JavaScript 数组 : how to group

Question

我有一些数据未排序的数组，我想创建新的数组宽度排序数据:

   var gridData = [], i = -1;
        while (i++ < 5) {
                gridData[i] = {
                    "ProjectOwner": ["Corporation1","Corporation2","Corporation3"][i % 2],
                    "ProjectId": i,
                    "ProjectName": String.fromCharCode("A".charCodeAt(0) + i % 3) + (1 + i % 3),
                    "ProjectType": ["Conference","Programming","Research","Conference"][i % 4],
                    "ProjectLeader": "Name1Name2Name3".substr(i % 3 * 5,5) + " " + "Surname1Surname2Surname3".substr(i % 3 * 8,8),
                    "StartDate": new Date(2009 + i % 4 % 2, (i + 3) % 12, 5 + i * 4 % 24).toLocaleDateString(),
                    "EndDate": new Date(2010 + i % 4 % 2, (i + 3) % 12, 5 + i * 4 % 24).toLocaleDateString(),
                    "Summary": "bla bla"
                };
            }

未排序的数据看起来像对象数组

我想把它分组

//First group by ProjectOwner
Object {Corporation1: Array[3], Corporation2: Array[3]}
   //Second group by ProjectName
   Corporation1: Array[x]
            A1 : Array [x]
                 Object1
                 Object2
                 ...
   Corporation2: Array[x]

等等... 我试过 array.map 和 reduce 但它们似乎在 IE 中不起作用 我也试过:

$.each(aggregation_content, function(i,aggr) {

    $.each(_newData, function(a, objA) {
        if(i > 0)
            o = _newData[a][aggregation_content[i - 1].cell];
        n = _newData[a][aggregation_content[i].cell];

        if(typeof o == "undefined") //For first row when new Data is empty array
        {
            if(! (n in newData))
                newData[n] = [];
            newData[n].push(objA);
        }
    })})
console.log(newData);

aggregation_content 在哪里

{cell:"ProjectOwner",value:""},{cell:"ProjectName",value:""},{cell:"ProjectLeader",value:""},{cell:"ProjectType",value:""}

新数据从一开始就是一个gridData。 这非常适合第一次聚合。但问题是当我聚合我需要使用的数组 newData 时

if(! (n in newData[o]))
    newData[o][n] = [];
 newData[o][n].push(objA);

并且 [o] 应该是 [n] 节点的父节点。 好的-第二组也可以使用该代码，但是当我想创建 5 个内部组时，我需要像

newData["firstGroup"]["secondGroup"]["thirdGroup"]...[n].push("Some content").

这如何以编程方式实现？如果我这样做

newData = newData[o] - no good
or
temp = newData
do something to temp
newData[o] = temp not good eather :'(

我希望我写的是可以理解的文字:D

____Edited_2012_12_13__________________________________________

所以输入数据是

[Object, Object, Object, Object, Object, Object]
  0: Object
   EndDate: "Monday, April 05, 2010"
   ProjectId: 0
   ProjectLeader: "Name1 Surname1"
   ProjectName: "A1"
   ProjectOwner: "Corporation1"
   ProjectType: "Conference"
   StartDate: "Sunday, April 05, 2009"
   Summary: "bla bla"
   __proto__: Object
 1: Object
   EndDate: "Monday, May 09, 2011"
   ProjectId: 1
   ProjectLeader: "Name2 Surname2"
   ProjectName: "B2"
   ProjectOwner: "Corporation2"
   ProjectType: "Programming"
   StartDate: "Sunday, May 09, 2010"
   Summary: "bla bla"
   __proto__: Object
 2: Object
   ...
 3: Object
    ...
 4: Object
    ...
 5: Object
    ....

输出数据应该是这样的

Object {Corporation1: Array[3], Corporation2: Array[3]}
 Corporation1: Array[2]
   0: A1:  Array[X]
         0: Object
            EndDate: "Monday, May 09, 2011"
            ProjectId: 1
            ProjectLeader: "Name2 Surname2"
            ProjectName: "A1"
            ProjectOwner: "Corporation2"
            ProjectType: "Programming"
            StartDate: "Sunday, May 09, 2010"
            Summary: "bla bla"
         1: Object
            ...
            ...
   1: B2:  Array[X]
         0: Object
            EndDate: "Monday, May 09, 2011"
            ProjectId: 1
            ProjectLeader: "Name2 Surname2"
            ProjectName: "B2"
            ProjectOwner: "Corporation2"
            ProjectType: "Programming"
            StartDate: "Sunday, May 09, 2010"
            Summary: "bla bla"
         1: Object
            ...
            ...
 Corporation2: Array[2]
   0: A1:  Array[X]
         0: Object
            EndDate: "Monday, May 09, 2011"
            ProjectId: 1
            ProjectLeader: "Name2 Surname2"
            ProjectName: "A1"
            ProjectOwner: "Corporation2"
            ProjectType: "Programming"
            StartDate: "Sunday, May 09, 2010"
            Summary: "bla bla"
         1: Object
            ...
            ...
   1: B2:  Array[X]
         0: Object
            EndDate: "Monday, May 09, 2011"
            ProjectId: 1
            ProjectLeader: "Name2 Surname2"
            ProjectName: "B2"
            ProjectOwner: "Corporation2"
            ProjectType: "Programming"
            StartDate: "Sunday, May 09, 2010"
            Summary: "bla bla"
         1: Object
            ...
            ...

TreeView 应该是两个或多个嵌套节点。

Answer 1

我之前也遇到过类似的情况。我写了一个库来做到这一点:

https://github.com/raghavv/array-mod

请检查。我不是在宣传我的图书馆，但我认为它符合您的要求。

此外，对于旧浏览器支持 IE < 9。您还需要包含另一个优秀的库:

https://github.com/kriskowal/es5-shim/

这里是你如何得到你的第一组数组:

a$(gridData).findAll("Corporation1", "ProjectOwner");
a$(gridData).findAll("Corporation2", "ProjectOwner");
a$(gridData).findAll("A1", "ProjectName");
a$(gridData).findAll("B2", "ProjectName");