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如何在JavaScript中交织两个数组?

loops javascript

totalnoob · 技术社区 · 8 年前

我对运行这个程序感到非常困惑,因为嵌套循环对我来说仍然有点滑稽。

使用以下两个数组:

let a = ['y','y','y'];
let z = ['x','x','x','x','x'];

我怎么把它打印出来 一个值来自 ,然后 z中的两个值 在那之后等等?

'y',
'x',
'x',
'y',
'x',
'x',
'y',
'x',
'x'

如果值改为:

let a = ['y','y'];
let z = ['x','x','x','x','x'];

它会打印:

'y',
'x',
'x',
'y',
'x',
'x',
'x'

这就是我目前所做的尝试:

for (let i = 0; i < a.length; i++) {
    console.log(a[i]);
    for (let j = 0; j < z.length; j++) {
      console.log(z[j], z[j+1]);
    // break?
    }
}

11 回复 | 直到 8 年前

protoproto 8 年前

我修复你的回路:

let a = ['y','y','y'];
let z = ['x','x','x','x','x'];
for (let i = 0; i < a.length; i++) {
  console.log(a[i]);
  console.log(z[i*2], z[i*2+1]);
}

Ahmet TaÅgöz 8 年前

`
let a = ['y','y'];
let z = ['x','x','x','x','x'];
let j = 0;
for (let i = 0; i < a.length; i++) {
  console.log(a[i]);
  for (j; j < z.length; j++) {
    console.log(z[j], z[j+1]);
    if(j % 2 == 1){
       break;
    }
  }
}`

试试这个。

Eddie 8 年前

你有一个选择就是 do / while 循环。使用 shift() 删除第一个第项,共项阵列

let a = ['y1', 'y2', 'y3'];
let z = ['x1', 'x2', 'x3', 'x4', 'x5', 'x6'];

//Copy the array a and z
let aTemp = [...a]; 
let zTemp = [...z];

do {
  if (aTemp.length) console.log(aTemp.shift());
  if (zTemp.length) console.log(zTemp.shift());
  if (zTemp.length) console.log(zTemp.shift());
} while (aTemp.length || zTemp.length); //Loop while both the temp variables have elements

Jonas Wilms 8 年前

  let aIndex = 0, zIndex = 0;

 while(aIndex < a.length || zIndex < z.length) {
    console.log(
      a[aIndex++],
      z[zIndex++],
      z[zIndex++]
   );
 }

WarPro 8 年前

这是一种非破坏性的方法

var pointer = 0;

    for (let i = 0; i < a.length; i++) {
      var count = 0;
      console.log(a[i]);
      for (let j = pointer; j < z.length; j++) {    
        console.log(z[j]);
        count++;

        if(count == 2){
            count = 0;
            if(i == a.length-1) { continue; }else { pointer = j+1; break; }
        }

      }
    }

Abdulfatai 8 年前

let c = j = 0;
z.map(function(item){
  if(c === 0 && a[j]){ console.log(a[j++]); }
  console.log(item);
  c = c > 0 ? 0 : c + 1;
});

htoniv 8 年前

请看一下下面的代码。我将您的预期输出结果推送到一个数组中。

    let a = ['y','y'];
    let z = ['x', 'x', 'x', 'x', 'x'];
    let arr = [];
    for (let i = 0; i < a.length; i++) {
        var count = 0
      arr.push(a[i]);
      for (let j = i * 2; j < z.length; j++) {
        arr.push(z[j]);
        count++
        if (count > 1) {
            if (z[j+1] !== undefined && a[i+1] === undefined) {
            for (let k = j+1; k < z.length; k++) {
              arr.push(z[k])
            }
          }
        break;
        }
      }
    }
console.log(arr);
// ["y", "x", "x", "y", "x", "x", "x"]

SNag 8 年前

您根本不需要嵌套循环。这可以通过一个带有两个计数器的循环来完成。

let a = ['y','y','y'];
let z = ['x','x','x','x','x'];
let result = [];

for(let i = 0, j = 0; i < a.length, j < z.length; i++, j+=2) {
	result.push(a[i], z[j], z[j+1]);
}

result = result.filter(item => item !== undefined);

console.log(result);

下面是压缩成一行核心的相同代码:

let a = ['y','y','y'];
let z = ['x','x','x','x','x'];
let result = [];

for(let i = 0, j = 0; i < a.length, j < z.length; i++, j+=2) {
	result.push.apply(result, [a[i], z[j], z[j+1]].filter(m => m !== undefined));
}

console.log(result);

Nina Scholz 8 年前

您可以获取长度并计算用于拼接和推送到结果集的数组的间隔/部分长度。

function getX(a, b) {
    var iA = Math.ceil(a.length / b.length),
        iB = Math.ceil(b.length / a.length),
        pA = 0,
        pB = 0,
        result = [];

    while (pA < a.length || pB < b.length) {
        result.push(
            ...a.slice(pA, pA += iA),
            ...b.slice(pB, pB += iB)
        );
    }
    return result;
}

console.log(getX(['y','y','y'], ['x','x','x','x','x']).join(' '));
console.log(getX(['y','y'], ['x','x','x','x','x']).join(' '));

dziraf 8 年前

下面是一个使用 Array.prototype.reduce() 以及 destructuring assignment :

let a = ["y", "y", "y"];
let z = ["x", "x", "x", "x", "x"];

let y = a.reduce((acc, current, index) => {
  return [...acc, current, ...z.splice(0, a[index + 1] ? 2 : z.length)];
}, []);

console.log(y);
console.log(z);

在每个 a 获取先前累积数组的数组元素(最初 [] ),添加 current 元素和两个元素 z 如果 一 有下一个元素或其他 z轴 元素如果 现在的 是 一 . 很遗憾,这会从 z轴 在这个过程中,因此您可能希望在运行之前复制它,也就是说,您可以在函数中包装reduce并传递 z轴 :

let a = ["y", "y", "y"];
let z = ["a", "b", "c", "d", "e", "f", "g"];

const interweave = (arr1, arr2) => {
  return arr1.reduce((acc, current, index) => {
    return [
      ...acc,
      current,
      ...arr2.splice(0, arr1[index + 1] ? 2 : arr2.length)
    ];
  }, []);
};

console.log(interweave(a, [...z]));
console.log(z);

Hassan Sadeghi 8 年前

你可以不用 for 或 while 我也是!试试这个:

let a = ['y','y','y'];
let z = ['x','x','x','x','x'];
console.log(
    [].concat.apply([], a.map(function(m, i){return [m].concat(z.slice(i*2, i*2+2));}))
);

现在,如果希望这样,可以加入结果:

let a = ['y','y','y'];
let z = ['x','x','x','x','x'];
console.log(
    [].concat.apply([], a.map(function(m, i){return [m].concat(z.slice(i*2, i*2+2));})).join(",")
);

或者如果你喜欢 reduce 函数,您可以尝试:

let a = ['y','y','y'];
let z = ['x','x','x','x','x'];
console.log(
  a.reduce(function(ac, m, i){return ac.push.apply(ac, [m].concat(z.slice(i*2, i*2+2))), ac;}, [])
);