Rank languages by number of users
Use the mediawiki API rather than web scraping since it is much faster and less resource intensive. Show languages with more than 25 users since that is still a pretty short list and to demonstrate how tied rankings are handled. Change the $minimum parameter to adjust what the cut-off point will be.
This is all done in a single pass; ties are not detected until a language has the same count as a previous one, so ties are marked by a T next to the count indicating that this language has the same count as the previous.
use HTTP::UserAgent;
use URI::Escape;
use JSON::Fast;
my $client = HTTP::UserAgent.new;
my $url = 'https://rosettacode.org/w';
my $start-time = now;
say "========= Generated: { DateTime.new(time) } =========";
my $lang = 1;
my $rank = 0;
my $last = 0;
my $tie = ' ';
my $minimum = 25;
.say for
mediawiki-query(
$url, 'pages',
:generator<categorymembers>,
:gcmtitle<Category:Language users>,
:gcmlimit<350>,
:rawcontinue(),
:prop<categoryinfo>
)
.map({ %( count => .<categoryinfo><pages> || 0,
lang => .<title>.subst(/^'Category:' (.+) ' User'/, ->$/ {$0}) ) })
.sort( { -.<count>, .<syntaxhighlight lang="text"> } )
.map( { last if .<count> < $minimum; display(.<count>, .<syntaxhighlight lang="text">) } );
say "========= elapsed: {(now - $start-time).round(.01)} seconds =========";
sub display ($count, $which) {
if $last != $count { $last = $count; $rank = $lang; $tie = ' ' } else { $tie = 'T' };
sprintf "#%3d Rank: %2d %s with %-4s users: %s", $lang++, $rank, $tie, $count, $which;
}
sub mediawiki-query ($site, $type, *%query) {
my $url = "$site/api.php?" ~ uri-query-string(
:action<query>, :format<json>, :formatversion<2>, |%query);
my $continue = '';
gather loop {
my $response = $client.get("$url&$continue");
my $data = from-json($response.content);
take $_ for $data.<query>.{$type}.values;
$continue = uri-query-string |($data.<query-continue>{*}».hash.hash or last);
}
}
sub uri-query-string (*%fields) {
join '&', %fields.map: { "{.key}={uri-escape .value}" }
}Output:
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