JavaScript 展平对象数组

Question

我有一个包含几个数组的数组，每个数组包含几个对象，类似于这个。

[[object1, object2],[object1],[object1,object2,object3]]

这是记录到控制台的对象的屏幕截图。

将其展平为对象数组的最佳方法是什么？

我试过这个没有运气：

console.log(searchData);  
  var m = [].concat.apply([],searchData);    
console.log(m);

searchData 注销了上面的截图，但是 m 注销了 [ ]

这里是searchData的实际内容：

[[{"_id":"55064111d06b96d974937a6f","title":"Generic Title","shortname":"generic-title","contents":"<p>The Healing Center offers practical, social, and spiritual support to individuals and families. Services include, but are not limited to: food and clothing, job skills training and job search assistance, auto repair (Saturdays only), mentoring, financial counseling, tutoring, prayer, life skills training, and helpful information about local community services.</p><p>Stay in touch with us:</p>","__v":0},{"_id":"5508e1405c621d4aad2d2969","title":"test english","shortname":"test-page","contents":"<h2>English Test</h2>","__v":0}],[{"_id":"550b336f33a326aaee84f883","shortname":"ok-url","title":"now english","contents":"<p>okokko</p>","category":"Transportation","__v":0}]]

Answer 1

你可以像下面这样使用Array.concat：-

var arr = [['object1', 'object2'],['object1'],['object1','object2','object3']];
var flattened = [].concat.apply([],arr);

flattened 将是您预期的数组。

ES 2020 提供 flat，如果您想迭代，还提供 flatMap 到平面列表：

[['object1'], ['object2']].flat() // ['object1', 'object2']

Answer 2

如果您只需要简单的展平，这可能有效：

var arr = [['object1', 'object2'],['object1'],['object1','object2','object3']];
var flatenned = arr.reduce(function(a,b){ return a.concat(b) }, []);

对于更复杂的展平，Lodash 有展平功能，这可能是你需要的：https://lodash.com/docs#flatten

//Syntax: _.flatten(array, [isDeep])

_.flatten([1, [2, 3, [4]]]);
// → [1, 2, 3, [4]];

// using `isDeep` to recursive flatten
_.flatten([1, [2, 3, [4]]], true);
// → [1, 2, 3, 4];

Answer 3

递归地展平一个数组：

function flatten(array) {
   return !Array.isArray(array) ? array : [].concat.apply([], array.map(flatten));
}
 
var yourFlattenedArray = flatten([[{"_id":"55064111d06b96d974937a6f","title":"Generic Title","shortname":"generic-title","contents":"<p>The Healing Center offers practical, social, and spiritual support to individuals and families. Services include, but are not limited to: food and clothing, job skills training and job search assistance, auto repair (Saturdays only), mentoring, financial counseling, tutoring, prayer, life skills training, and helpful information about local community services.</p><p>Stay in touch with us:</p>","__v":0},{"_id":"5508e1405c621d4aad2d2969","title":"test english","shortname":"test-page","contents":"<h2>English Test</h2>","__v":0}],[{"_id":"550b336f33a326aaee84f883","shortname":"ok-url","title":"now english","contents":"<p>okokko</p>","category":"Transportation","__v":0}]]
);

log(yourFlattenedArray);

function log(data) {
  document.write('<pre>' + JSON.stringify(data, null, 2) + '</pre><hr>');
}

* {font-size: 12px; }

Answer 4

深度（嵌套）展平的递归解决方案：

function flatten(a) {
  return Array.isArray(a) ? [].concat.apply([], a.map(flatten)) : a;
}

使用 ES6 更紧凑：

var flatten = a => Array.isArray(a) ? [].concat(...a.map(flatten)) : a;

为了好玩，使用名为 F 的生成器“flatten”，懒惰地生成扁平值：

function *F(a) {
  if (Array.isArray(a)) for (var e of a) yield *F(e); else yield a;
}

>> console.log(Array.from(F([1, [2], 3])));
<< [ 1, 2, 3 ]

对于那些不熟悉生成器的人，yield * 语法从另一个生成器生成值。 Array.from 接受一个迭代器（例如调用生成器函数的结果）并将其转换为一个数组。

Answer 5

let functional = {
    flatten (array) {
        if (Array.isArray(array)) {
            return Array.prototype.concat(...array.map(this.flatten, this));
        }

        return array;
    }
};

functional.flatten([0, [1, 2], [[3, [4]]]]); // 0, 1, 2, 3, 4

Answer 6

使用 ES6 扩展运算符

Array.prototype.concat(...searchData)

或

[].concat(...searchData)

Answer 7

let nestedArray = [[1, 2], [3, 4], [5, 6]];

let flattenArray = function(nestedArray) {

	let flattenArr = [];
  
	nestedArray.forEach(function(item) {
  	flattenArr.push(...item);
  });
  
  return flattenArr;
};

console.log(flattenArray(nestedArray)); // [1, 2, 3, 4, 5, 6]

Answer 8

var arr = [1,[9,22],[[3]]];
var res = [];

function flatten(arr){
for(let i=0;i<arr.length;i++){
if(typeof arr[i] == "number"){
res.push(arr[i]);
}
else if(typeof arr[i] == "object"){
fatten(arr[i]);
}
}
}

调用函数

flatten(arr);
console.log(res);

结果

[1, 9, 22, 3]

Answer 9

我注意到人们正在使用成本不友好的递归，尤其是新的 ES6 标准赋予了我们传播运算符的力量。当您将项目推入主阵列时，只需使用 ... 它会自动添加展平对象。类似的东西

array.push(...subarray1)    // subarray1 = [object1, object2]
array.push(...subarray2)    // subarray2 = [object3]
array.push(...subarray3)    // subarray3 = [object4,object5, object6]
// output -> array = [object1, object2, object3, object4, object5, object6]

Answer 10

你可以使用flat()：

const data = [ [{id:1}, {id:2}], [{id:3}] ];
const result = data.flat();
console.log(result);

// you can specify the depth

const data2 = [ [ [ {id:1} ], {id:2}], [{id:3}] ];
const result2 = data2.flat(2);

console.log(result2);

在你的情况下：

const data = [[{"_id":"55064111d06b96d974937a6f","title":"Generic Title","shortname":"generic-title","contents":"<p>The Healing Center offers practical, social, and spiritual support to individuals and families. Services include, but are not limited to: food and clothing, job skills training and job search assistance, auto repair (Saturdays only), mentoring, financial counseling, tutoring, prayer, life skills training, and helpful information about local community services.</p><p>Stay in touch with us:</p>","__v":0},{"_id":"5508e1405c621d4aad2d2969","title":"test english","shortname":"test-page","contents":"<h2>English Test</h2>","__v":0}],[{"_id":"550b336f33a326aaee84f883","shortname":"ok-url","title":"now english","contents":"<p>okokko</p>","category":"Transportation","__v":0}]]

const result = data.flat();

console.log(result);

Answer 11

我将对象数组展平并返回单个数组的解决方案。

flattenArrayOfObject = (arr) => {
  const flattened = {};

  arr.forEach((obj) => {
    Object.keys(obj).forEach((key) => {
      flattened[key] = obj[key];
    });
  });

  return flattened;
};

例子

const arr = [
  {
    verify: { '0': 'xyzNot verified', '1': 'xyzVerified' },
    role_id: { '1': 'xyzMember', '2': 'xyzAdmin' },
    two_factor_authentication: { '0': 'No', '1': 'Yes' }
  },
  { status: { '0': 'xyzInactive', '1': 'Active', '2': 'xyzSuspend' } }
]

flattenArrayOfObject(arr)

// {
//   verify: { '0': 'xyzNot verified', '1': 'xyzVerified' },
//   status: { '0': 'xyzInactive', '1': 'Active', '2': 'xyzSuspend' },
//   role_id: { '1': 'xyzMember', '2': 'xyzAdmin' },
//   two_factor_authentication: { '0': 'No', '1': 'Yes' }
// }

Answer 12

如果每个对象都有一个数组并以同样的方式继续嵌套：

function flatten(i,arrayField){
  if(Array.isArray(i)) return i.map(c=>flatten(c,arrayField));
  if(i.hasOwnProperty(arrayField)) return [{...i,[arrayField]:null},...i[arrayField].map(c=>flatten(c,arrayField))];
  return {...i,[arrayField]:null};
}

let data=flatten(myData,'childs');

我的数据是这样的：

[
{
    "id": 1,
    "title": "t1",
    "sort_order": 200,
    "childs": [
        {
            "id": 2,
            "title": "t2",
            "sort_order": 200,
            "childs": []
        },
        {
            "id": 3,
            "title":"mytitle",
            "sort_order": 200,
            "childs": []
        },
        {
            "id": 4,
            "title":"mytitle",
            "sort_order": 200,
            "childs": []
        },
        {
            "id": 5,
            "title":"mytitle",
            "sort_order": 200,
            "childs": []
        },
        {
            "id": 6,
            "title":"mytitle",
            "sort_order": 200,
            "childs": []
        }
    ]
},
{
    "id": 7,
    "title": "راهنما",
    "sort_order":"mytitle",
    "childs": [
        {
            "id": 8,
            "title":"mytitle",
            "sort_order": 200,
            "childs": []
        },
        {
            "id": 9,
            "title":"mytitle",
            "sort_order": 200,
            "childs": []
        },
        {
            "id": 10,
            "title":"mytitle",
            "sort_order": 200,
            "childs": []
        }
    ]
}

]

Answer 13

// Polyfill flat method

var flatten = a => Array.isArray(a) ? [].concat(...a.map(flatten)) : a;

var deepFlatten = (arr, depth = 1) => {
   return depth > 0 ? arr.reduce((acc, val) => acc.concat(Array.isArray(val) ? deepFlatten(val, depth - 1) : val), [])
                : arr.slice();
}

console.log(deepFlatten([0, 1, 2, [[[3, 4]]]], Infinity));

// You can pass label in place of 'Infinity'