JavaScript 展平对象数组的数组

JavaScript flattening an array of arrays of objects

提问by byrdr

I have an array which contains several arrays, each containing several objects, similar to this.

我有一个数组，其中包含多个数组，每个数组包含多个对象，与此类似。

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

Here is a screenhot of the object logged to the console.

这是记录到控制台的对象的屏幕截图。

What would be the best approach to flattening this out so it just an array of objects?

将其展平以便它只是一组对象的最佳方法是什么？

I've tried this with no luck:

我试过这个没有运气：

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

searchData logs out the screenshot above, but m logs out [ ]

searchData 登出上面的截图，但 m 登出 [ ]

Here is the actual contents of searchData:

这是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}]]

回答by Mritunjay

You can use Array.concatlike bellow:-

您可以像下面这样使用Array.concat：-

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

flattenedwill be your expected array.

flattened将是您预期的数组。

回答by byrdr

A recursive solution for deep (nested) flattening:

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;
}

A bit more compactly with ES6:

使用 ES6 更紧凑一点：

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

For fun, using a generator named Ffor "flatten", to lazily generate flattened values:

为了好玩，使用一个名为F“flatten”的生成器来懒惰地生成扁平化的值：

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 ]

For those not familiar with generators the yield *syntax yields values from another generator. Array.fromtakes an iterator (such as results from invoking the generator function) and turns it into an array.

对于那些不熟悉生成器的人来说，yield *语法从另一个生成器产生值。Array.from获取一个迭代器（例如调用生成器函数的结果）并将其转换为数组。

回答by Huy Hoang Pham

If you only need simple flatten, this may works:

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) }, []);

For more complex flattening, Lodash has the flatten function, which maybe what you need: https://lodash.com/docs#flatten

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];

回答by Shreyas

Using ES6 Spread Operator

使用 ES6 扩展运算符

Array.prototype.concat(...searchData)

Array.prototype.concat(...searchData)

OR

或者

[].concat(...searchData)

[].concat(...searchData)

回答by Adnan Khan

I've noticed that people are using recursions which are not cost friendly, especially with new ES6 standards giving us the power of spread operators. When you're pushing the items into the master array just use ... and it will automatically add flattened objects. Something like

我注意到人们正在使用成本不友好的递归，尤其是新的 ES6 标准赋予我们扩展运算符的力量。当您将项目推入主阵列时，只需使用 ... 它会自动添加扁平对象。就像是

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]

回答by Miguel Mota

Recursively flatten an array:

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; }

回答by Alexander Abashkin

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

回答by Taki

you can use flat():

您可以使用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);

in your case :

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);

回答by Tarun Majumder

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]

回答by Gajender Singh

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]);
}
}
}

Calling function

调用函数

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

Result ?

结果 ？

[1, 9, 22, 3]