使用Object.fromEntries（）

Question

有时候，我读到了有关新方法Object.fromEntries()的建议，该方法在某些浏览器（reference）的较新版本中受支持。在阅读有关内容时，我想到了使用此方法deep-clone和object来代替使用JSON.parse(JSON.stringify(obj))的想法。到目前为止，从我的角度来看，我已经提出了下一种递归方法。

const obj = {
  key1: {key11: "key11", key12: "key12", key13: {key131: 22}},
  key2: {key21: "key21", key22: "key22"},
  key3: "key3",
  key4: [1,2,3,4]
}

const cloneObj = (obj) =>
{
    if (typeof obj !== "object")
       return obj;
    else if (Array.isArray(obj))
       return obj.slice();

    return Object.fromEntries(Object.entries(obj).map(
        ([k,v]) => ([k, cloneObj(v)])
    ));
}

// Clone the original object.
let newObj = cloneObj(obj);

// Make changes on the original object.
obj.key1.key11 = "TEST";
obj.key3 = "TEST";
obj.key1.key13.key131 = "TEST";
obj.key4[1] = "TEST";

// Display both objects on the console.
console.log("Original object: ", obj);
console.log("Cloned object: ", newObj);

.as-console {background-color:black !important; color:lime;}
.as-console-wrapper {max-height:100% !important; top:0;}

但是，对于是否要完成某个对象的deep-clone，我是否有所遗漏，我感到怀疑。所以我的问题是：

1）我缺少一些重要的东西来完成对象的deep-clone？

2）在这种方法还可以的情况下，您是否认为这样做比使用JSON.parse()和JSON.stringify()更好？

谢谢！

---

更新1

这是更新版本，其中包含答案的反馈：

const obj = {
  key1: {key11: "key11", key12: "key12", key13: {key131: 22}},
  key2: {key21: "key21", key22: "key22"},
  key3: "key3",
  key4: [1,2,3,{key: "value"}]
}

const cloneObj = (obj) =>
{
    if (Object(obj) !== obj)
       return obj;
    else if (Array.isArray(obj))
       return obj.map(cloneObj);

    return Object.fromEntries(Object.entries(obj).map(
        ([k,v]) => ([k, cloneObj(v)])
    ));
}

// Clone the original object.
let newObj = cloneObj(obj);

// Make changes on the original object.
obj.key1.key11 = "TEST";
obj.key3 = "TEST";
obj.key1.key13.key131 = "TEST";
obj.key4[1] = "TEST";
obj.key4[3].key = "TEST";

// Display both objects on the console.
console.log("Original object: ", obj);
console.log("Cloned object: ", newObj);

.as-console {background-color:black !important; color:lime;}
.as-console-wrapper {max-height:100% !important; top:0;}

更新2

针对第二个问题添加了效果比较：

Tested on Firefox 66.0.3 (64bits):
Test_JSON: 1923.000ms
Test_cloneObj: 2047.000ms

Tested on Chrome 73.0.3683.103 (64 bits):
Test_JSON: 2276.560ms
Test_cloneObj: 1903.675ms

const cloneObj = (obj) =>
{
    if (Object(obj) !== obj)
       return obj;
    else if (Array.isArray(obj))
       return obj.map(cloneObj);

    return Object.fromEntries(Object.entries(obj).map(
        ([k,v]) => ([k, cloneObj(v)])
    ));
}

// Generate an object.

const getRandom = (min, max) => Math.floor(Math.random() * (max - min) + min);

let obj = {};

for (let i = 0; i < 100000; i++)
{
    obj["Array" + i] = Array.from({length: 100}, () => getRandom(0, 1000));
    obj["Obj" + i] = {"key": getRandom(0, 1000)};
    obj["Const" + i] = "some_string";
}

// Test performance on JSON.parse()/stringify()

console.time("Test_JSON");
let obj1 = JSON.parse(JSON.stringify(obj));
console.timeEnd("Test_JSON");

// Test performance on cloneObj().

console.time("Test_cloneObj");
let obj2 = cloneObj(obj);
console.timeEnd("Test_cloneObj");

.as-console {background-color:black !important; color:lime;}
.as-console-wrapper {max-height:100% !important; top:0;}

Answer 1

您缺少一件事：

else if (Array.isArray(obj))
  return obj.slice();

这将返回数组的浅表副本。如果数组包含对象，则不会克隆这些基础对象：

const obj = [
  ['foo']
];

const cloneObj = (obj) =>
{
    if (typeof obj !== "object")
       return obj;
    else if (Array.isArray(obj))
       return obj.slice();

    return Object.fromEntries(Object.entries(obj).map(
        ([k,v]) => ([k, cloneObj(v)])
    ));
}

// Clone the original object.
let newObj = cloneObj(obj);

// Make changes on the original object.
obj[0][0] = 'bar';

// Display both objects on the console.
console.log("Original object: ", obj);
console.log("Cloned object: ", newObj);

.as-console {background-color:black !important; color:lime;}
.as-console-wrapper {max-height:100% !important; top:0;}

要解决此问题，请改用return obj.map(cloneObj);：

const obj = [
  ['foo']
];

const cloneObj = (obj) =>
{
    if (typeof obj !== "object")
       return obj;
    else if (Array.isArray(obj))
       return obj.map(cloneObj);

    return Object.fromEntries(Object.entries(obj).map(
        ([k,v]) => ([k, cloneObj(v)])
    ));
}

// Clone the original object.
let newObj = cloneObj(obj);

// Make changes on the original object.
obj[0][0] = 'bar';

// Display both objects on the console.
console.log("Original object: ", obj);
console.log("Cloned object: ", newObj);

.as-console {background-color:black !important; color:lime;}
.as-console-wrapper {max-height:100% !important; top:0;}

Answer 2

1. 深层克隆很好，因为深层嵌套的对象仍然保留其属性-但是，阵列也需要被克隆。将obj.slice()替换为obj.map(o => cloneObj(o))。
2. 此方法实际上比JSON.parse(JSON.stringify(obj))更快-在JSBench和JSON上进行的三个测试每次的运行速度都降低了10％以上。

Answer 3

如上所述，您的数组不是深度克隆-请使用obj.map(cloneObj)而不是obj.slice()。

但是另一个疏忽是typeof obj !== "object"，它不适用于null。最好使用Object(obj) !== obj。