通过两个键对对象数组进行排序和分组，并使用新键创建分组项数组

Question

我有以下数组：

const items = [
  { name: 'john', class: 'one', section: 'a' },
  { name: 'john2', class: 'one', section: 'b' },
  { name: 'john3', class: 'one', section: 'a' },
  { name: 'john4', class: 'two', section: 'a' },
  { name: 'john5', class: 'two', section: 'b' },
  { name: 'john6', class: 'two', section: 'b' },
  { name: 'john7', class: 'three', section: 'a' },
  { name: 'john8', class: 'four', section: 'a' },
  { name: 'john9', class: 'four', section: 'b' }
];

我希望它以这种方式分组：

[
    {
        'oneA': [
            { name: 'john', class: 'one', section: 'a' },
            { name: 'john3', class: 'one', section: 'a' }
        ]
    },
    {
        'oneB': [
            { name: 'john2', class: 'one', section: 'b' }
        ]
    },
    {
        'twoA': [
            { name: 'john4', class: 'two', section: 'a' }
        ]
    },
    {
        'twoB': [
            { name: 'john5', class: 'two', section: 'b' },
            { name: 'john6', class: 'two', section: 'b' }
        ]
    },
    {
        'threeA': [
            { name: 'john7', class: 'three', section: 'a' }
        ]
    },
    {
        'fourA': [
            { name: 'john8', class: 'four', section: 'a' }
        ]
    },
    {
        'fourB': [
            { name: 'john9', class: 'four', section: 'b' }
        ]
    }
]

我尝试过这样的事情：

items.sort(function (a, b) {
  if (a.class > b.class) return 1;
  if (a.class < b.class) return -1;

  if (a.section > b.section) return 1;
  if (a.section < b.section) return -1;
})

这可以按我的意愿对数组进行排序，但它没有像我上面描述的那样分组。 你知道达到那个目的的方法吗？

Answer 1

您可以通过 Map 将它们分组。

它生成一个class 和大写section 的键，尝试从映射中获取值或获取一个空数组以使用实际对象传播一个新数组。然后它将这个数组设置为地图中的新值。

Array.from 从映射中获取所有键/值对并使用computed property name 构建新对象。

const
    getKey = o => `${o.class}${o.section.toUpperCase()}`,
    items = [{ name: 'john', class: 'one', section: 'a' }, { name: 'john2', class: 'one', section: 'b' }, { name: 'john3', class: 'one', section: 'a' }, { name: 'john4', class: 'two', section: 'a' }, { name: 'john5', class: 'two', section: 'b' }, { name: 'john6', class: 'two', section: 'b' }, { name: 'john7', class: 'three', section: 'a' }, { name: 'john8', class: 'four', section: 'a' }, { name: 'john9', class: 'four', section: 'b' }],
    result = Array.from(
        items.reduce(
            (m, o) => m.set(getKey(o), [...(m.get(getKey(o)) || []), o]),
            new Map
        ),
        ([k, v]) => ({ [k]: v })
    );

console.log(result);

.as-console-wrapper { max-height: 100% !important; top: 0; }

Answer 2

const items = [
    { name: 'john', class: 'one', section: 'a' },
    { name: 'john2', class: 'one', section: 'b' },
    { name: 'john3', class: 'one', section: 'a' },
    { name: 'john4', class: 'two', section: 'a' },
    { name: 'john5', class: 'two', section: 'b' },
    { name: 'john6', class: 'two', section: 'b' },
    { name: 'john7', class: 'three', section: 'a' },
    { name: 'john8', class: 'four', section: 'a' },
    { name: 'john9', class: 'four', section: 'b' }
];

let newArray = items.reduce((main, curr) => {
    let index = main.findIndex(obj => Object.keys(obj).includes(curr.class + curr.section))
    if (index == -1) main.push({ [curr.class + curr.section]: [curr] })
    else main[index][curr.class + curr.section].push(curr)
    return main
}, [])
console.log(newArray)

Answer 3

您可以使用reduce 函数，例如：

const items = [
  { name: 'john', class: 'one', section: 'a' },
  { name: 'john2', class: 'one', section: 'b' },
  { name: 'john3', class: 'one', section: 'a' },
  { name: 'john4', class: 'two', section: 'a' },
  { name: 'john5', class: 'two', section: 'b' },
  { name: 'john6', class: 'two', section: 'b' },
  { name: 'john7', class: 'three', section: 'a' },
  { name: 'john8', class: 'four', section: 'a' },
  { name: 'john9', class: 'four', section: 'b' }
];

const groupedItems = items.reduce((result, current) => {
  const key = current.class + current.section.toUpperCase();
  
  if (Array.isArray(result[key])) {
    result[key].push(current);
  } else {
    result[key] = [current];
  }
  
  return result;
}, {});

const expectedResult = Array.from(Object.keys(groupedItems), (elem) => ({[elem]: groupedItems[elem]}));

console.log(expectedResult);

Answer 4

我建议省略结果的外部数组，因为对我来说，如果你想创建一组对象，那么拥有它是没有意义的。

要解决您的问题，只需遍历所有项目并检查您的结果对象是否已经具有所需的键。如果没有，则创建它，然后将当前项插入到相应的数组中。

const items = [
  { name: 'john', class: 'one', section: 'a' },
  { name: 'john2', class: 'one', section: 'b' },
  { name: 'john3', class: 'one', section: 'a' },
  { name: 'john4', class: 'two', section: 'a' },
  { name: 'john5', class: 'two', section: 'b' },
  { name: 'john6', class: 'two', section: 'b' },
  { name: 'john7', class: 'three', section: 'a' },
  { name: 'john8', class: 'four', section: 'a' },
  { name: 'john9', class: 'four', section: 'b' }
];

let result = {};

items.forEach(item => {
  let key = `${item.class}${item.section.toUpperCase()}`;
  if(!result.hasOwnProperty(key)) {
    result[key] = [];
  }
  result[key].push(item);
});

console.info(result);

Answer 5

如果您希望在不使用内置函数的情况下使用纯 javascript 获得答案以便更好地理解。

var items = [
    { name: 'john', class: 'one', section: 'a' },
    { name: 'john2', class: 'one', section: 'b' },
    { name: 'john3', class: 'one', section: 'a' },
    { name: 'john4', class: 'two', section: 'a' },
    { name: 'john5', class: 'two', section: 'b' },
    { name: 'john6', class: 'two', section: 'b' },
    { name: 'john7', class: 'three', section: 'a' },
    { name: 'john8', class: 'four', section: 'a' },
    { name: 'john9', class: 'four', section: 'b' }
];
var resultArray =[];
for(var i=0;i<items.length;i++){
    if(resultArray[items[i]['class']+items[i]['section'].toUpperCase()]){
        resultArray[items[i]['class']+items[i]['section'].toUpperCase()].push(items[i]);
    } else {
        var a ="";
        a = items[i]['class']+items[i]['section'].toUpperCase();
        resultArray[a]=[];
        resultArray[a].push(items[i])
    }
}

console.log(resultArray);