Question

我的数组为空

locations = []

我还有另一个具有位置和城市的数组

Data = [
    { location: "01" , city: "01"},
    { location: "01" , city: "02"},
    { location: "03" , city: "03"},
    { location: "04" , city: "04"},
    { location: "01" , city: "01"},
    { location: "01" , city: "01"}
]

如何遍历数组Data[]并在locations[]中添加对象而又不像第一个对象和最后两个对象那样重复

locations = [
    { location: "01" , city: "01"},
    { location: "01" , city: "02"},
    { location: "02" , city: "02"},
    { location: "03" , city: "03"},
    { location: "04" , city: "04"}
]

Answer 1

您可以使用Array.reduce()来过滤出重复项，并获得唯一的对象数组locations：

var Data = [
{ location: "01" , city: "01"},
{ location: "01" , city: "02"},
{ location: "03" , city: "03"},
{ location: "04" , city: "04"},
{ location: "01" , city: "01"},
{ location: "01" , city: "01"}
];
var locations = Data.reduce((acc, obj)=>{
  var exist = acc.find(({location, city}) => obj.location === location && obj.city === city);
  if(!exist){
    acc.push(obj);
  }
  return acc;
},[]);
console.log(locations);

Answer 2

您可以简单地使用docker service create --limit-cpu 2 nginx并结合forEach和location并以city分隔的值来推送值

"_"

Answer 3

可以做

Data = [
  { location: "01", city: "01" },
  { location: "01", city: "02" },
  { location: "03", city: "03" },
  { location: "04", city: "04" },
  { location: "01", city: "01" },
  { location: "01", city: "01" }
]

const locations = Data.reduce( ( acc, d ) => {
  if ( !acc.some( ( a ) => a.city == d.city && a.location == d.location ) ) {
    acc.push( d )
  }
  return acc
}, [] )

console.log( locations )

Answer 4

您可以结合使用.filter()和.map()方法，例如：

var locations = Data.filter((d,i) => Data.map(dt => dt.location).indexOf(d.location) == i);

演示：

Data = [
{ location: "01" , city: "01"},
{ location: "01" , city: "02"},
{ location: "03" , city: "03"},
{ location: "04" , city: "04"},
{ location: "01" , city: "01"},
{ location: "01" , city: "01"}
];

var locations = Data.filter((d,i) => Data.map(dt => dt.location).indexOf(d.location) == i);
console.log(locations);

Answer 5

function unique(a, key) {
            var seen = {};
            return a.filter(function (item) {
                item = key ? item[key] : item;
                return seen.hasOwnProperty(item) ? false : (seen[item] = true);
            });
        }
//var locations= unique(Data , "city");
var locations= unique(unique(Data , "city") , "location");

Answer 6

您可以简单地使用Array.filter()和Set()（每个条目均设置城市和位置的组合）。如果集合中不存在某个元素，则添加该元素并将其添加到集合中。请尝试以下操作：

let Data = [ { location: "01" , city: "01"}, { location: "01" , city: "02"}, { location: "03" , city: "03"}, { location: "04" , city: "04"}, { location: "01" , city: "01"}, { location: "01" , city: "01"} ];
let set = new Set();

let result = Data.filter((e)=>{
   var key = e.location + "_" + e.city;
  if(!set.has(key)){
    set.add(key);
    return true;
  }
  return false;
});
console.log(result);

Answer 7

您可以将位置名称作为键添加到对象以进行唯一性检查，然后在使用Array.reduce完成数组迭代后返回对象值。

var data = [{
    location: "01",
    city: "01"
  },
  {
    location: "01",
    city: "02"
  },
  {
    location: "03",
    city: "03"
  },
  {
    location: "04",
    city: "04"
  },
  {
    location: "01",
    city: "01"
  },
  {
    location: "01",
    city: "01"
  }
]

var result = data.reduce((mem, cur) => {
  var key = `{cur.location}--${cur.city}`
  if (!mem[key]) {
    mem[key] = cur;
  }
  return mem
}, {})

console.log(Object.values(result));

仅当对象不存在时才如何向该对象注入Array？

7 个答案: