JavaScript：将对象中的数据分组 [英] Javascript: group data in objects

问题描述

我正在尝试找出用Javascript组织此数据的最佳方法，因此，如果这个问题比平时更广泛，我深表歉意。

I'm trying to figure out the best way to organize this data in Javascript, so I apologize if this question is broader than usual.

我有数据：

State, Zip
"CA", "945"
"CA", "934"
"MA", "934"
"MA", "021"
"MA", "021"
 etc.

我正在尝试创建某种Javascript数组，可以引用该Javascript数组以获取所有 Zip 某个州

I am trying to create some sort of Javascript Array that I can reference to get all the Zip in a certain State

此引用可能类似于：

State: "CA", Zip:["945", "934"]
State: "MA", Zip:["934", "021", "021"]

这有意义吗？

我想这样做是因为在极少数情况下3位数字 Zip 将存在于两个 State s中（如上例所示）。

I want to do it this way because on rare occasion a 3-digit Zip would exist in two States (as in the example above).

我正在使用下面的代码，但是它不起作用。

I was playing with the below code, but it does not work.

var stateArray = [],
var zipMapper = [];

for (var i = 0; i < data.length; i++){
    if (_.contains(stateArray,data[i].state) == false) {
        stateArray.push(data[i].state)
        zipMapper.push({state: data[i].state, zips: []})
        zipMapper[data[i].state].zips.push(data[zip])
    } else {
        zipMapper[data[i].state].zips.push(data[zip])
    }
}

推荐答案

我建议将数据结构更改为

I suggest to change the data structure to version with fast access to the state and their Zips.

{
    "CA": [
        "945",
        "934"
    ],
    "MA": [
        "934",
        "021",
        "021"
    ]
}

要从 CA ，只需采用 grouped.CA 或 grouped ['CA'] 。

您建议解决方案始终需要进行一次交互，直到找到状态的数据。

The solution, you suggested, need always an interation until the data for a state is found. It makes access slow.

var data = [{ State: "CA", Zip: "945" }, { State: "CA", Zip: "934" }, { State: "MA", Zip: "934" }, { State: "MA", Zip: "021" }, { State: "MA", Zip: "021" }],
    grouped = function (array) {
        var r = {};
        array.forEach(function (a) {
            r[a.State] = r[a.State] || [];
            r[a.State].push(a.Zip);
        });
        return r;
    }(data);


document.write('<pre>' + JSON.stringify(grouped, 0, 4) + '</pre>');

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