您可以使用reduce()创建一个包含所有已恢复的cgi值的集合。然后，您可以使用filter()删除此集合中所有不需要的值。

• 除了Set，你也可以简单地使用一个数组。我使用了一个Set，这样我们就不会得到重复的值。

const data = [{"cgi":"abc-123","tag":"revert","name":"Piyush"},{"cgi":"abc-123","tag":null,"name":"Piyush"},{"cgi":"abc-456","tag":null,"name":"Piyush"},{"cgi":"abc-789","tag":null,"name":"Piyush"}];

// Get all `cgi` where tag is `revert`.
const revertedTags = data.reduce((acc, item) => {
  if(item.tag === 'revert') {
    acc.add(item.cgi);
  }

  return acc;
}, new Set());

// Filter out all elements who are present in `revertedTags` collection.
const filteredTags = data.filter((item) => !revertedTags.has(item.cgi));

// Result.
console.log(filteredTags);

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一行程序：你可以使用一个memoized函数作为过滤器来过滤项目：

const data = [{"cgi":"abc-123","tag":"revert","name":"Piyush"},{"cgi":"abc-123","tag":null,"name":"Piyush"},{"cgi":"abc-456","tag":null,"name":"Piyush"},{"cgi":"abc-789","tag":null,"name":"Piyush"}];

const filteredTags = data.filter((map => e => map[e.cgi] ??= !data.find(e2 => e2.tag === 'revert' && e.cgi === e2.cgi))({}));

console.log(filteredTags);

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