将计数值添加到具有相同键值的对象（对象数组）

Question

您好，我要对一系列对象进行复杂的迭代。 我有这样的数组：

[ 
  { name: 'Jacob', lastName: 'Smith', dob: '1995-11-29' },
  { name: 'Jacob', lastName: 'Smith', dob: '1991-08-21' },
  { name: 'Ann', lastName: 'Smith', dob: '1995-11-29' },
  { name: 'Ann', lastName: 'Nansen', dob: '1983-01-01' },
  { name: 'Jacob', lastName: 'Smith', dob: '1985-06-15' },
  { name: 'Jacob', lastName: 'Smith', dob: '1995-11-29' },
  { name: 'Ann', lastName: 'Smith', dob: '2010-11-29' },
]

我想将count属性添加到对具有相同名称和姓氏的对象进行计数的每个对象中，因此现在应该是：

[ 
  { name: 'Jacob', lastName: 'Smith', count: 4 },
  { name: 'Ann', lastName: 'Smith', count: 2 },
  { name: 'Ann', lastName: 'Nansen', count: 1' },
]

Answer 1

您可以使用Array.reduce和Object.values

• 使用键作为名称和姓氏组合（其中值是结果对象）转换对象中的数组。
• 从对象获取所有值作为最终结果

 let arr = [{ name: 'Jacob', lastName: 'Smith', dob: '1995-11-29' },{ name: 'Jacob', lastName: 'Smith', dob: '1991-08-21' },{ name: 'Ann', lastName: 'Smith', dob: '1995-11-29' },{ name: 'Ann', lastName: 'Nansen', dob: '1983-01-01' },{ name: 'Jacob', lastName: 'Smith', dob: '1985-06-15' },{ name: 'Jacob', lastName: 'Smith', dob: '1995-11-29' },{ name: 'Ann', lastName: 'Smith', dob: '2010-11-29' }]; let result = Object.values(arr.reduce((a,{name, lastName}) => { let key = `${name}_${lastName}`; a[key] = a[key] || {name, lastName, count : 0}; a[key].count++; return a; }, {})); console.log(result); 

Answer 2

  const hash = [];

  for(const { name, lastName } of persons) {
    const key = name + "/" + lastName;
    if(!hash[key]) hash[key] = {
      name,
      lastName,
      count: 0,
    };

    hash[key].count++;
 }

  const result = Object.values(hash);

Answer 3

您可以使用JSON.stringify以安全的方式组合名称和姓氏。 我喜欢使用Map将具有相同键的记录分组在一起：

 const data = [{ name: 'Jacob', lastName: 'Smith', dob: '1995-11-29' },{ name: 'Jacob', lastName: 'Smith', dob: '1991-08-21' },{ name: 'Ann', lastName: 'Smith', dob: '1995-11-29' },{ name: 'Ann', lastName: 'Nansen', dob: '1983-01-01' },{ name: 'Jacob', lastName: 'Smith', dob: '1985-06-15' },{ name: 'Jacob', lastName: 'Smith', dob: '1995-11-29' },{ name: 'Ann', lastName: 'Smith', dob: '2010-11-29' }]; const keyed = data.map(o => [JSON.stringify([o.name, o.lastName]), o]); const map = new Map(keyed.map(([key, {name, lastName}]) => [key, {name, lastName, count: 0}])); keyed.forEach(([key, o]) => map.get(key).count++); const result = Array.from(map.values()); console.log(result); 

Answer 4

 let arr=[ { name: 'Jacob', lastName: 'Smith', dob: '1995-11-29' }, { name: 'Jacob', lastName: 'Smith', dob: '1991-08-21' }, { name: 'Ann', lastName: 'Smith', dob: '1995-11-29' }, { name: 'Ann', lastName: 'Nansen', dob: '1983-01-01' }, { name: 'Jacob', lastName: 'Smith', dob: '1985-06-15' }, { name: 'Jacob', lastName: 'Smith', dob: '1995-11-29' }, { name: 'Ann', lastName: 'Smith', dob: '2010-11-29' }, ]; let outerArr=[]; for(arrValue of arr) { delete arrValue.dob let index=outerArr.findIndex(item=> item.name==arrValue.name && item.lastName==arrValue.lastName); if(index==-1) { let arrFind=arr.filter(item=> item.name==arrValue.name && item.lastName==arrValue.lastName) arrValue.count=arrFind.length outerArr.push(arrValue) } } console.log('result',outerArr) 

Answer 5

您可以通过减少原始数组来实现。 在遍历people ，可以检查是否已使用Array.some对它们进行“分组”-如果尚未将其构建的对象推入先前返回的Array。

 const getInstances = ({ name, lastName }, data) => data.filter(d => d.name === name && d.lastName === lastName).length const people = [ { name: 'Jacob', lastName: 'Smith', dob: '1995-11-29' }, { name: 'Jacob', lastName: 'Smith', dob: '1991-08-21' }, { name: 'Ann', lastName: 'Smith', dob: '1995-11-29' }, { name: 'Ann', lastName: 'Nansen', dob: '1983-01-01' }, { name: 'Jacob', lastName: 'Smith', dob: '1985-06-15' }, { name: 'Jacob', lastName: 'Smith', dob: '1995-11-29' }, { name: 'Ann', lastName: 'Smith', dob: '2010-11-29' }, ] const groupedPeople = people.reduce((group, person, i, people) => { const alreadyBeenGrouped = group.some(({ name, lastName }) => name === person.name && lastName === person.lastName) if (!alreadyBeenGrouped) { group.push({ name: person.name, lastName: person.lastName, count: getInstances(person, people) }) } return group }, []) console.log(groupedPeople)