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[英]How to insert all elements from one immutable list into another at given index
[英]Insert elements from one array to another after each 'n' index
原生JS。 例如,有2个数组
arr1 = ['orange', 'blue', 'red', 'black', 'white', 'magenta', 'cyan']
arr2 = [1, 2, 3]
我需要一个函数,将这两个数组结合起来,像这样
function arrInsertAfter(arr1, arr2, afterElement)
// for example make afterElement = 2 - insert element of second array after each 2 elements of first array
....
结果应该是
['orange', 'blue', 1, 'red', 'black', 2, 'white', 'magenta', 3, 'cyan']
例如,如果afterElement = 3
,则应返回
['orange', 'blue', 'red', 1, 'black', 'white', 'magenta', 2, 'cyan', 3] // append remaining elements of second array at the end simply
不要使用任何第三方连接的库,这一点很重要。
这是我的功能
let arr1 = ['orange', 'blue', 'red', 'black', 'white', 'magenta', 'cyan'],
arr2 = [1, 2, 3];
function arrInsertAfter(arr1, arr2, afterElement) {
let curPos = afterElement;
arr2.forEach(function(e) {
arr1.splice(curPos, 0, e);
curPos = afterElement+curPos+1;
});
return arr1;
}
与传统循环
function arrInsertAfter(arr1, arr2, afterElement) {
let curPos = afterElement;
for (let i = 0; i < arr2.length; i++) {
arr1.splice(curPos, 0, arr2[i]);
curPos = (curPos+afterElement)+1;
}
return true;
}
arrInsertAfter(arr1, arr2, 2);
alert(arr1);
这应该做:)
function arrInsertAfter(arr1, arr2, afterElement){ result = []; for(var i = 1; arr2.length>0 || arr1.length>0;i++){ if(arr1.length>0){ result.push(arr1.shift()); } if(i%afterElement==0 && arr2.length>0){ result.push(arr2.shift()); } } return result; } const arr1 = ['orange', 'blue', 'red', 'black', 'white', 'magenta', 'cyan']; const arr2 = [1, 2, 3]; result = arrInsertAfter(arr1, arr2, 3); console.log(result);
在数组中使用forEach
在回调函数中进行变异可能很棘手。 下面是使用传统的循环和不同诱变方法splice
将元素添加到a
在特定索引位置数组:
function arrInsertAfter(a, b, after) { let j = 0; for (let i = after; i < a.length && j < b.length; i += after) { a.splice(i++, 0, b[j++]); } while (j < b.length) { a.push(b[j++]); } } const arr1 = ['orange', 'blue', 'red', 'black', 'white', 'magenta', 'cyan']; const arr2 = [1, 2, 3]; arrInsertAfter(arr1, arr2, 3); console.log(arr1);
这是保留两个参数的非变异版本:
function arrInsertAfter(a, b, after) { const res = a.slice(0); let j = 0; for (let i = after; i < res.length && j < b.length; i += after) { res.splice(i++, 0, b[j++]); } while (j < b.length) { res.push(b[j++]); } return res; } const arr1 = ['orange', 'blue', 'red', 'black', 'white', 'magenta', 'cyan']; const arr2 = [1, 2, 3]; console.log(arrInsertAfter(arr1, arr2, 2)); console.log(arrInsertAfter(arr1, arr2, 3));
而且,如果您不介意将b
突变:
const arrInsertAfter = (a, b, after) => a.reduce((r, e, i) => r.concat(i && i % after === 0 && b.length ? [b.shift(), e] : [e]) , []).concat(b) ; const arr1 = ['orange', 'blue', 'red', 'black', 'white', 'magenta', 'cyan']; const arr2 = [1, 2, 3]; console.log(arrInsertAfter(arr1, arr2.slice(), 2)); console.log(arrInsertAfter(arr1, arr2, 3));
您可以复制并在正确位置拼接元素。
function insertAfter(array1, array2, n) { var temp = array1.slice(); array2.forEach((v, i) => temp.splice((i + 1) * n + i, 0, v)); return temp; } console.log(insertAfter(['orange', 'blue', 'red', 'black', 'white', 'magenta', 'cyan'], [1, 2, 3], 2)); console.log(insertAfter(['orange', 'blue', 'red', 'black', 'white', 'magenta', 'cyan'], [1, 2, 3], 3));
.as-console-wrapper { max-height: 100% !important; top: 0; }
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