9

我试图通过总和（查询时间）计算平均查询时间，然后将它们除以计数。我怎样才能得到计数？

var querytimeByMonthGroup = moveMonths.group().reduceSum(function (d) {
    return d.querytime;
});

var querytimeByMonthGroup = moveMonths.group().reduceSum(function (d) {
    return d.querytime / d.count; ???
});
4

2 回答 2

11

我认为更好的（也是预期的）方法是定义自己的 reduce 函数（添加、删除、初始）。然后，您可以将运行总和、计数等存储在 reduce 函数中，并在过滤器从组中添加和删除数据时适当地调整它们。

在这个类似的问题中给出了使用平均值和最小值和最大值执行此操作的示例：使用交叉过滤器，是否可以在分组时跟踪最大值/最小值？

于 2013-07-19T11:09:45.410 回答
2

我不熟悉交叉过滤器，只是刚刚开始使用它。可能有更好的方法，但这提供了一种计算用于分组的维度计数的方法（我不是 100% 清楚 d.count 是指用于分组的维度的计数，使用如果需要，另一个分组）。

示例源自以下代码：https ://github.com/square/crossfilter/wiki/API-Reference

var payments = crossfilter([
    {date: "2011-11-14T16:17:54Z", quantity: 2, total: 190, tip: 100, type: "tab"},
    {date: "2011-11-14T16:20:19Z", quantity: 2, total: 190, tip: 100, type: "tab"},
    {date: "2011-11-14T16:28:54Z", quantity: 1, total: 300, tip: 200, type: "visa"},
    {date: "2011-11-14T16:30:43Z", quantity: 2, total: 90, tip: 0, type: "tab"},
    {date: "2011-11-14T16:48:46Z", quantity: 2, total: 90, tip: 0, type: "tab"},
    {date: "2011-11-14T16:53:41Z", quantity: 2, total: 90, tip: 0, type: "tab"},
    {date: "2011-11-14T16:54:06Z", quantity: 1, total: 100, tip: 0, type: "cash"},
    {date: "2011-11-14T16:58:03Z", quantity: 2, total: 90, tip: 0, type: "tab"},
    {date: "2011-11-14T17:07:21Z", quantity: 2, total: 90, tip: 0, type: "tab"},
    {date: "2011-11-14T17:22:59Z", quantity: 2, total: 90, tip: 0, type: "tab"},
    {date: "2011-11-14T17:25:45Z", quantity: 2, total: 200, tip: 0, type: "cash"},
    {date: "2011-11-14T17:29:52Z", quantity: 1, total: 200, tip: 100, type: "visa"}
]);

var paymentsByType = payments.dimension(function(d) { return d.type; }),
        paymentVolumeByType = paymentsByType.group(),
        counts = paymentVolumeByType.reduceCount().all(),
        countByType = {}; 

// what is returned by all is a pseudo-array. An object that behaves like an array. 
// Trick to make it a proper array
Array.prototype.slice.call(counts).forEach(function(d) { countByType[d.key] = d.value; })
var paymentVolumeByType = paymentVolumeByType.reduceSum(function(d, i) { 
    console.log(d.total, d.type, countByType[d.type])
    return d.total / countByType[d.type]; 
});
// accessing parentVolumeByType to cause the reduceSum function to be called
var topTypes = paymentVolumeByType.top(1);
于 2013-03-22T00:35:37.207 回答