假设obj为 * 数组对象 *。
函数对对象进行分组

const groupByKey = (list: any[], key: string, {omitKey = false}: { omitKey: true; }) => list.reduce((hash, {[key]: value, ...rest}) => ({
...hash, [value]: (hash[value] || []).concat(omitKey ? {...rest} : {...rest})
}), {});

const groupByKey1 = groupByKey(obj, 'AirCode', {omitKey: true});

const x: any[] = [];
Object.entries(groupByKey1).forEach(
    ([key, value]) => x.push(value)
);
const z: { ABaseFare: any; ATax: any; RfNo: any; YBaseFare: any; }[] = [];
x.forEach(value => {
  const p = {
    ABaseFare: value[0].ABaseFare,
    ATax: value[0].ATax,
    RfNo: value[0].RfNo,
    YBaseFare: value[0].YBaseFare,
   };
 z.push(p);
});
console.log(z);

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我正在寻找和你一样的 * 过滤 *。The other answer工作（谢谢！），但是我和你都不需要分组。我为我们的用例找到了一个更简单，更快的解决方案：

this.faresData.reduce((r, a) => r.some(x => x.AirCode == a.AirCode) ? r : [...r, a], []);

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在else的例子中，我使用spread语法来“推送”每个第一个 fare。结果包含了你所期望的json：

[
 {
  "RfNo": "500",
  "ABaseFare": "110",
  "ATax": "267.22",
  "YBaseFare": "0",
  "AirCode": "AF"
 },
 {
  "RfNo": "502",
  "ABaseFare": "110",
  "ATax": "267.22",
  "YBaseFare": "0",
  "AirCode": "DD"
 }
]

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