我有一个JavaScript数组,如:

[["$6"], ["$12"], ["$25"], ["$25"], ["$18"], ["$22"], ["$10"]]

如何将单独的内部数组合并为一个,例如:

["$6", "$12", "$25", ...]

当前回答

我提出了两个没有递归的简短解决方案。从计算复杂性的角度来看,它们不是最优的,但在一般情况下工作良好:

let a = [1, [2, 3], [[4], 5, 6], 7, 8, [9, [[10]]]];

// Solution #1
while (a.find(x => Array.isArray(x)))
    a = a.reduce((x, y) => x.concat(y), []);

// Solution #2
let i = a.findIndex(x => Array.isArray(x));
while (i > -1)
{
    a.splice(i, 1, ...a[i]);
    i = a.findIndex(x => Array.isArray(x));
}

其他回答

/**
* flatten an array first level
* @method flatten
* @param array {Array}
* @return {Array} flatten array
*/
function flatten(array) {
  return array.reduce((acc, current) => acc.concat(current), []);
}


/**
* flatten an array recursively
* @method flattenDeep
* @param array {Array}
* @return {Array} flatten array
*/
function flattenDeep(array) {
  return array.reduce((acc, current) => {
    return Array.isArray(current) ? acc.concat(flattenDeep(current)) : acc.concat([current]);
  }, []);
}

/**
* flatten an array recursively limited by depth
* @method flattenDepth
* @param array {Array}
* @return {Array} flatten array
*/
function flattenDepth(array, depth) {
  if (depth === 0) {
    return array;
  }
  return array.reduce((acc, current) => {
    return Array.isArray(current) ? acc.concat(flattenDepth(current, --depth)) : acc.concat([current]);
  }, []);
}

使用那里的代码。

我会写:

myArray.enumerable().selectMany(function(x) { return x; }).array()

我使用这个方法来展开混合数组:(这对我来说似乎最简单)。用较长的版本来解释步骤。

function flattenArray(deepArray) {
    // check if Array
    if(!Array.isArray(deepArray)) throw new Error('Given data is not an Array')

    const flatArray = deepArray.flat() // flatten array
    const filteredArray = flatArray.filter(item => !!item) // filter by Boolean
    const uniqueArray = new Set(filteredArray) // filter by unique values
    
    return [...uniqueArray] // convert Set into Array
}

//较短版本:

const flattenArray = (deepArray) => [...new Set(deepArray.flat().filter(item=>!!item))]
flattenArray([4,'a', 'b', [3, 2, undefined, 1], [1, 4, null, 5]])) // 4,'a','b',3,2,1,5

Codesandbox链接

现代方法

使用[].flat(Infinity)方法

const nestedArray = [1,[2,[3],[4,[5,[6,[7]]]]]]
const flatArray = nestedArray.flat(Infinity)
console.log(flatArray)

最好使用javascript reduce函数。

var arrays = [["$6"], ["$12"], ["$25"], ["$25"], ["$18"], ["$22"], ["$10"], ["$0"], ["$15"],["$3"], ["$75"], ["$5"], ["$100"], ["$7"], ["$3"], ["$75"], ["$5"]];

arrays = arrays.reduce(function(a, b){
     return a.concat(b);
}, []);

或者,使用ES2015:

arrays = arrays.reduce((a, b) => a.concat(b), []);

js小提琴

Mozilla文档

适用于所有数据类型的递归版本

 /*jshint esversion: 6 */

// nested array for testing
let nestedArray = ["firstlevel", 32, "alsofirst", ["secondlevel", 456,"thirdlevel", ["theinnerinner", 345, {firstName: "Donald", lastName: "Duck"}, "lastinner"]]];

// wrapper function to protect inner variable tempArray from global scope;
function flattenArray(arr) {

  let tempArray = [];

  function flatten(arr) {
    arr.forEach(function(element) {
      Array.isArray(element) ? flatten(element) : tempArray.push(element);     // ternary check that calls flatten() again if element is an array, hereby making flatten() recursive.
    });
  }

  // calling the inner flatten function, and then returning the temporary array
  flatten(arr);
  return tempArray;
}

// example usage:
let flatArray = flattenArray(nestedArray);