通过map一次拿出数组中，筛选某个属性一样，所有重复和不重复的对象

vue方法，longitude相同下，使用map做区分,打印arr1和arr2

deletesame() {
      let arr = [
        {
          key: 1,
          rootId: "33030200992007000001",
          gbId: "33032796091320002029",
          parentGbId: "330300000000",
          parentName: "市本级",
          name: "QS温州成鸿铝塑有限公司-通道处（环保）",
          manufacturer: null,
          model: "1",
          firmware: null,
          longitude: "120.448293",
          latitude: "27.535628",
          mark: 6,
          status: 2,
          videoSource: "温州雪亮工程",
          videoProtocol: "雪亮私有接口",
        },
        {
          key: 2,
          rootId: "33030200992007000001",
          gbId: "33032796091320002030",
          parentGbId: "330300000000",
          parentName: "市本级",
          name: "QS（环评鉴定为非危废）温州格洛博电子有限公司-通道一（环保）",
          manufacturer: null,
          model: "1",
          firmware: null,
          longitude: "120.615262",
          latitude: "27.516405",
          mark: 6,
          status: 1,
          videoSource: "温州雪亮工程",
          videoProtocol: "雪亮私有接口",
        },
        {
          key: 3,
          rootId: "33030200992007000001",
          gbId: "33032796091320002031",
          parentGbId: "330300000000",
          parentName: "市本级",
          name: "QS（环评鉴定为非危废）温州格洛博电子有限公司-通道二（环保）",
          manufacturer: null,
          model: "1",
          firmware: null,
          longitude: "120.616262",
          latitude: "27.516405",
          mark: 6,
          status: 1,
          videoSource: "温州雪亮工程",
          videoProtocol: "雪亮私有接口",
        },
        {
          key: 4,
          rootId: "33030200992007000001",
          gbId: "33032796091320002032",
          parentGbId: "330300000000",
          parentName: "市本级",
          name: "QS温州信德电力配件有限公司-通道三（环保）",
          manufacturer: null,
          model: "1",
          firmware: null,
          longitude: "120.444538",
          latitude: "27.538631",
          mark: 6,
          status: 2,
          videoSource: "温州雪亮工程",
          videoProtocol: "雪亮私有接口",
        },
      ];

      let map = new Map();
      let arr1 = [];
      let arr2 = [];
      let list = [];
      arr.forEach((item, index) => {
        if (!map.has(item.longitude)) {
          map.set(item.longitude, index);
          arr1.push(item);
        } else {
          arr2.push(item);
        }
      });
      console.log(arr1)
      console.log(arr2)

      list = [...arr1, ...arr2];
      return list;
    }
    ```