javascript 展平/取消展平嵌套 JSON 对象的最快方法
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Fastest way to flatten / un-flatten nested JSON objects
提问by Louis Ricci
I threw some code together to flatten and un-flatten complex/nested JSON objects. It works, but it's a bit slow (triggers the 'long script' warning).
我将一些代码放在一起来展平和取消展平复杂/嵌套的 JSON 对象。它有效，但有点慢（触发“长脚本”警告）。
For the flattened names I want "." as the delimiter and [INDEX] for arrays.
对于扁平化的名字，我想要“。” 作为数组的分隔符和 [INDEX]。
Examples:
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}
I created a benchmark that ~simulates my use case http://jsfiddle.net/WSzec/
我创建了一个基准测试，模拟我的用例http://jsfiddle.net/WSzec/
- Get a nested JSON object
- Flatten it
- Look through it and possibly modify it while flattened
- Unflatten it back to it's original nested format to be shipped away
- 获取嵌套的 JSON 对象
- 压平它
- 查看它并可能在展平时对其进行修改
- 将其解压回原始嵌套格式以运走
I would like faster code: For clarification, code that completes the JSFiddle benchmark (http://jsfiddle.net/WSzec/) significantly faster (~20%+ would be nice) in IE 9+, FF 24+, and Chrome 29+.
我想要更快的代码：为了澄清起见，在 IE 9+、FF 24+ 和 Chrome 29中完成 JSFiddle 基准测试（http://jsfiddle.net/WSzec/）的代码要快得多（~20%+ 会更好） +。
Here's the relevant JavaScript code: Current Fastest: http://jsfiddle.net/WSzec/6/
这是相关的 JavaScript 代码：当前最快：http: //jsfiddle.net/WSzec/6/
JSON.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[""];
}
JSON.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;
}
EDIT 1Modified the above to @Bergi 's implementation which is currently the fastest. As an aside, using ".indexOf" instead of "regex.exec" is around 20% faster in FF but 20% slower in Chrome; so I'll stick with the regex since it's simpler (here's my attempt at using indexOf to replace the regex http://jsfiddle.net/WSzec/2/).
编辑 1将上述修改为 @Bergi 的实现，这是目前最快的。顺便说一句，在 FF 中使用“.indexOf”而不是“regex.exec”大约快 20%，但在 Chrome 中慢 20%；所以我会坚持使用正则表达式，因为它更简单（这是我尝试使用 indexOf 替换正则表达式http://jsfiddle.net/WSzec/2/）。
EDIT 2Building on @Bergi 's idea I managed to created a faster non-regex version (3x faster in FF and ~10% faster in Chrome). http://jsfiddle.net/WSzec/6/In the this (the current) implementation the rules for key names are simply, keys cannot start with an integer or contain a period.
编辑 2基于@Bergi 的想法，我设法创建了一个更快的非正则表达式版本（FF 中快 3 倍，Chrome 中快 10%）。http://jsfiddle.net/WSzec/6/在这个（当前）实现中，键名的规则很简单，键不能以整数开头或包含句点。
Example:
例子：
- {"foo":{"bar":[0]}} => {"foo.bar.0":0}
- {"foo":{"bar":[0]}} => {"foo.bar.0":0}
EDIT 3Adding @AaditMShah 's inline path parsing approach (rather than String.split) helped to improve the unflatten performance. I'm very happy with the overall performance improvement reached.
编辑 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。我对达到的整体性能改进感到非常满意。
The latest jsfiddle and jsperf:
最新的jsfiddle和jsperf：
回答by Bergi
Here's my much shorter implementation:
这是我更短的实现：
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;
};
flattenhasn't changed much (and I'm not sure whether you really need those isEmptycases):
flatten没有太大变化（我不确定您是否真的需要这些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;
}
Together, they run your benchmarkin about the half of the time (Opera 12.16: ~900ms instead of ~ 1900ms, Chrome 29: ~800ms instead of ~1600ms).
它们一起在大约一半的时间内运行您的基准测试（Opera 12.16：~900 毫秒而不是~1900 毫秒，Chrome 29：~800 毫秒而不是~1600 毫秒）。
Note:This and most other solutions answered here focus on speed and are susceptible to prototype pollutionand shold not be used on untrusted objects.
