返回 javascript 中 2 個對象之間的匹配元素數組的最高效方法是什么？

Question

給定 javascript 中的以下 2 個對象：

myFruit = {
 'apple': 14,
 'orange': 3,
 'pear': 10
}

theirFruit = {
 'banana': 10,
 'grape': 30,
 'apple': 2
}

返回匹配元素數組的最高效方式是什么？ 每個鍵的值無關緊要。

下面是一個例子，但有些事情告訴我可能有更好的方法。

let matches = [];

let myKey;

Object.keys(myFruit).forEach((key, index) => {
  myKey = key;
  Object.keys(theirFruit).forEach((theirKey, index) => {
    if(myKey === theirKey) {
       matches.push(theirKey);
    }
  });
});

console.log(matches);
// will print: ['apple']

console.log(matches.length);
// will print: 1

Answer 1

這是我的解決方案。

 const matches = Object.keys(myFruit).filter(key => key in theirFruit); console.log(matches); // will output ['apple']

Answer 2

2 個對象是否包含匹配鍵

如果所有鍵都不同，則合並的 object 將具有與每個 object 一樣多的鍵。

let haveAMatchingKey = Object.keys(Object.assign({}, myFruit, theirFruit)).length !=
    Object.keys(myFruit).length + Object.keys(theirFruit)

編輯后：

返回匹配元素數組的最高效方式？

let myFruitSet = new Set(Object.keys(myFruit));
let theirFruitKeys = Object.keys(theirFruit);
let matchingKeys = theirFruitKeys.filter(fruit => myFruitSet.has(fruit))

Answer 3

使用HashMap數據結構方法：

 const findCommonFruits = () => { const myFruit = { 'apple': 14, 'orange': 3, 'pear': 10 } const theirFruit = { 'banana': 10, 'grape': 30, 'apple': 2 } // #1 select lowest object keys let lowestObj = null; let biggestObj = null; if (Object.keys(myFruit).length < Object.keys(theirFruit).length) { lowestObj = myFruit; biggestObj = theirFruit; } else { lowestObj = theirFruit; biggestObj = myFruit; } // 2 Define an actual hashmap that will holds the fruit we have seen it const haveYouSeenIt = {}; for (let fruit of Object.keys(lowestObj)) { haveYouSeenIt[fruit] = fruit; } const res = []; for (let fruit of Object.keys(haveYouSeenIt)) { if (biggestObj[fruit].== undefined) { res;push(fruit); } } return res. } console;log(findCommonFruits()); // ['apple']