如何对唯一键值进行排序和操作

Question

我想迭代一系列不同的评级，按每个唯一的 id 排序，求和并计算每个 id 评级的平均值。然后将平均值保存在一个新数组中，我可以在其中调用类似 averageRating[i] 的东西，其中每个条目将是每个 id 的评分。

原始对象数组如下所示，其中 id 可以是任意数字。

data = [{id: 1, rating: 1}, {id: 1, rating: 3}, {id: 1, rating: 1}, {id: 1, rating: 4}, {id: 1, rating}, {id: 2, rating: 3}, {id: 3, rating: 5}, {id: 1, rating: 5}, {id: 1, rating: 5}, {id: 1, rating: 5, {id: 1, rating: 1}, {id: 2, rating: 4}, {id: 1, rating: 3}, {id: 1, rating: 2}]

我只用一个特定的 id 就可以完成这项工作，执行如下操作，但是在弄清楚如何处理动态数量的 id 时遇到了一些麻烦。

var [average, updateAverage] = useState(0);

let ratings = data.map((item) => item.rating);

// Calculate average of the array
let sum = ratings.reduce((a, b) => a + b, 0);
let avg = sum / ratings.length || 0;
let avgRounded = Math.round(avg); // Round to nearest whole number

updateAverage = avgRounded;

Answer 1

这可能超过了最高性能，但如果你喜欢一点数学，你实际上可以在一个循环中完成它。见this mathematical solution

const avgCount = {} // This is a temp var needed for accumulative averaging mathematical formula

// THIS VAR HAS THE ANSWER IN
const averages = data.reduce((averages, individualRating) => {
    // We need a counter of the number of times this ID has been seen already. This is needed by the algorithm.
    // Now we have seen a new rating with this id, increase the counter to reflect that.
    // If it's not there we start from 1.
    avgCount[individualRating.id] =  (avgCount[individualRating.id] ?? 0) + 1

    // Get the current rolling average. If there isn't one (this is first rating with this id we have seen), then its just the rating of this current item.
    const currAccumulatedAverage = averages[individualRating.id] ?? individualRating.rating

    // Calculate the new rolling average from the previous one
    averages[individualRating.id] = ((currAccumulatedAverage * (avgCount[individualRating.id] - 1)) + individualRating.rating) / avgCount[individualRating.id]
    
    return averages
}, {}) 

这应该是高性能的，因为没有多个循环或中间结构。

对于此输入：

let data = [{id: 1, rating: 1}, {id: 1, rating: 3}, {id: 1, rating: 1}, {id: 1, rating: 4}, {id: 1, rating: 1}, {id: 2, rating: 3}, {id: 3, rating: 5}, {id: 1, rating: 5}, {id: 1, rating: 5}, {id: 1, rating: 5}, {id: 1, rating: 1}, {id: 2, rating: 4}, {id: 1, rating: 3}, {id: 1, rating: 2}]

它返回

{1: 2.8181818181818183, 2: 3.5, 3: 5}

Answer 2

您可以使用常规循环并进行所有处理。读取内联 cmets：

// Rating data
const data = [
  {id: 1, rating: 1},
  {id: 1, rating: 3},
  {id: 1, rating: 1},
  {id: 1, rating: 4},
  {id: 1, rating: 2},
  {id: 2, rating: 3},
  {id: 3, rating: 5},
  {id: 1, rating: 5},
  {id: 1, rating: 5},
  {id: 1, rating: 5},
  {id: 1, rating: 1},
  {id: 2, rating: 4},
  {id: 1, rating: 3},
  {id: 1, rating: 2}
];

// Create new object for id -> average ratings
const avr = {};

// Iterate data array
for(const item of data) {
  // Create object of arrays
  // for calculate avarage later
  if(avr[item.id]) avr[item.id].push(item.rating);
  else avr[item.id] = [item.rating];
}

// Iterate object and calculate average values
for(const el in avr) {
  // Reduce
  const sum = avr[el].reduce((a, b) => a + b, 0);
  // Calculate average and update object
  avr[el] = Math.round(sum / avr[el].length || 0);
}

// Test
console.log(avr);