给定一个对象数组,如何通过深度为n的属性对对象进行排序

Cod*_*ate 6 javascript sorting

我有一个对象数组,我希望通过一些分组属性数据和一个字符串告诉我要分组的属性(例如:'Organization'或'Organization.Name')

我需要编写一个函数,它接收看起来像beforeData并返回的数据afterData

输入:

beforeData = [
{'name':'John Doe', 'Id':1, 'Organizations':[{'Id':12, 'LongName': 'Group A'},{'Id':13, 'LongName': 'Group B'}]},
{'name':'FooBar', 'Id':2, 'Organizations':[{'Id':13, 'LongName': 'Group B'},{'Id':14, 'LongName': 'Group C'}]},
{'name':'Kristine Bell', 'Id':3, 'Organizations':[{'Id':12, 'LongName': 'Group A'}]},
{'name':'Adrian P', 'Id':4, 'Organizations':[{'Id':12, 'LongName': 'Group A'}]}
]
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输出:

    afterData = [
    {   
        'Group': 'Group A', 
        'entities':[
            {'name':'Adrian P', 'Id':4, 'Organizations':[{'Id':12, 'LongName': 'Group A'}]},
            {'name':'Kristine Bell', 'Id':3, 'Organizations':[{'Id':12, 'LongName': 'Group A'}]},
            {'name':'John Doe', 'Id':1, 'Organizations':[{'Id':12, 'LongName': 'Group A'},{'Id':13, 'LongName': 'Group B'}]}]
    },
    {   
        'Group': 'Group B', 
        'entities':[
            {'name':'John Doe', 'Id':1, 'Organizations':[{'Id':12, 'LongName': 'Group A'},{'Id':13, 'LongName': 'Group B'}]},
            {'name':'FooBar', 'Id':2, 'Organizations':[{'Id':13, 'LongName': 'Group B'},{'Id':13, 'LongName': 'Group C'}]},]
    },
    {   
        'Group': 'Group C', 
        'entities':[
            {'name':'FooBar', 'Id':2, 'Organizations':[{'Id':13, 'LongName': 'Group B'},{'Id':13, 'LongName': 'Group C'}]},]
    }
]
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我将如何实现这一目标?我目前的尝试非常臃肿,并且在给定大量数据的情况下永远都是如此.

特别踢球者!:解决此问题的函数需要能够在不事先知道" 属性分组 "是否为深度1或2(例如:'Organization'或'Organization.LongName')的情况下解决它.

zer*_*kms 4

我的一些事情:

// this function performs data extraction from an object
// the first argument is a name of the property to be extracted
// it might be just a 1st level deep value like `name`
// or nested like `foo.bar.baz`
// in case if one of intermediate items is an array - an array of
// results is returned
function dot(name, obj) {
    if (!name) {
        return obj;
    }

    var match = name.match(/^([^.]+)(?:\.(.*))?$/),
        head = match[1],
        tail = match[2];

    if (Array.isArray(obj)) {
        return obj.map(function(item) {
            return dot(name, item);
        });
    }

    if (obj === null || typeof obj != 'object') {
        return null;
    }

    return dot(tail, obj[head]);
}

// this function accepts an array of data and a key to group by
// as a result it returns an object with keys equal to a group by key
// and values that hold that key
function groupBy(data, key) {
    return data.reduce(function(result, item) {
        var keys = dot(key, item);
        if (!Array.isArray(keys)) {
            keys = [keys];
        }

        keys.forEach(function(key) {
            if (!(key in result)) {
                result[key] = [];
            }

            result[key].push(item);
        });

        return result;
    }, {});
}

console.log(groupBy(beforeData, 'Organizations.LongName'));
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JSFiddle: http: //jsfiddle.net/w8N4j/

现在可以轻松地将其重新格式化为您想要的任何其他格式。

例如,要从这里的问题中获取确切的格式,需要一个微型转换器:

function transformerExample(hash) {
    var result = [];
    for (var key in hash) if (hash.hasOwnProperty(key)) {
        result.push({
            Group: key,
            entities: hash[key]
        });
    }

    return result;
}
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PS:主要实现显然可能无法处理所有可能的错误。根据实际需求进行改进并不难。