我将一些代码放在一起,以平抑和反平抑复杂/嵌套的JavaScript对象。它可以工作,但有点慢(触发“长脚本”警告)。

对于扁平的名称,我希望用“.”作为分隔符,用[INDEX]作为数组。

例子:

un-flattened | flattened
---------------------------
{foo:{bar:false}} => {"foo.bar":false}
{a:[{b:["c","d"]}]} => {"a[0].b[0]":"c","a[0].b[1]":"d"}
[1,[2,[3,4],5],6] => {"[0]":1,"[1].[0]":2,"[1].[1].[0]":3,"[1].[1].[1]":4,"[1].[2]":5,"[2]":6}

我创建了一个基准测试,用于模拟我的用例http://jsfiddle.net/WSzec/

获得一个嵌套对象 压平它 查看它,并可能修改它,而扁平 将其平放回原始的嵌套格式,然后运走

我想要更快的代码:为了澄清,代码完成JSFiddle基准测试(http://jsfiddle.net/WSzec/)显著更快(~20%+会很好)在IE 9+, FF 24+和Chrome 29+。

以下是相关JavaScript代码:当前最快速度:http://jsfiddle.net/WSzec/6/

var unflatten = function(data) {
    "use strict";
    if (Object(data) !== data || Array.isArray(data))
        return data;
    var result = {}, cur, prop, idx, last, temp;
    for(var p in data) {
        cur = result, prop = "", last = 0;
        do {
            idx = p.indexOf(".", last);
            temp = p.substring(last, idx !== -1 ? idx : undefined);
            cur = cur[prop] || (cur[prop] = (!isNaN(parseInt(temp)) ? [] : {}));
            prop = temp;
            last = idx + 1;
        } while(idx >= 0);
        cur[prop] = data[p];
    }
    return result[""];
}
var flatten = function(data) {
    var result = {};
    function recurse (cur, prop) {
        if (Object(cur) !== cur) {
            result[prop] = cur;
        } else if (Array.isArray(cur)) {
             for(var i=0, l=cur.length; i<l; i++)
                 recurse(cur[i], prop ? prop+"."+i : ""+i);
            if (l == 0)
                result[prop] = [];
        } else {
            var isEmpty = true;
            for (var p in cur) {
                isEmpty = false;
                recurse(cur[p], prop ? prop+"."+p : p);
            }
            if (isEmpty)
                result[prop] = {};
        }
    }
    recurse(data, "");
    return result;
}

编辑1修改了上面的@Bergi的实现,这是目前最快的。顺便说一句,使用“。”用indexOf代替正则表达式。exec”在FF中快了20%左右,但在Chrome中慢了20%;所以我将坚持使用正则表达式,因为它更简单(这里是我尝试使用indexOf来取代正则表达式http://jsfiddle.net/WSzec/2/)。

在@Bergi的想法的基础上,我设法创建了一个更快的非正则表达式版本(在FF快3倍,在Chrome快10%)。http://jsfiddle.net/WSzec/6/在这个(当前)实现中,密钥名称的规则很简单,密钥不能以整数开头或包含句点。

例子:

{"foo":{"bar":[0]}} => {"foo.bar.0":0}

EDIT 3添加@AaditMShah的内联路径解析方法(而不是String.split)有助于提高unflatten性能。我对整体性能的提升非常满意。

最新版本的jsfiddle和jsperf:

http://jsfiddle.net/WSzec/14/

http://jsperf.com/flatten-un-flatten/4


当前回答

这段代码递归地展开JSON对象。

我在代码中包含了我的计时机制,它给了我1毫秒,但我不确定这是否是最准确的。

            var new_json = [{
              "name": "fatima",
              "age": 25,
              "neighbour": {
                "name": "taqi",
                "location": "end of the street",
                "property": {
                  "built in": 1990,
                  "owned": false,
                  "years on market": [1990, 1998, 2002, 2013],
                  "year short listed": [], //means never
                }
              },
              "town": "Mountain View",
              "state": "CA"
            },
            {
              "name": "qianru",
              "age": 20,
              "neighbour": {
                "name": "joe",
                "location": "opposite to the park",
                "property": {
                  "built in": 2011,
                  "owned": true,
                  "years on market": [1996, 2011],
                  "year short listed": [], //means never
                }
              },
              "town": "Pittsburgh",
              "state": "PA"
            }]

            function flatten(json, flattened, str_key) {
                for (var key in json) {
                  if (json.hasOwnProperty(key)) {
                    if (json[key] instanceof Object && json[key] != "") {
                      flatten(json[key], flattened, str_key + "." + key);
                    } else {
                      flattened[str_key + "." + key] = json[key];
                    }
                  }
                }
            }

        var flattened = {};
        console.time('flatten'); 
        flatten(new_json, flattened, "");
        console.timeEnd('flatten');

        for (var key in flattened){
          console.log(key + ": " + flattened[key]);
        }

输出:

flatten: 1ms
.0.name: fatima
.0.age: 25
.0.neighbour.name: taqi
.0.neighbour.location: end of the street
.0.neighbour.property.built in: 1990
.0.neighbour.property.owned: false
.0.neighbour.property.years on market.0: 1990
.0.neighbour.property.years on market.1: 1998
.0.neighbour.property.years on market.2: 2002
.0.neighbour.property.years on market.3: 2013
.0.neighbour.property.year short listed: 
.0.town: Mountain View
.0.state: CA
.1.name: qianru
.1.age: 20
.1.neighbour.name: joe
.1.neighbour.location: opposite to the park
.1.neighbour.property.built in: 2011
.1.neighbour.property.owned: true
.1.neighbour.property.years on market.0: 1996
.1.neighbour.property.years on market.1: 2011
.1.neighbour.property.year short listed: 
.1.town: Pittsburgh
.1.state: PA

其他回答

三年半后……

对于我自己的项目,我想在mongoDB点表示法中平坦JSON对象,并提出了一个简单的解决方案:

/**
 * Recursively flattens a JSON object using dot notation.
 *
 * NOTE: input must be an object as described by JSON spec. Arbitrary
 * JS objects (e.g. {a: () => 42}) may result in unexpected output.
 * MOREOVER, it removes keys with empty objects/arrays as value (see
 * examples bellow).
 *
 * @example
 * // returns {a:1, 'b.0.c': 2, 'b.0.d.e': 3, 'b.1': 4}
 * flatten({a: 1, b: [{c: 2, d: {e: 3}}, 4]})
 * // returns {a:1, 'b.0.c': 2, 'b.0.d.e.0': true, 'b.0.d.e.1': false, 'b.0.d.e.2.f': 1}
 * flatten({a: 1, b: [{c: 2, d: {e: [true, false, {f: 1}]}}]})
 * // return {a: 1}
 * flatten({a: 1, b: [], c: {}})
 *
 * @param obj item to be flattened
 * @param {Array.string} [prefix=[]] chain of prefix joined with a dot and prepended to key
 * @param {Object} [current={}] result of flatten during the recursion
 *
 * @see https://docs.mongodb.com/manual/core/document/#dot-notation
 */
function flatten (obj, prefix, current) {
  prefix = prefix || []
  current = current || {}

  // Remember kids, null is also an object!
  if (typeof (obj) === 'object' && obj !== null) {
    Object.keys(obj).forEach(key => {
      this.flatten(obj[key], prefix.concat(key), current)
    })
  } else {
    current[prefix.join('.')] = obj
  }

  return current
}

特性和/或注意事项

它只接受JSON对象。因此,如果你传递类似{a:() =>{}}的东西,你可能得不到你想要的! 它删除空数组和对象。所以这个{a: {}, b:[]}被平化为{}。

您可以试用jpflat包。

它平抑、膨胀、解析承诺、平抑数组,具有可定制的路径创建和可定制的值序列化。

简化器和序列化器将整个路径作为其各部分的数组接收,因此可以对路径进行更复杂的操作,而不是修改单个键或更改分隔符。

Json路径是默认的,因此是“jp”flat。

https://www.npmjs.com/package/jpflat

let flatFoo = await require('jpflat').flatten(foo)

下面是我更简短的实现:

Object.unflatten = function(data) {
    "use strict";
    if (Object(data) !== data || Array.isArray(data))
        return data;
    var regex = /\.?([^.\[\]]+)|\[(\d+)\]/g,
        resultholder = {};
    for (var p in data) {
        var cur = resultholder,
            prop = "",
            m;
        while (m = regex.exec(p)) {
            cur = cur[prop] || (cur[prop] = (m[2] ? [] : {}));
            prop = m[2] || m[1];
        }
        cur[prop] = data[p];
    }
    return resultholder[""] || resultholder;
};

flatten没有改变太多(我不确定你是否真的需要这些isEmpty案例):

