将数组转换为子数组的数组

Question

我正在寻找将数组转换为子数组的数组。

例如：

let arr = [1, 2, 3, 4, 5, 6, 7, 8]
arr = [[1, 2], [3, 4], [5, 6], [7, 8]]

现在如果这个数组的长度不能被2整除

let arr = [1, 2, 3, 4, 5, 6, 7]
arr = [[1, 2], [3, 4], [5, 6], [7]]

我对这个问题的解决方案，IMO 不是很干净

const isPrime = num => {
   for (let i = 2; i < num; i++)
      if (num % i === 0) return false;
    return num > 1;
}
        
const howToGroupArray = (array) => {
    if (array.length === 1) {
        return [1, 0] // group of 1 remainder 0
    }
    else if (!isPrime(array.length) || array.length === 2) {
        return [2, 0] // groups of 2 remainder 0
    }
    else {
        return [2, 1] // groups of 2 and a group of 1
    }

}
        
let values = [1,2,3,4,5,6,7]
let newArray = []
let groups = values.forEach((x, i) => {
    if ((i === 0 || i % 2 === 0) && (i !== (values.length - 1))) {
        newArray.push([x, values[i + 1]])
    }
    else if ((i % 2 === 1) && (i !== (values.length - 1))) {
        null
    }
    else {
        howToGroupArray(values)[1] === 0 ? null : newArray.push([x])
    }
})

Answer 1

您可以只使用索引进行迭代并将其递增 2。

然后取索引进行切片，将部分放入结果数组

const
    getGrouped = array => {
        let result = [],
            i = 0;
        
        while (i < array.length) result.push(array.slice(i, i += 2));

        return result;
    };

console.log(getGrouped([1, 2, 3, 4, 5, 6, 7, 8]));
console.log(getGrouped([1, 2, 3, 4, 5, 6, 7]));

Answer 2

Ngoc Vuong 写了一篇关于分块数组的 nice blog post。他在他的 post 中提供了一些解决方案，我已将其粘贴在下方。

遍历数组

const myArray = [1, 2, 3, 4, 5, 6, 7, 8, 9];
const chunkedArray = chunk(myArray, 2);
console.log(chunkedArray);

function chunk(array, size) {
    const chunkedArray = [];
    for (let i = 0; i < array.length; i++) {
        const last = chunkedArray[chunkedArray.length - 1];
        if (!last || last.length === size) {
            chunkedArray.push([array[i]]);
        } else {
            last.push(array[i]);
        }
    }
    return chunkedArray;
}

遍历块数

const myArray = [1, 2, 3, 4, 5, 6, 7, 8, 9];
const chunkedArray = chunk(myArray, 2);
console.log(chunkedArray);

function chunk(array, size) {
    const chunkedArray = [];
    const copied = [...array];
    const numOfChild = Math.ceil(copied.length / size);
    for (let i = 0; i < numOfChild; i++) {
        chunkedArray.push(copied.splice(0, size));
    }
    return chunkedArray;
}

使用 slice()

const myArray = [1, 2, 3, 4, 5, 6, 7, 8, 9];
const chunkedArray = chunk(myArray, 2);
console.log(chunkedArray);

function chunk(array, size) {
    const chunkedArray = [];
    let index = 0;
    while (index < array.length) {
        chunkedArray.push(array.slice(index, size + index));
        index += size;
    }
    return chunkedArray;
}

递归方法

function chunk(array, size) {
    if (!array) return [];
    const firstChunk = array.slice(0, size); // create the first chunk of the given array
    if (!firstChunk.length) {
        return array; // this is the base case to terminal the recursive
    }
    return [firstChunk].concat(chunk(array.slice(size, array.length), size)); 
}

Answer 3

您可以使用 .reduce() 而不是 .forEach()，如下所示：

        
let values = [1,2,3,4,5,6,7]

let newArray = values.reduce((res, x, i, orig) =>
  i%2 ? res : [...res, [x, ...orig.slice(i+1, i+2)]]
, [])

console.log(newArray);

它使用扩展语法在每个偶数索引上构建结果。奇数仅 return 当前累积。

将数组转换为子数组的数组

convert array into array of subarrays

javascript

performance

refactoring

code-readability