如何将数组垂直映射到列?


15

如果我有一系列物品,例如,

const array = [ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18 ]

我该如何映射它,以便呈现屏幕/页面,

1     6     11    16
2     7     12    17
3     8     13    18
4     9     14
5     10    15

我能够使它与

    const chunkSize = 5;

    array
      .map((e, i) => {
        return i % chunkSize === 0 ?
           selected.slice(i, i + chunkSize)
         : null;
      })
      .filter(e => e);

但是我无法使其垂直工作。我怎样才能做到这一点?

编辑:

来自另一个答案的建议解决方案将返回子数组,这不是我在此问题中提出的。

Answers:


3

这是一个紧凑但可读的版本。

const columnize = (items, rows) => {
  const toColumns = (table, item, iteration) => {
    let row = iteration % rows;

    table[row] = table[row] || [];
    table[row].push(item);

    return table;
  };

  return items.reduce(toColumns, []);
};

可以用作:

const numbers = [ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18 ];

console.log(columnize(numbers, 5));

https://jsfiddle.net/69fshprq/

这是一种根据问题输出的方法。我没有严格注意间距,我将其留给字符串填充函数或实现的技术。

const numbers = [ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18 ]

// <pre id="out"></pre>
let out = document.getElementById('out')
let list = columnize(numbers, 5)

for (var column in list) {
    var item = list[column]
    var line = item.reduce((line, item) => line + item + '    ', '')

    out.textContent += line + ' \n'
}

https://jsfiddle.net/t60rfcpe/


11

您可以计算该行的索引。

var array = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18],
    chunk = 5,
    result = array.reduce((r, v, i) => {
        (r[i % chunk] = r[i % chunk] || []).push(v);
        return r;
    }, []);
    

result.forEach(a => console.log(...a));


很快。我花了一段时间才解决它,但是我发现你更快
Wei

6

const array = [ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18 ]
const chunkSize = 5;

let result = [];
for (let i = 0; i < chunkSize; i++) {
    result[i] = [];
}

array.forEach((e,i) => {
    result[i % chunkSize].push(e);
});

console.log(result); 
/* 
Result : 
[ [ 1, 6, 11, 16 ],
  [ 2, 7, 12, 17 ],
  [ 3, 8, 13, 18 ],
  [ 4, 9, 14 ],
  [ 5, 10, 15 ] ]
*/


您不需要for循环。我假设您使用它来设置密钥?您在forEach中执行的原理完全相同,只是模数稍微复杂一点
Martijn
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