroup$ Commented Feb 5, 2016 at 19:11
  • $\begingroup$ I've edited it... If it doesn't work, try using Google you search "data stack exchange" and it sounds be the first link. $\endgroup$
    – Kyle Kanos
    Commented Feb 5, 2016 at 19:15
  • $\begingroup$ @Kyle Kanos Got it. but site shows no such data. Should I query it there. $\endgroup$ Commented Feb 5, 2016 at 19:19
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    $\begingroup$ That site is the Stack Exchange Data Explorer. It is where you go to write a SQL script to parse what data is stored & publicly available. You may try the search feature to see if someone has written such a script already. $\endgroup$
    – Kyle Kanos
    Commented Feb 5, 2016 at 19:22
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    $\begingroup$ @Kyle Kanos Can we use simple mysql commands we learn in school there. Language they use is not of my level. $\endgroup$ Commented Feb 5, 2016 at 19:28
  • $\begingroup$ When I click Run query it shows. {"captcha":true} only $\endgroup$ Commented Feb 5, 2016 at 19:37
  • $\begingroup$ @Kyle Kanos: Thanks got it. 86489 from India. $\endgroup$ Commented Feb 5, 2016 at 20:04

1 Answer 1


There is a limited amount of site analytics available to high-rep users but this does not include any geographical information. Geographical data on visitors (not users) is presumably available to the SE team but it is not made available through that channel; you'd have to ask the SE team directly to get it.

Self-declared user location data is available through the Data Explorer, but as e.g. this query shows, being user-provided it is a huge mixed bag and you'd need some processing to make statistical sense of it. This query makes things a bit easier to handle, and the corresponding code described here can help draw some inferences if that's where you're headed.

On a similar track, StackMaps can help provide a geographical view (but not quite a statistical one) on this information.

Generally, though, this information is hard to come by and to handle. Visitor geographic data is not available, and user geographic data is self-reported, so for one you should take its accuracy with a grain of salt, and there is an enormous selection bias you need to take into account: 48k of 105k users have not provided any location, and some fraction have provided fictional location. Moreover, it is at least likely that whether a user reports their location is correlated with the location itself (so users from e.g. Africa might feel that declaring their location is not advantageous, so that location gets (severely?) under-represented in the self-reported data).

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    $\begingroup$ In query just add count(location) where location like '%India%' or '%Germany%' or France or which so ever country you want $\endgroup$ Commented Feb 6, 2016 at 18:16
  • $\begingroup$ Mmmmhmmm. But it requires lots of case-by-case handling of the multiple variants. If you want to do it yourself, the data is there for you to download and process - no one is going to do it for you. And, as I said, any results from that analysis are hugely suspect, to the point of complete uselessness, by the selection bias. $\endgroup$ Commented Feb 6, 2016 at 18:22
  • $\begingroup$ I had made some queries thanks to you and Kyle. $\endgroup$ Commented Feb 6, 2016 at 18:25
  • $\begingroup$ why do you think that users from e.g. Africa might feel that declaring their location is not advantageous ? $\endgroup$
    – user46925
    Commented Feb 23, 2016 at 2:15

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