Object.flatten = function(data) {
    var result = {};
    function recurse (cur, prop) {
        if (Object(cur) !== cur) {
            result[prop] = cur;
        } else if (Array.isArray(cur)) {
             for(var i=0, l=cur.length; i<l; i++)
                 recurse(cur[i], prop + "[" + i + "]");
            if (l == 0)
                result[prop] = [];
        } else {
            var isEmpty = true;
            for (var p in cur) {
                isEmpty = false;
                recurse(cur[p], prop ? prop+"."+p : p);
            }
            if (isEmpty && prop)
                result[prop] = {};
        }
    }
    recurse(data, "");
    return result;
}

它们一起运行你的基准测试只需要一半的时间(Opera 12.16: ~900ms而不是~ 1900ms, Chrome 29: ~800ms而不是~1600ms)。

注意:这个解决方案和这里回答的大多数其他解决方案侧重于速度,容易受到原型污染,不应该用于不可信的对象。

这里有另一个方法,比上面的答案运行得慢(大约1000毫秒),但有一个有趣的想法:-)

它不是遍历每个属性链,而是只选择最后一个属性,并使用一个查找表来存储其余属性,从而存储中间结果。这个查找表将被迭代,直到没有剩余的属性链,并且所有值都驻留在未协同的属性上。

JSON.unflatten = function(data) {
    "use strict";
    if (Object(data) !== data || Array.isArray(data))
        return data;
    var regex = /\.?([^.\[\]]+)$|\[(\d+)\]$/,
        props = Object.keys(data),
        result, p;
    while(p = props.shift()) {
        var m = regex.exec(p),
            target;
        if (m.index) {
            var rest = p.slice(0, m.index);
            if (!(rest in data)) {
                data[rest] = m[2] ? [] : {};
                props.push(rest);
            }
            target = data[rest];
        } else {
            target = result || (result = (m[2] ? [] : {}));
        }
        target[m[2] || m[1]] = data[p];
    }
    return result;
};

它目前使用表的数据输入参数,并在上面放置了许多属性-一个非破坏性的版本也应该是可能的。也许巧妙地使用lastIndexOf会比正则表达式执行得更好(取决于正则表达式引擎)。

在这里看到它的行动。

这段代码递归地展开JSON对象。

我在代码中包含了我的计时机制,它给了我1毫秒,但我不确定这是否是最准确的。

            var new_json = [{
              "name": "fatima",
              "age": 25,
              "neighbour": {
                "name": "taqi",
                "location": "end of the street",
                "property": {
                  "built in": 1990,
                  "owned": false,
                  "years on market": [1990, 1998, 2002, 2013],
                  "year short listed": [], //means never
                }
              },
              "town": "Mountain View",
              "state": "CA"
            },
            {
              "name": "qianru",
              "age": 20,
              "neighbour": {
                "name": "joe",
                "location": "opposite to the park",
                "property": {
                  "built in": 2011,
                  "owned": true,
                  "years on market": [1996, 2011],
                  "year short listed": [], //means never
                }
              },
              "town": "Pittsburgh",
              "state": "PA"
            }]

            function flatten(json, flattened, str_key) {
                for (var key in json) {
                  if (json.hasOwnProperty(key)) {
                    if (json[key] instanceof Object && json[key] != "") {
                      flatten(json[key], flattened, str_key + "." + key);
                    } else {
                      flattened[str_key + "." + key] = json[key];
                    }
                  }
                }
            }

        var flattened = {};
        console.time('flatten'); 
        flatten(new_json, flattened, "");
        console.timeEnd('flatten');

        for (var key in flattened){
          console.log(key + ": " + flattened[key]);
        }

输出:

flatten: 1ms
.0.name: fatima
.0.age: 25
.0.neighbour.name: taqi
.0.neighbour.location: end of the street
.0.neighbour.property.built in: 1990
.0.neighbour.property.owned: false
.0.neighbour.property.years on market.0: 1990
.0.neighbour.property.years on market.1: 1998
.0.neighbour.property.years on market.2: 2002
.0.neighbour.property.years on market.3: 2013
.0.neighbour.property.year short listed: 
.0.town: Mountain View
.0.state: CA
.1.name: qianru
.1.age: 20
.1.neighbour.name: joe
.1.neighbour.location: opposite to the park
.1.neighbour.property.built in: 2011
.1.neighbour.property.owned: true
.1.neighbour.property.years on market.0: 1996
.1.neighbour.property.years on market.1: 2011
.1.neighbour.property.year short listed: 
.1.town: Pittsburgh
.1.state: PA