MeteorJS和UnderscoreJS:将结果分组以在Google Map上绘制点 [英] MeteorJS and UnderscoreJS: grouping results to plot points on a Google Map

问题描述

我正在开发一个小型MeteorJS应用程序，该应用程序根据受欢迎的工作区域在地图上绘制点.

I am working on a small MeteorJS app that plots points on a map based on popular areas for work.

我有这个:

Template.list.jobs = function() {
  if(Session.get('currentIndustryOnet')) {
    jobs = Jobs.find({onet: Session.get('currentIndustryOnet')}).fetch();
    // Session.set('jobCount', jobs.count());

    var cnt = _.groupBy(jobs, 'address');
    console.log(cnt);

    return Pagination.collection(jobs);
  } else {
    jobs = Jobs.find()
    Session.set('jobCount', jobs.count());
    return Pagination.collection(jobs.fetch());
  }
}

cnt 变量返回一个正确分组的数组(该数组的键是一个地址，例如 Allentown，PA ).我收集了一些美国曾经有过的城市及其LAT/LONG，可以在Google地图上进行绘制.因此，我将从分组数组中获得前100名，在Cities集合中找到经度/纬度，然后在地图上绘制这些点.

The cnt variable returns a properly grouped array (the key of the array is an address like Allentown, PA). I have a collection of Cities which have ever city in the USA along with their LAT/LONGs to plot on a Google Map. So I will take the top 100 from the grouped array, find the lat/long in the Cities collection and plot those points on a map.

我不熟悉使用 groupedBy 方法根据长度对列表进行排序，然后拉出用作我的搜索键的方法.

I am not familiar with working with a groupedBy method to sort the list based on the length and then pull out the key to use as my search.

推荐答案

我不确定100％的数据结构...但是假设作业具有 address 字段，而您想要它们按出现频率排序并设置上限，您可以执行以下操作:

I'm not 100% certain about the data structure... but assuming jobs have an address field, and you want them sorted by frequency of occurrence and capped, you could do something like this:

var addresses = _.chain(jobs)
  .countBy('address')
  .pairs()
  .sortBy(function(j) {return -j[1];})
  .map(function(j) {return j[0];})
  .first(100)
  .value();

请注意，使用下划线可能会有更巧妙的方法来得出此结果.一旦获得了有上限的地址列表，您就可以通过类似以下的查找来获取经度/纬度值:

Note there may be a more clever way to use underscore to arrive at this result. Once you have the capped, sorted list of addresses, you can probably get the lat/long values via a find like:

Cities.find({address: {$in: addresses}}).fetch();

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