遍历对象并按指定顺序映射新对象，编写数量构建函数

Hey there! Let's break down and solve these two practical coding tasks for you.

问题分析 #

You want to create a new object from an original one, strictly following a custom order of keys. The core idea is to prioritize extracting key-value pairs in your specified sequence, with optional handling for keys that aren't in your order list.

实现代码示例 #

function mapObjectByOrder(originalObj, keyOrder) {
  const newObj = {};
  // Iterate through the specified key order to build the new object
  keyOrder.forEach(key => {
    if (originalObj.hasOwnProperty(key)) {
      newObj[key] = originalObj[key];
    }
    // Uncomment below if you want to keep keys from the original object that aren't in the order list
    // else {
    //   newObj[key] = originalObj[key]; // Or set a default value here
    // }
  });
  return newObj;
}

// Example usage
const originalUser = { name: 'Alice', age: 30, city: 'New York' };
const targetKeyOrder = ['age', 'name', 'city'];
const orderedUser = mapObjectByOrder(originalUser, targetKeyOrder);
console.log(orderedUser); // { age: 30, name: 'Alice', city: 'New York' }

Quick Notes #

• The function takes the original object and your desired key order array, then builds the new object in that exact sequence
• hasOwnProperty ensures we only work with properties directly on the original object (avoids prototype chain properties)
• If you need to retain extra keys from the original object that aren't in your order list, add an extra loop to append those after processing the specified order

问题分析 #

You need a function that takes a number (from 1 up to a total sum, like 7) and returns an array of that length, where elements follow the repeating sequence: variantId:1111 → 2222 → 3333 → 4444 → 1111 (second instance) → 2222 (second instance) → .... Modulo arithmetic is perfect here to handle the repeating pattern.

实现代码示例 #

function buildVariantArray(count) {
  // Define the base repeating sequence
  const baseSequence = [1111, 2222, 3333, 4444];
  const resultArray = [];
  
  for (let i = 0; i < count; i++) {
    // Use modulo to cycle through the base sequence
    const currentVariantId = baseSequence[i % baseSequence.length];
    resultArray.push({ variantId: currentVariantId });
  }
  
  return resultArray;
}

// Example usage: input 7
const variantList = buildVariantArray(7);
console.log(variantList);
// Output:
// [
//   { variantId: 1111 },
//   { variantId: 2222 },
//   { variantId: 3333 },
//   { variantId: 4444 },
//   { variantId: 1111 },
//   { variantId: 2222 },
//   { variantId: 3333 }
// ]

Quick Notes #

• The baseSequence array holds the values we want to repeat
• i % baseSequence.length calculates which index of the base sequence to use for each position in the result array, creating the repeating pattern
• Each iteration creates an object with the variantId and pushes it to the result array, ensuring we end up with exactly count elements

内容的提问来源于stack exchange，提问作者chris cozzens