注意：这里回答的大多数其他解决方案都侧重于速度，并且容易受到原型污染的影响，并且不得用于不受信任的对象。
回答by Aadit M Shah
I wrote two functions to flattenand unflattena JSON object.
我写了两个函数flatten和unflatten一个 JSON 对象。
var flatten = (function (isArray, wrapped) {
return function (table) {
return reduce("", {}, table);
};
function reduce(path, accumulator, table) {
if (isArray(table)) {
var length = table.length;
if (length) {
var index = 0;
while (index < length) {
var property = path + "[" + index + "]", item = table[index++];
if (wrapped(item) !== item) accumulator[property] = item;
else reduce(property, accumulator, item);
}
} else accumulator[path] = table;
} else {
var empty = true;
if (path) {
for (var property in table) {
var item = table[property], property = path + "." + property, empty = false;
if (wrapped(item) !== item) accumulator[property] = item;
else reduce(property, accumulator, item);
}
} else {
for (var property in table) {
var item = table[property], empty = false;
if (wrapped(item) !== item) accumulator[property] = item;
else reduce(property, accumulator, item);
}
}
if (empty) accumulator[path] = table;
}
return accumulator;
}
}(Array.isArray, Object));
Performance:
性能:
- It's faster than the current solution in Opera. The current solution is 26% slower in Opera.
- It's faster than the current solution in Firefox. The current solution is 9% slower in Firefox.
- It's faster than the current solution in Chrome. The current solution is 29% slower in Chrome.
- 它比 Opera 中的当前解决方案更快。当前的解决方案在 Opera 中慢了 26%。
- 它比 Firefox 中的当前解决方案更快。当前的解决方案在 Firefox 中慢了 9%。
- 它比 Chrome 中的当前解决方案更快。当前的解决方案在 Chrome 中慢了 29%。
function unflatten(table) {
var result = {};
for (var path in table) {
var cursor = result, length = path.length, property = "", index = 0;
while (index < length) {
var char = path.charAt(index);
if (char === "[") {
var start = index + 1,
end = path.indexOf("]", start),
cursor = cursor[property] = cursor[property] || [],
property = path.slice(start, end),
index = end + 1;
} else {
var cursor = cursor[property] = cursor[property] || {},
start = char === "." ? index + 1 : index,
bracket = path.indexOf("[", start),
dot = path.indexOf(".", start);
if (bracket < 0 && dot < 0) var end = index = length;
else if (bracket < 0) var end = index = dot;
else if (dot < 0) var end = index = bracket;
else var end = index = bracket < dot ? bracket : dot;
var property = path.slice(start, end);
}
}
cursor[property] = table[path];
}
return result[""];
}
Performance:
性能:
- It's faster than the current solution in Opera. The current solution is 5% slower in Opera.
- It's slower than the current solution in Firefox. My solution is 26% slower in Firefox.
- It's slower than the current solution in Chrome. My solution is 6% slower in Chrome.
- 它比 Opera 中的当前解决方案更快。当前的解决方案在 Opera 中慢了 5%。
- 它比 Firefox 中的当前解决方案慢。我的解决方案在 Firefox 中慢了 26%。
- 它比 Chrome 中的当前解决方案慢。我的解决方案在 Chrome 中慢了 6%。
Flatten and unflatten a JSON object:
Overall my solution performs either equally well or even better than the current solution.
总的来说，我的解决方案与当前的解决方案一样好，甚至更好。
Performance:
性能:
- It's faster than the current solution in Opera. The current solution is 21% slower in Opera.
- It's as fast as the current solution in Firefox.
- It's faster than the current solution in Firefox. The current solution is 20% slower in Chrome.
- 它比 Opera 中的当前解决方案更快。当前的解决方案在 Opera 中慢了 21%。
- 它与 Firefox 中的当前解决方案一样快。
- 它比 Firefox 中的当前解决方案更快。当前的解决方案在 Chrome 中慢了 20%。
Output format:
输出格式：
A flattened object uses the dot notation for object properties and the bracket notation for array indices:
扁平对象使用点表示法表示对象属性，使用方括号表示法表示数组索引：
{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}
{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}
In my opinion this format is better than only using the dot notation:
在我看来，这种格式比仅使用点表示法要好：
{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}
{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}
Advantages:
优点：
- Flattening an object is faster than the current solution.
- Flattening and unflattening an object is as fast as or faster than the current solution.
- Flattened objects use both the dot notation and the bracket notation for readability.
- 展平对象比当前解决方案更快。
- 展平和取消展平对象的速度与当前解决方案一样快或快于当前解决方案。
- 扁平对象同时使用点符号和括号符号以提高可读性。
Disadvantages:
缺点：
- Unflattening an object is slower than the current solution in most (but not all) cases.
- 在大多数（但不是全部）情况下，展开对象比当前的解决方案慢。
The current JSFiddle demogave the following values as output:
当前的JSFiddle 演示给出了以下值作为输出：
Nested : 132175 : 63
Flattened : 132175 : 564
Nested : 132175 : 54
Flattened : 132175 : 508
My updated JSFiddle demogave the following values as output:
我更新的JSFiddle 演示给出了以下值作为输出：
Nested : 132175 : 59
Flattened : 132175 : 514
Nested : 132175 : 60
Flattened : 132175 : 451
I'm not really sure what that means, so I'll stick with the jsPerf results. After all jsPerf is a performance benchmarking utility. JSFiddle is not.
我不太确定这意味着什么，所以我会坚持使用 jsPerf 结果。毕竟 jsPerf 是一个性能基准测试工具。JSFiddle 不是。
回答by Yan Foto
3 ? Years later...
3 ? 多年后...
For my own project I wanted to flatten JSON objects in mongoDB dot notationand came up with a simple solution:
对于我自己的项目，我想用mongoDB 点表示法扁平化 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
}
Features and/or caveats
功能和/或警告
- It only accepts JSON objects. So if you pass something like
{a: () => {}}you might not get what you wanted! - It removes empty arrays and objects. So this
{a: {}, b: []}is flattened to{}.
- 它只接受 JSON 对象。所以如果你通过了类似的东西，
{a: () => {}}你可能得不到你想要的东西！ - 它删除空数组和对象。所以这
{a: {}, b: []}被扁平化为{}.
回答by Guy
ES6 version:
ES6版本：
const flatten = (obj, path = '') => {
if (!(obj instanceof Object)) return {[path.replace(/\.$/g, '')]:obj};
return Object.keys(obj).reduce((output, key) => {
return obj instanceof Array ?
{...output, ...flatten(obj[key], path + '[' + key + '].')}:
{...output, ...flatten(obj[key], path + key + '.')};
}, {});
}
Example:
例子：
console.log(flatten({a:[{b:["c","d"]}]}));
console.log(flatten([1,[2,[3,4],5],6]));
回答by Bergi
Here's another approach that runs slower (about 1000ms) than the above answer, but has an interesting idea :-)
这是另一种运行速度比上述答案慢（大约 1000 毫秒）的方法，但有一个有趣的想法:-)
Instead of iterating through each property chain, it just picks the last property and uses a look-up-table for the rest to store the intermediate results. This look-up-table will be iterated until there are no property chains left and all values reside on uncocatenated properties.
它不是遍历每个属性链，而是选择最后一个属性并使用查找表来存储其余的中间结果。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连接的属性上。
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;
};
It currently uses the datainput parameter for the table, and puts lots of properties on it - a non-destructive version should be possible as well. Maybe a clever lastIndexOfusage performs better than the regex (depends on the regex engine).
它目前使用data表的输入参数，并在其上放置许多属性 - 非破坏性版本也应该是可能的。也许巧妙的lastIndexOf用法比正则表达式表现更好（取决于正则表达式引擎）。
回答by Tom Esterez
You can use https://github.com/hughsk/flat
您可以使用https://github.com/hughsk/flat
Take a nested Javascript object and flatten it, or unflatten an object with delimited keys.
获取嵌套的 Javascript 对象并将其展平，或使用分隔键取消展平对象。
Example from the doc
文档中的示例
var flatten = require('flat')
flatten({
key1: {
keyA: 'valueI'
},
key2: {
keyB: 'valueII'
},
key3: { a: { b: { c: 2 } } }
})
// {
// 'key1.keyA': 'valueI',
// 'key2.keyB': 'valueII',
// 'key3.a.b.c': 2
// }
var unflatten = require('flat').unflatten
unflatten({
'three.levels.deep': 42,
'three.levels': {
nested: true
}
})
// {
// three: {
// levels: {
// deep: 42,
// nested: true
// }
// }
// }
回答by sfrizvi6
This code recursively flattens out JSON objects.
此代码递归地展平 JSON 对象。
I included my timing mechanism in the code and it gives me 1ms but I'm not sure if that's the most accurate one.
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]);
}
Output:
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
回答by jtrumbull
I added +/- 10-15% efficiency to the selected answer by minor code refactoring and moving the recursive function outside of the function namespace.
我通过次要代码重构和将递归函数移到函数命名空间之外，为所选答案增加了 +/- 10-15% 的效率。
See my question: Are namespaced functions reevaluated on every call?for why this slows nested functions down.
请参阅我的问题：是否在每次调用时重新评估命名空间函数？为什么这会减慢嵌套函数的速度。
function _flatten (target, obj, path) {
var i, empty;
if (obj.constructor === Object) {
empty = true;
for (i in obj) {
empty = false;
_flatten(target, obj[i], path ? path + '.' + i : i);
}
if (empty && path) {
target[path] = {};
}
}
else if (obj.constructor === Array) {
i = obj.length;
if (i > 0) {
while (i--) {
_flatten(target, obj[i], path + '[' + i + ']');
}
} else {
target[path] = [];
}
}
else {
target[path] = obj;
}
}
function flatten (data) {
var result = {};
_flatten(result, data, null);
return result;
}
See benchmark.
参见基准。
回答by paulwal222
Here's mine. It runs in <2ms in Google Apps Script on a sizable object. It uses dashes instead of dots for separators, and it doesn't handle arrays specially like in the asker's question, but this is what I wanted for my use.
这是我的。它在 Google Apps 脚本中在一个相当大的对象上运行 <2 毫秒。它使用破折号而不是点作为分隔符，并且它不像提问者的问题那样特别处理数组，但这是我想要的。
function flatten (obj) {
var newObj = {};
for (var key in obj) {
if (typeof obj[key] === 'object' && obj[key] !== null) {
var temp = flatten(obj[key])
for (var key2 in temp) {
newObj[key+"-"+key2] = temp[key2];
}
} else {
newObj[key] = obj[key];
}
}
return newObj;
}
Example:
例子：
var test = {
a: 1,
b: 2,
c: {
c1: 3.1,
c2: 3.2
},
d: 4,
e: {
e1: 5.1,
e2: 5.2,
e3: {
e3a: 5.31,
e3b: 5.32
},
e4: 5.4
},
f: 6
}
Logger.log("start");
Logger.log(JSON.stringify(flatten(test),null,2));
Logger.log("done");
Example output:
示例输出：
[17-02-08 13:21:05:245 CST] start
[17-02-08 13:21:05:246 CST] {
"a": 1,
"b": 2,
"c-c1": 3.1,
"c-c2": 3.2,
"d": 4,
"e-e1": 5.1,
"e-e2": 5.2,
"e-e3-e3a": 5.31,
"e-e3-e3b": 5.32,
"e-e4": 5.4,
"f": 6
}
[17-02-08 13:21:05:247 CST] done
回答by o.z
Use this library:
使用这个库：
npm install flat
Usage (from https://www.npmjs.com/package/flat):
用法（来自https://www.npmjs.com/package/flat）：
Flatten:
压平：
var flatten = require('flat')
flatten({
key1: {
keyA: 'valueI'
},
key2: {
keyB: 'valueII'
},
key3: { a: { b: { c: 2 } } }
})
// {
// 'key1.keyA': 'valueI',
// 'key2.keyB': 'valueII',
// 'key3.a.b.c': 2
// }
Un-flatten:
取消展平：
var unflatten = require('flat').unflatten
unflatten({
'three.levels.deep': 42,
'three.levels': {
nested: true
}
})
// {
// three: {
// levels: {
// deep: 42,
// nested: true
// }
// }
// }
