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", ...]
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
我已经使用递归和闭包完成了
function flatten(arr) {
var temp = [];
function recursiveFlatten(arr) {
for(var i = 0; i < arr.length; i++) {
if(Array.isArray(arr[i])) {
recursiveFlatten(arr[i]);
} else {
temp.push(arr[i]);
}
}
}
recursiveFlatten(arr);
return temp;
}
其他回答
在javascript中定义一个名为foo的数组数组,并使用javascript的arrayconcat内置方法将该数组展平为单个数组:
const foo = [["$6"], ["$12"], ["$25"], ["$25"], ["$18"], ["$22"], ["$10"]]
console.log({foo});
const bar = [].concat(...foo)
console.log({bar});
应打印:
{ foo:
[ [ '$6' ],
[ '$12' ],
[ '$25' ],
[ '$25' ],
[ '$18' ],
[ '$22' ],
[ '$10' ] ] }
{ bar: [ '$6', '$12', '$25', '$25', '$18', '$22', '$10' ] }
以下代码将压平深度嵌套的数组:
/**
* [Function to flatten deeply nested array]
* @param {[type]} arr [The array to be flattened]
* @param {[type]} flattenedArr [The flattened array]
* @return {[type]} [The flattened array]
*/
function flattenDeepArray(arr, flattenedArr) {
let length = arr.length;
for(let i = 0; i < length; i++) {
if(Array.isArray(arr[i])) {
flattenDeepArray(arr[i], flattenedArr);
} else {
flattenedArr.push(arr[i]);
}
}
return flattenedArr;
}
let arr = [1, 2, [3, 4, 5], [6, 7]];
console.log(arr, '=>', flattenDeepArray(arr, [])); // [ 1, 2, [ 3, 4, 5 ], [ 6, 7 ] ] '=>' [ 1, 2, 3, 4, 5, 6, 7 ]
arr = [1, 2, [3, 4], [5, 6, [7, 8, [9, 10]]]];
console.log(arr, '=>', flattenDeepArray(arr, [])); // [ 1, 2, [ 3, 4 ], [ 5, 6, [ 7, 8, [Object] ] ] ] '=>' [ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 ]
以下是Typescript中最快的解决方案,它也适用于具有多层嵌套的数组:
export function flatten<T>(input: Array<any>, output: Array<T> = []): Array<T> {
for (const value of input) {
Array.isArray(value) ? flatten(value, output) : output.push(value);
}
return output;
}
以及:
const result = flatten<MyModel>(await Promise.all(promises));
只是为了增加伟大的解决方案。我用递归来解决这个问题。
const flattenArray = () => {
let result = [];
return function flatten(arr) {
for (let i = 0; i < arr.length; i++) {
if (!Array.isArray(arr[i])) {
result.push(arr[i]);
} else {
flatten(arr[i])
}
}
return result;
}
}
测试结果:https://codepen.io/ashermike/pen/mKZrWK
最好是以递归的方式执行,这样如果另一个数组中还有另一个,就可以很容易地过滤。。。
const flattenArray = arr =>
arr.reduce(
(res, cur) =>
!Array.isArray(cur)
? res.concat(cur)
: res.concat(flattenArray(cur)), []);
你可以这样称呼它:
flattenArray([[["Alireza"], "Dezfoolian"], ["is a"], ["developer"], [[1, [2, 3], ["!"]]]);
结果如下:
["Alireza", "Dezfoolian", "is a", "developer", 1, 2, 3, "!"]