使用 JavaScript 混合数组

Question

像下图那样混合多个数组的最佳方法是什么，

PS：

• 不知道每个数组的长度是多少

• 数组将包含 +10000 个元素

• 会有3个以上的数组

我为它制定了解决方案，但我正在寻找更好的解决方案

这是我自己的解决方案，我正在寻找更好的想法

import { compact, flattenDeep } from "lodash/array";

export const doTheMagic = master => {
  const longest = master.reduce((p, c, i, a) => (a[p].length > c.length ? p : i), 0);
  const mixed = master[longest].map((i, k) => {
    return master.map((o, a) => {
      if (master[a][k]) return master[a][k];
    });
  });
  const flaten = flattenDeep(mixed);
  return  compact(flaten);// to remove falsey values
};
const one = [1,2,3];
const two = ['a','b','c','d','e'];
const three = ['k','l','m','n'];
const mix = doTheMagic([one,two,three]);
console.log('mix: ', mix);

Answer 1

您可以使用 lodash 作为您的解决方案。

const { flow, zip, flatten, filter} = _

const doTheMagic = flow(
  zip,
  flatten,
  filter
)

const one = [1, 2, 3]
const two = ['?', '?', '?', '?', '?', '?']
const three = ['foo', 'bar', 'wow', 'okay']

const result = doTheMagic(one, two, three)

console.log(result)

<script src="https://cdn.jsdelivr.net/npm/lodash@4.17.15/lodash.min.js"></script>

适用于不同长度的数组，并利用函数式编程来编写优雅的代码。

这是一个要运行的代码笔：https://codepen.io/matteodem/pen/mddQrwe

Answer 2

这是我自己的解决方案，我正在寻找更好的想法

import { compact, flattenDeep } from "lodash/array";

export const doTheMagic = master => {
  const longest = master.reduce((p, c, i, a) => (a[p].length > c.length ? p : i), 0);
  const mixed = master[longest].map((i, k) => {
    return master.map((o, a) => {
      if (master[a][k]) return master[a][k];
    });
  });
  const flaten = flattenDeep(mixed);
  return  compact(flaten);// to remove falsey values
};
const one = [1,2,3];
const two = ['a','b','c','d','e'];
const three = ['k','l','m','n'];
const mix = doTheMagic([one,two,three]);
console.log('mix: ', mix);

Answer 3

let a1 = [1, 2, 3, 4, 5];
let a2 = ["?", "?", "?", "?", "?", "?"];
let a3 = ['one', 'two', 'three', 'four', 'five'];

const doTheMagic = arrayOfArrays => {
  let maxLength = 0;
  let result = [];
  for (arr in arrayOfArrays) {
    maxLength = Math.max(maxLength, arrayOfArrays[arr].length);
  }
  for (let i = 0; i < maxLength; i++) {
    for (let j = 0; j < arrayOfArrays.length; j++) {
      if (arrayOfArrays[j][i]) {
        result.push(arrayOfArrays[j][i]);
      }
    }
  }
  return result;
}

console.log(doTheMagic([a1, a2, a3]));

Answer 4

这适用于未知数量的数组，每个数组的长度未知：

const arrays = [
  [1, 2, 3, 4],
  ["a", "b", "c", "d", "e"],
  ["@", "#", "?"]
];

let output = [];

while (arrays.some(a => a.length)) { // While any of the arrays still has an element in it, keep going
  for (let a of arrays) {
    if (!a.length) continue;
    output.push(a.shift()); // remove the first element of the array and add it to output
  }
}

console.log(output)

Answer 5

这是我实现这一目标的方法，一个for loop 可以做到。如果您也不知道数组的数量和数组长度，这将起作用。

function doTheMagic(arr){
  let ans = [];
  let maxLength = -1;
  
  arr.forEach((tempArr)=>{
    if(tempArr.length > maxLength){
      maxLength = tempArr.length;
    }
  })
  
  let l1=0,l2=0,l3=0;
  
  for(let i=0;i<maxLength;i++){
    arr.forEach((tempArr)=>{
      if(tempArr[i]){
        ans.push(tempArr[i]);
      }
    })

  }
  return ans;
}


let a1 = [1,2,3,4,5];
let a2 = ["?","?","?","?","?","?"];
let a3 = ['1','2','3','4','5'];


console.log(doTheMagic([a1,a2,a3]));

Answer 6

不确定这是否更好，但是如何编写代码来处理传入的任意数量的数组。

const weave = (...args) =>  // convert arguments to an array
  args.reduce((res, arr, offset) => {  // loop over the arrays 
    arr.forEach((v, i) => res[offset + i * args.length] = v) // loop over array and add items to their indexes
    return res
  }, []).filter(x => x !== undefined) // remove the unused indexes


const one = [1, 2, 3]
const two = ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i']
const three = ['w', 'x', 'y', 'z']
const result = weave(one, two, three)
console.log(result)

const result2 = weave(one, two)
console.log(result2)

const result3 = weave(one, two, three, ['*', '&'])
console.log(result3)