如何过滤对象数组并根据它们在数组中出现的次数对它们进行排序

Question

我有一组用户对象，其中一些出现多次。我正在尝试计算每个唯一用户对象在数组中出现的次数，并对它们重新排序，以使出现频率最高的用户在数组中排在第一位，而在数组中出现最少的用户在数组中显示在最后。

const users = 
[
   {id: "0f933bbd-d1fb-4ad5-80f0-661c3c0aa2f8", handle: "lisa", email: "lisa@gmail.com", createdAt: 1593894321997},
 
   {id: "bbfc927c-f3d4-4cdd-b872-9cb233a194aa", handle: "jisoo", email: "jisoo@gmail.com", createdAt: 1592452421714},

   {id: "be942039-6a59-46a4-9f92-1f7808b20c2f", handle: "unnieJennie", email: "jennie@gmail.com", createdAt: 1593894227232},

   {id: "0f933bbd-d1fb-4ad5-80f0-661c3c0aa2f8", handle: "lisa", email: "lisa@gmail.com", createdAt: 1593894321997},

   {id: "1d1e31cd-eff2-47de-b46d-4d45bb2dd97f", handle: "kawikaLovesKpop", email: "kawika@gmail.com", createdAt: 1593894550566},

   {id: "be942039-6a59-46a4-9f92-1f7808b20c2f", handle: "unnieJennie", email: "jennie@gmail.com", createdAt: 1593894227232},

   {id: "0f933bbd-d1fb-4ad5-80f0-661c3c0aa2f8", handle: "lisa", email: "lisa@gmail.com", createdAt: 1593894321997},

   {id: "1d1e31cd-eff2-47de-b46d-4d45bb2dd97f", handle: "kawikaLovesKpop", email: "kawika@gmail.com", createdAt: 1593894550566},

   {id: "be942039-6a59-46a4-9f92-1f7808b20c2f", handle: "unnieJennie", email: "jennie@gmail.com", createdAt: 1593894227232}
]

我尝试过将 countBy 方法与 lodash 一起使用

const bestMatched = countBy(users, "handle");

返回

{lisa: 3, jisoo: 1, unnieJennie: 3, kawikaLovesKpop: 2}

我仍然需要访问包含所有对象键和值的用户数组。

Answer 1

您可以按位使用NOT 来执行此操作，因为~undefined = -1 并且我们需要一个对象来存储我们的结果：

const result = users.reduce((m,{handle})=>({...m, [handle]:-~m[handle]}), {})

试试看：

const users = 
[
   {id: "0f933bbd-d1fb-4ad5-80f0-661c3c0aa2f8", handle: "lisa", email: "lisa@gmail.com", createdAt: 1593894321997},
 
   {id: "bbfc927c-f3d4-4cdd-b872-9cb233a194aa", handle: "jisoo", email: "jisoo@gmail.com", createdAt: 1592452421714},

   {id: "be942039-6a59-46a4-9f92-1f7808b20c2f", handle: "unnieJennie", email: "jennie@gmail.com", createdAt: 1593894227232},

   {id: "0f933bbd-d1fb-4ad5-80f0-661c3c0aa2f8", handle: "lisa", email: "lisa@gmail.com", createdAt: 1593894321997},

   {id: "1d1e31cd-eff2-47de-b46d-4d45bb2dd97f", handle: "kawikaLovesKpop", email: "kawika@gmail.com", createdAt: 1593894550566},

   {id: "be942039-6a59-46a4-9f92-1f7808b20c2f", handle: "unnieJennie", email: "jennie@gmail.com", createdAt: 1593894227232},

   {id: "0f933bbd-d1fb-4ad5-80f0-661c3c0aa2f8", handle: "lisa", email: "lisa@gmail.com", createdAt: 1593894321997},

   {id: "1d1e31cd-eff2-47de-b46d-4d45bb2dd97f", handle: "kawikaLovesKpop", email: "kawika@gmail.com", createdAt: 1593894550566},

   {id: "be942039-6a59-46a4-9f92-1f7808b20c2f", handle: "unnieJennie", email: "jennie@gmail.com", createdAt: 1593894227232}
]

const result = users.reduce((m,{handle})=>({...m, [handle]:-~m[handle]}), {})

console.log({result})

由于对象是无序的，所以如果你想有一个有序的集合，你需要使用数组。

const result = users.reduce((m,{handle})=>({...m, [handle]:-~m[handle]}), {})
Object.keys(result).sort((a,b)=>result[a]-result[b]).map(key=>({[key]:result[key]}))

Answer 2

试试这个：

let counterObj = {}

for (user of users){
    counterObj[user.handle] = 1 + (counterObj[user.handle] || 0)
}

console.log(counterObj)

