是否可以在crossfilter中按多个维度进行分组？

Question

例如,如果我们有书籍,作者和日期信息的数据.我们可以为每月为作者提供多少本书构建一个交叉过滤器吗？

Answer 1

在伪sql术语中,您要做的是:

SELECT COUNT(book)
GROUP BY author, month

我处理这类问题的方法是将字段"组合"为一个维度.因此,在您的情况下,我会将月份和作者信息连接在一起.

让这成为我们的测试数据:

var cf = crossfilter([
{ date:"1 jan 2014", author: "Mr X", book: "Book 1" },
{ date:"2 jan 2014", author: "Mr X", book: "Book 2" },
{ date:"3 feb 2014", author: "Mr X", book: "Book 3" },
{ date:"1 mar 2014", author: "Mr X", book: "Book 4" },
{ date:"2 apr 2014", author: "Mr X", book: "Book 5" },
{ date:"3 apr 2014", author: "Mr X", book: "Book 6"},
{ date:"1 jan 2014", author: "Ms Y", book: "Book 7" },
{ date:"2 jan 2014", author: "Ms Y", book: "Book 8" },
{ date:"3 jan 2014", author: "Ms Y", book: "Book 9" },
{ date:"1 mar 2014", author: "Ms Y", book: "Book 10" },
{ date:"2 mar 2014", author: "Ms Y", book: "Book 11" },
{ date:"3 mar 2014", author: "Ms Y", book: "Book 12" },
{ date:"4 apr 2014", author: "Ms Y", book: "Book 13" }
]);  

维度定义如下:

var dimensionMonthAuthor = cf.dimension(function (d) {
  var thisDate = new Date(d.date);
  return 'month='+thisDate.getMonth()+';author='+d.author;
});

现在我们可以简单地做一个减少计数来计算每个作者每月有多少本书(即每个维度单位):

var monthAuthorCount = dimensionMonthAuthor.group().reduceCount(function (d) { return d.book; }).all();

结果如下:

{"key":"month=0;author=Mr X","value":2}
{"key":"month=0;author=Ms Y","value":3}
{"key":"month=1;author=Mr X","value":1}
{"key":"month=2;author=Mr X","value":1}
{"key":"month=2;author=Ms Y","value":3}
{"key":"month=3;author=Mr X","value":2}
{"key":"month=3;author=Ms Y","value":1}

Answer 2

我没有找到所有有用的答案.

我使用以下代替.

我第一次做了一个关键组(在你的情况下是一个月)

   var authors = cf.dimension(function (d) {
     return +d['month'];
   })

接下来,我在键控数据集上使用了map reduce方法来计算平均值

分组辅助函数:

var monthsAvg = authors.group().reduce(reduceAddbooks, reduceRemovebooks, reduceInitialbooks).all();

map-reduce函数:

function reduceAddbooks(p, v) {
    p.author = v['author'];
    p.books = +v['books'];
    return p;
}

function reduceRemovebooks(p, v) {
    p.author = v['author'];
    p.books = +v['books'];
    return p;
}

function reduceInitialbooks() {
    return {
        author:0,
        books:0
    };
}

Answer 3

我想用以下描述的新工作更新旧答案：https : //github.com/dc-js/dc.js/pull/91

此性能尚未在大型数据集上进行测试

  var cf = crossfilter([
  { date:"1 jan 2014", author: "Mr X", book: "Book 1" },
  { date:"2 jan 2014", author: "Mr X", book: "Book 2" },
  { date:"3 feb 2014", author: "Mr X", book: "Book 3" },
  { date:"1 mar 2014", author: "Mr X", book: "Book 4" },
  { date:"2 apr 2014", author: "Mr X", book: "Book 5" },
  { date:"3 apr 2014", author: "Mr X", book: "Book 6"},
  { date:"1 jan 2014", author: "Ms Y", book: "Book 7" },
  { date:"2 jan 2014", author: "Ms Y", book: "Book 8" },
  { date:"3 jan 2014", author: "Ms Y", book: "Book 9" },
  { date:"1 mar 2014", author: "Ms Y", book: "Book 10" },
  { date:"2 mar 2014", author: "Ms Y", book: "Book 11" },
  { date:"3 mar 2014", author: "Ms Y", book: "Book 12" },
  { date:"4 apr 2014", author: "Ms Y", book: "Book 13" }
  ]);

  var dimensionMonthAuthor = cf.dimension(function (d) {
    var thisDate = new Date(d.date);
    //stringify() and later, parse() to get keyed objects
    return JSON.stringify ( { date: thisDate.getMonth() , author: d.author } ) ;
  });

  group = dimensionMonthAuthor.group();
  //this forEach method could be very expensive on write.
  group.all().forEach(function(d) {
    //parse the json string created above
    d.key = JSON.parse(d.key);
  });

  return group.all()

结果是：

[ { key: { date: 0, author: 'Mr X' },
    value: 2 },
  { key: { date: 0, author: 'Ms Y' },
    value: 3 },
  { key: { date: 1, author: 'Mr X' },
    value: 1 },
  { key: { date: 2, author: 'Mr X' },
    value: 1 },
  { key: { date: 2, author: 'Ms Y' },
    value: 3 },
  { key: { date: 3, author: 'Mr X' },
    value: 2 },
  { key: { date: 3, author: 'Ms Y' },
    value: 1 } ]