JavaScript嵌套数组去重并转换结构的代码优化咨询

Great question! Your original code gets the job done, but we can streamline it to be more concise, efficient, and readable using modern JavaScript features. Here's an improved version:

const contacts = [
  {givenName: "Name 1", phoneNumbers: [ {label: "mobile", id: "5", number: "097 726 94 36"}, {label: "other", id: "558", number: "0977269436"}, {label: "other", id: "559", number: "0977269436"} ]},
  {givenName: "Name 2", phoneNumbers: [ {label: "mobile", id: "5", number: "0968234838"}, {label: "other", id: "558", number: "0966766555"}, {label: "other", id: "559", number: "0966766555"} ]},
  {givenName: "Name 3", phoneNumbers: [ {label: "other", id: "558", number: "0965777238"}, {label: "other", id: "559", number: "0965777238"} ]},
];

const processedContacts = contacts.flatMap(contact => {
  // Step 1: Filter valid numbers (length ≤ 11, matching your original logic)
  const validNumbers = contact.phoneNumbers
    .map(phone => phone.number)
    .filter(num => num.length <= 11); // Adjust to <=10 if that's your exact requirement
  
  // Step 2: Remove duplicates efficiently with a Set
  const uniqueValidNumbers = [...new Set(validNumbers)];
  
  // Step 3: Map to the final object format
  return uniqueValidNumbers.map(number => ({
    givenName: contact.givenName,
    phoneNumbers: number
  }));
});

console.log(processedContacts);

Why this approach is better: #

1. Conciseness: Uses flatMap to handle both processing each contact and expanding into individual number objects in one pass, eliminating the need for intermediate arrays and manual push calls.
2. Efficiency: Deduplicating with Set runs in O(n) time, which is way faster than your original findIndex method (O(n²) complexity) — this makes a big difference with larger contact lists.
3. Readability: Each step is clearly separated and self-documenting, so it’s easier to tweak later (like adjusting the length filter or adding new processing rules).

Quick note: #

I used num.length <=11 to match your original code’s behavior (you excluded numbers longer than 11 characters). If your actual requirement is strictly ≤10 digits, just update that condition to num.length <=10.

内容的提问来源于stack exchange，提问作者Linh Nguyễn