Answer 3

const users = 
[
   {id: "0f933bbd-d1fb-4ad5-80f0-661c3c0aa2f8", handle: "lisa", email: "lisa@gmail.com", createdAt: 1593894321997},
 
   {id: "bbfc927c-f3d4-4cdd-b872-9cb233a194aa", handle: "jisoo", email: "jisoo@gmail.com", createdAt: 1592452421714},

   {id: "be942039-6a59-46a4-9f92-1f7808b20c2f", handle: "unnieJennie", email: "jennie@gmail.com", createdAt: 1593894227232},

   {id: "0f933bbd-d1fb-4ad5-80f0-661c3c0aa2f8", handle: "lisa", email: "lisa@gmail.com", createdAt: 1593894321997},

   {id: "1d1e31cd-eff2-47de-b46d-4d45bb2dd97f", handle: "kawikaLovesKpop", email: "kawika@gmail.com", createdAt: 1593894550566},

   {id: "be942039-6a59-46a4-9f92-1f7808b20c2f", handle: "unnieJennie", email: "jennie@gmail.com", createdAt: 1593894227232},

   {id: "0f933bbd-d1fb-4ad5-80f0-661c3c0aa2f8", handle: "lisa", email: "lisa@gmail.com", createdAt: 1593894321997},

   {id: "1d1e31cd-eff2-47de-b46d-4d45bb2dd97f", handle: "kawikaLovesKpop", email: "kawika@gmail.com", createdAt: 1593894550566},

   {id: "be942039-6a59-46a4-9f92-1f7808b20c2f", handle: "unnieJennie", email: "jennie@gmail.com", createdAt: 1593894227232}
]

ob={lisa: 3, jisoo: 1, unnieJennie: 3, kawikaLovesKpop: 2}
map=new Map()
users.map(x=>map.set(x.handle,x))
res=Object.entries(ob).sort((a,b)=>b[1]-a[1])
.flatMap(o=>Array(o[1]).fill(map.get(o[0])))


console.log(res)

Answer 4

我相信这可能就是您想要的。 JSON 中也有一些错误，但现在应该没问题了。

const users = [
  {
    id: '0f933bbd-d1fb-4ad5-80f0-661c3c0aa2f8',
    handle: 'lisa',
    email: 'lisa@gmail.com',
    createdAt: 1593894321997,
  },
  {
    id: 'bbfc927c-f3d4-4cdd-b872-9cb233a194aa',
    handle: 'jisoo',
    email: 'jisoo@gmail.com',
    createdAt: 1592452421714,
  },
  {
    id: 'be942039-6a59-46a4-9f92-1f7808b20c2f',
    handle: 'unnieJennie',
    email: 'jennie@gmail.com',
    createdAt: 1593894227232,
  },
  {
    id: '0f933bbd-d1fb-4ad5-80f0-661c3c0aa2f8',
    handle: 'lisa',
    email: 'lisa@gmail.com',
    createdAt: 1593894321997,
  },
  {
    id: '1d1e31cd-eff2-47de-b46d-4d45bb2dd97f',
    handle: 'kawikaLovesKpop',
    email: 'kawika@gmail.com',
    createdAt: 1593894550566,
  },
  {
    id: 'be942039-6a59-46a4-9f92-1f7808b20c2f',
    handle: 'unnieJennie',
    email: 'jennie@gmail.com',
    createdAt: 1593894227232,
  },
  {
    id: '0f933bbd-d1fb-4ad5-80f0-661c3c0aa2f8',
    handle: 'lisa',
    email: 'lisa@gmail.com',
    createdAt: 1593894321997,
  },
  {
    id: '1d1e31cd-eff2-47de-b46d-4d45bb2dd97f',
    handle: 'kawikaLovesKpop',
    email: 'kawika@gmail.com',
    createdAt: 1593894550566,
  },
  {
    id: 'be942039-6a59-46a4-9f92-1f7808b20c2f',
    handle: 'unnieJennie',
    email: 'jennie@gmail.com',
    createdAt: 1593894227232,
  },
];

function sort(jsonArray) {
  var containerArray = [];
  jsonArray.map((item, index) => {
    if (item === 'null') {
    } else {
      var newArray = [];
      var itemHandle = item.handle;
      jsonArray.map((i, index) => {
        if (itemHandle === i.handle && itemHandle !== 'null') {
          newArray.push(i);
          jsonArray.splice(index, 1, 'null');
        }
      });
      jsonArray.splice(index, 1, 'null');
      containerArray.push(newArray);
    }
  });
  return containerArray;
}

var finalArray = sort(users);

//Sorts the finalArray by highest occurrence to lowest
finalArray.sort(function(a,b){
  return b.length - a.length
})

console.log(finalArray)

将对象放在一起，以便您仍然可以通过点表示法访问它们的属性。