我正在使用以 GeoJSON 作为输入的 javascript 地图来可视化年度英国电影放映。你可以在这里看到它适用于 2011 年的数据:http: //screened2011.herokuapp.com
GeoJSON 的生成效率非常低 - 每个“图块”通常需要 5 秒以上。
我有一个 Ruby 应用程序,它在 MongoDB 中查询边界框内的一组“筛选”(由 JS 请求),然后生成一个二维数组,表示在每个 16x16 中发生的筛选总数。这显然是瓶颈——它正在攻击服务器并拉下所有这些放映。
我想用一个 map/reduce 查询替换它,该查询将边界框中的所有筛选计数聚合到一个 16x16 的值数组中,但我没有取得太大的成功——这是我第一个 map/reduce 的任务!
这是我的代码的简化版本,其中删除了不相关的内容(这很糟糕,如果这不是即将结束的黑客攻击,我会重构):
get :boxed, :map => "/screenings/year/:year/quadrant/:area/bbox/:bbox", :provides => [:json, :jsonp], :cache => false do
box = parameter_box(params[:bbox]) # returns [[minx,miny],[maxx,maxy]]
year = params[:year].to_i
screenings = Screening.where(:location.within(:box) => box).where(:year => year)
jsonp Screening.heatmap(screenings, box, 16)
end
def self.heatmap screenings, box, scale
grid = []
min_x = box[0][0]
min_y = box[0][1]
max_x = box[1][0]
max_y = box[1][1]
box_width = max_x.to_f - min_x.to_f
box_height = max_y.to_f - min_y.to_f
return [] if box_width == 0 || box_height == 0
# Set up an empty GeoJSON-style array to hold the results
scalef = scale.to_f
(0..(scale - 1)).each do |i|
grid[i] = []
(0..(scale - 1)).each do |j|
box_min_x = min_x + (i * ( box_width / scalef ))
box_max_x = min_x + ((i + 1) * ( box_width / scalef ))
box_min_y = min_y + (j * ( box_height / scalef ))
box_max_y = min_y + ((j + 1) * ( box_height / scalef ))
grid[i][j] = {
:count => 0,
#:id => "#{box_min_x}_#{box_max_x}_#{box_min_y}_#{box_max_y}",
:coordinates => [
[
[box_min_x,box_min_y], [box_min_x, box_max_y], [box_max_x, box_max_y], [box_max_x, box_min_y], [box_min_x,box_min_y]
]
]
}
end
end
# This loop is the bottleneck and I'd like to replace with a map-reduce
screenings.only(:location, :total_showings).each do |screening|
x = (scale * ((screening.location[0] - min_x) / box_width)).floor
y = (scale * ((screening.location[1] - min_y) / box_height)).floor
raise if x > (scale - 1)
raise if y > (scale - 1)
grid[x][y][:count] += screening.total_showings.to_i
end
# This converts the resulting 16x16 into GeoJSON
places = []
grid.each do |x|
x.each do |p|
if p[:count].to_i > 0
properties = {}
properties[:total_showings] = p[:count]
places << {
"id" => p[:id],
"type" => "Feature",
"geometry" => {
"type" => "Polygon",
"coordinates" => p[:coordinates]
},
"properties"=> properties
}
end
end
end
{
"type" => "FeatureCollection",
"features" => places
}
end
我正在使用 Mongoid,所以我可以将 mapreduce 链接到筛选查询,我希望这会大大加快这个过程 - 但是我应该如何让类似下面的东西传递给这个函数呢?:
[
[1,20000,30,3424,53,66,7586,54543,76764,4322,7664,43242,43,435,32,643],
...
]
...基于此结构中的几百万条记录(基本上是对边界框内的每个记录求和):
{"_id"=>BSON::ObjectId('50e481e653e6dfbc92057e8d'),
"created_at"=>2013-01-02 18:52:22 +0000,
"ended_at"=>Thu, 07 Jun 2012 00:00:00 +0100,
"events"=>["10044735484"],
"film_id"=>BSON::ObjectId('4f96a91153e6df5ebc001afe'),
"genre_ids"=>[],
"location"=>[-2.003309596016, 52.396317185921],
"performance_id"=>"9001923080",
"specialised"=>false,
"started_at"=>Fri, 01 Jun 2012 00:00:00 +0100,
"total_showings"=>1,
"updated_at"=>2013-01-02 18:52:22 +0000,
"venue_id"=>BSON::ObjectId('4f9500bf53e6df004000034d'),
"week"=>nil,
"year"=>2012}
先谢谢各位了!