Question

我知道有很多方法可以过滤数组中的唯一值，但是对于给定字段具有唯一值的对象的过滤数组呢？

例如，我有[obj1, obj2, obj3, ...]，其中每个对象的格式如下：

{
firstName: "...",
lastName: "..."
}

如何过滤数组以最终数组为止所有对象都有唯一的名字？单行会更好，但不会以可读性为代价。

Answer 1

仅过滤那些先前未在数组中找到的项目。我们将cond定义为返回是否应该考虑两个项目＆＃34;相等＆＃34;。

＆＃13;

function uniqueBy(a, cond) {
  return a.filter((e, i) => a.findIndex(e2 => cond(e, e2)) === i);
}

const test = [
  { firstname: "John", lastname: "Doe" },
  { firstname: "Jane", lastname: "Doe" },
  { firstname: "John", lastname: "Smith" }
];

console.log(uniqueBy(test, (o1, o2) => o1.firstname === o2.firstname));

＆＃13;

Answer 2

您可以将数组缩减为Map，然后将地图值传播回数组。

如果您想要保持遇到第一个对象，您必须检查地图是否已将firstName作为关键字，并且只有在它尚不存在时才设置。

＆＃13;

const arr = [{"firstname":"Robert","lastname":"Smith"},{"firstname":"Francis","lastname":"Collins"},{"firstname":"Robert","lastname":"Ludlum"},{"firstname":"Francis","lastname":"bacon"}];
              
const result = [...arr.reduce((map, obj) => map.has(obj.firstname) ? map : map.set(obj.firstname, obj), new Map()).values()];

console.log(result);

＆＃13;

如果您想保留最后一个对象，可以跳过检查，只需设置它们：

＆＃13;

const arr = [{"firstname":"Robert","lastname":"Smith"},{"firstname":"Francis","lastname":"Collins"},{"firstname":"Robert","lastname":"Ludlum"},{"firstname":"Francis","lastname":"bacon"}];

const result = [...arr.reduce((map, obj) => map.set(obj.firstname, obj), new Map()).values()];

console.log(result);

＆＃13;

Answer 3

您可以使用reduce方法，并将初始值设置为数组，有条件地将对象推送到新数组，即如果第一个名称尚不存在：

var arr = [ { firstname: "John",
              lastname: "Doe" },
            { firstname: "Jane",
              lastname: "Doe" },
            { firstname: "John",
              lastname: "Doe" }];
             

console.log(
  arr.reduce(
    function(unique_arr, obj) {
      if(!unique_arr.some(x => x.firstname === obj.firstname)) {
        unique_arr.push(obj)
      }
      return unique_arr; 
    }, [])
);

Answer 4

2班轮怎么样。

var ar = [{firstName: 'Jo'}, {firstName: 'Bill'}, {firstName: 'Bill'}];

var firstNames = [];
var uniqueFirstName = ar.filter(obj =>
  firstNames.indexOf(obj.firstName) > -1
      ? false : firstNames.push(obj.firstName))
console.log( JSON.stringify(uniqueFirstName));

Answer 5

答案似乎是要检查结果数组是否具有值，但是对于唯一性，我发现使用对象并始终设置值更容易，更易于管理，而无需检查它。这依赖于对象可能只有唯一键的前提：

＆＃13;

var arr = [{
    firstName: "Amy",
    lastName: "Adams"
  },
  {
    firstName: "Bryan",
    lastName: "Heart"
  },
  {
    firstName: "Amy",
    lastName: "Adams"
  }
];


var unique = {};

for (var i = 0, n = arr.length; i < n; i++) {
  // Key is your unique condition
  var key = [arr[i].firstName, arr[i].lastName].join(' ');
  unique[key] = arr[i];
}

console.log('Keys:', Object.keys(unique));     // only shown for austerity
console.log('Values:', Object.values(unique));

＆＃13;

Answer 6

我将过滤并保留一组已经出现的thr名：

 const names = new Set();

 const result = array.filter(({ firstName }) => !names.has(firstName) && names.add(firstName));

筛选唯一字段值的数组

6 个答案: