如何對唯一鍵值進行排序和操作

我想迭代一系列不同的評級，按每個唯一的 id 排序，求和并計算每個 id 評級的平均值。然后將平均值保存在一個新陣列中，我可以在其中呼叫類似 averageRating[i] 的東西，其中每個條目將是每個 id 的評分。

原始物件陣列如下所示，其中 id 可以是任意數字。

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}]

我只用一個特定的 id 就可以完成這項作業，執行如下操作，但是在弄清楚如何處理動態數量的 id 時遇到了一些麻煩。

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;

uj5u.com熱心網友回復：

這可能超過了最高性能，但如果你喜歡一點數學，你實際上可以在一個回圈中完成它。看到這個數學解決方案

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.id

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

這應該是高性能的，因為沒有多個回圈或中間結構。

對于此輸入：

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}]

它回傳

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

uj5u.com熱心網友回復：

您可以使用常規回圈并進行所有處理。閱讀行內評論：

// 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);

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