Question

我需要一些帮助来整理这些数据，我有一系列产品，我需要按设置配置进行排序和显示。如果显示为真，输出必须与设置数组（索引）具有相同的顺序。提前致谢。这是我尝试过的：

var products = [
    {id: 0, name: 'Chocolate', category: 'Sweet'},
    {id: 1, name: 'Almendras', category: 'Fruit'},
    {id: 2, name: 'Nueces', category: 'Fruit'},
    {id: 3, name: 'Mermelada', category: 'Jam'},
    {id: 4, name: 'Alfajor', category: 'Sweet'},
    {id: 5, name: 'Queso', category: 'UwU'},
    {id: 6, name: 'Arandanos', category: 'Fruit'},
    {id: 7, name: 'Maracuya', category: 'Fruit'}
];

let settings = [
    {
        name: 'Fruit',
        display: true
    },
    {
        name: 'Jam',
        display: false
    },
    {
        name: 'Sweet',
        display: true
    },
    {
        name: 'UwU',
        display: true
    }
]


let group = products.reduce((r, a) => {
      r[a.category] = [...r[a.category] || [], a];
      return r;
    }, {});

let arrangedProducts = Object.keys(group); 

console.log(group);
console.log(arrangedProducts);

这是我的预期输出：

/*
expected result = [
    [
        {id: 1, name: 'Almendras', category: 'Fruit'}, 
        {id: 2, name: 'Nueces', category: 'Fruit'}, 
        {id: 6, name: 'Arandanos', category: 'Fruit'},
        {id: 7, name: 'Maracuya', category: 'Fruit'}
    ],
    [
        {id: 0, name: 'Chocolate', category: 'Sweet'},
        {id: 4, name: 'Alfajor', category: 'Sweet'}
    ],
    [
        {id: 5, name: 'Queso', category: 'UwU'}
    ]
]
*/

Answer 1

解决方案

组建团体
应用设置并检索结果

const products = [
  { id: 0, name: "Chocolate", category: "Sweet" },
  { id: 1, name: "Almendras", category: "Fruit" },
  { id: 2, name: "Nueces", category: "Fruit" },
  { id: 3, name: "Mermelada", category: "Jam" },
  { id: 4, name: "Alfajor", category: "Sweet" },
  { id: 5, name: "Queso", category: "UwU" },
  { id: 6, name: "Arandanos", category: "Fruit" },
  { id: 7, name: "Maracuya", category: "Fruit" },
];

const productsGroup = products.reduce((r, a) => {
  r[a.category] = [...(r[a.category] || []), a];
  return r;
}, {});

function applySettings(settings) {
  return settings.filter((s) => s.display).map((s) => productsGroup[s.name]);
}

console.log(
  applySettings([
    {
      name: "Fruit",
      display: true,
    },
    {
      name: "Jam",
      display: false,
    },
  ])
);

console.log(
  applySettings([
    {
      name: "Fruit",
      display: true,
    },
    {
      name: "Sweet",
      display: true,
    },
    {
      name: "UwU",
      display: true,
    },
  ])
);

Answer 2

您可以根据 settings 属性filter display 列表，然后使用 Array.map 返回 products 中与类别匹配的对象列表:

const products = [
    {id: 0, name: 'Chocolate', category: 'Sweet'},
    {id: 1, name: 'Almendras', category: 'Fruit'},
    {id: 2, name: 'Nueces', category: 'Fruit'},
    {id: 3, name: 'Mermelada', category: 'Jam'},
    {id: 4, name: 'Alfajor', category: 'Sweet'},
    {id: 5, name: 'Queso', category: 'UwU'},
    {id: 6, name: 'Arandanos', category: 'Fruit'},
    {id: 7, name: 'Maracuya', category: 'Fruit'}
];

const settings = [
    { name: 'Fruit', display: true },
    { name: 'Jam', display: false },
    { name: 'Sweet', display: true },
    { name: 'UwU', display: true }
];

const result = settings
  .filter(c => c.display)
  .map(c => products.filter(o => o.category == c.name));

console.log(result);

请注意，此代码确实为每个具有 products 的 settings 值过滤了 display:true 数组，因此对于大型数组可能会很慢。然而，filter 的开销非常低，使用 OP 的示例数据进行的测试表明，它的运行速度是 reduce 版本的 3 倍；并使用更大的 products 数组（99 个条目）使运行速度提高 10 倍。

Answer 3

这应该很快，因为当 continue 为 display 时，它会false 进入下一次迭代而不执行内循环：

var products = [
    {id: 0, name: 'Chocolate', category: 'Sweet'},
    {id: 1, name: 'Almendras', category: 'Fruit'},
    {id: 2, name: 'Nueces', category: 'Fruit'},
    {id: 3, name: 'Mermelada', category: 'Jam'},
    {id: 4, name: 'Alfajor', category: 'Sweet'},
    {id: 5, name: 'Queso', category: 'UwU'},
    {id: 6, name: 'Arandanos', category: 'Fruit'},
    {id: 7, name: 'Maracuya', category: 'Fruit'}
];

let settings = [
    {
        name: 'Fruit',
        display: true
    },
    {
        name: 'Jam',
        display: false
    },
    {
        name: 'Sweet',
        display: true
    },
    {
        name: 'UwU',
        display: true
    }
];
function sortProducts(){
  const r = [];
  let i = -1;
  for(let s of settings){
    if(!s.display){
      continue;
    }
    i++;
    for(let o of products){
      if(s.name === o.category){
        if(r[i]){
          r[i].push(o);
        }
        else{
          r.push([o]);
        }
      }
    }
  }
  return r;
}
console.log(sortProducts());

按另一个对象数组引用对对象数组进行排序

3 个答案:

解决方案