英文:
JavaScript/TypeScript: count duplicate Objects in an Array of Objects to have an amount property on Item
问题
你可以尝试以下代码来实现你的目标：
const items = [{"id": "fishing_rod","emoji": "🎣","description": "Used to go fishing.","category": "fishing",},{"id": "fishing_rod","emoji": "🎣","description": "Used to go fishing.","category": "fishing",},{"id": "pickaxe","emoji": "⛏️","description": "Used to go mining.","category": "mining",},];// 创建一个空的对象来存储每个项目的数量const itemCounts = {};// 遍历数组并计算每个项目的数量items.forEach(item => {const itemId = item.id;if (itemCounts[itemId]) {itemCounts[itemId]++;} else {itemCounts[itemId] = 1;}});// 将结果放入一个新的数组const result = Object.keys(itemCounts).map(itemId => ({...items.find(item => item.id === itemId),amount: itemCounts[itemId]}));console.log(result);
这段代码会输出与你所期望的格式相匹配的结果。
英文:
So I have an array of items. There are duplicate items in that array and I am trying to count these duplicates. The count of those duplicates should be put into one of the duplicate items and the other item should be discarded.
My input:
(simplified)
[{"id": "fishing_rod","emoji": "🎣","description": "Used to go fishing.","category": "fishing",},{"id": "fishing_rod","emoji": "🎣","description": "Used to go fishing.","category": "fishing",},{"id": "pickaxe","emoji": "⛏️","description": "Used to go mining.","category": "mining",},],
What I'm looking for:
[{"id": "fishing_rod","emoji": "🎣","description": "Used to go fishing.","category": "fishing","amount": 2},{"id": "pickaxe","emoji": "⛏️","description": "Used to go mining.","category": "mining","amount": 1},]
What I tried so far:
I have tried a few things but couldn't get there.
I've only gotten so far that I was able to only get the Item Id and the count. That looked a like this:
const itemsArray = items.map((i) => i.id); // items being my input array (see top of this post)const itemsUnique = [...new Set(itemsArray)];const itemsCount = itemsUnique.map((item) => [item, itemsArray.filter((i) => i === item).length])
Output:
[[ "fishing_rod", 2 ],[ "pickaxe", 1 ],]
But that's not really what I am looking for...
A little help would be appreciated. <3
答案1
得分: 0
只需迭代并将具有相同id的项分组：
（以下代码段中有更简洁的版本。）
function mergeItems(items) {const grouped = items.reduce((record, item) => {if (Object.hasOwn(record, item.id)) {record[item.id].count++;} else {record[item.id] = { ...item, count: 1 };}return record;}, {});return Object.values(grouped);}
尝试一下：
console.config({ maximize: true });function mergeItems(items) {const grouped = items.reduce((record, item) => {record[item.id] ??= { ...item, count: 0 };record[item.id].count++;return record;}, {});return Object.values(grouped);}const input = [{"id": "fishing_rod","emoji": "🎣","description": "用于钓鱼。","category": "钓鱼",},{"id": "fishing_rod","emoji": "🎣","description": "用于钓鱼。","category": "钓鱼",},{"id": "pickaxe","emoji": "⛏️","description": "用于采矿。","category": "采矿",},];console.log(mergeItems(input));
<script src="https://gh-canon.github.io/stack-snippet-console/console.min.js"></script>
希望这有帮助。
英文:
Simply iterate over and group those with the same id:
(A terser version can be found in the snippet below.)
function mergeItems(items) {const grouped = items.reduce((record, item) => {if (Object.hasOwn(record, item.id)) {record[item.id].count++;} else {record[item.id] = { ...item, count: 1 };}return record;}, {});return Object.values(grouped);}
Try it:
<!-- begin snippet: js hide: true console: false babel: false -->
<!-- language: lang-js -->
console.config({ maximize: true });function mergeItems(items) {const grouped = items.reduce((record, item) => {record[item.id] ??= { ...item, count: 0 };record[item.id].count++;return record;}, {});return Object.values(grouped);}const input = [{"id": "fishing_rod","emoji": "🎣","description": "Used to go fishing.","category": "fishing",},{"id": "fishing_rod","emoji": "🎣","description": "Used to go fishing.","category": "fishing",},{"id": "pickaxe","emoji": "⛏️","description": "Used to go mining.","category": "mining",},];console.log(mergeItems(input));
<!-- language: lang-html -->
<script src="https://gh-canon.github.io/stack-snippet-console/console.min.js"></script>
<!-- end snippet -->
答案2
得分: 0
我们可以使用 filter 来移除重复项，同时增加计数。
const input = [{"id": "fishing_rod", "emoji": "🎣", "description": "Used to go fishing.", "category": "fishing"},{"id": "fishing_rod", "emoji": "🎣", "description": "Used to go fishing.", "category": "fishing"},{"id": "pickaxe", "emoji": "⛏️", "description": "Used to go mining.", "category": "mining"}]const output = input.filter((val, i, arr) => {const first = arr.findIndex(({ id }) => val.id === id);if (first === i)return val.amount = 1; // 当前值是第一次出现else arr[first].amount++; // 当前值不是第一次出现});console.log(output);
这是代码的翻译部分，没有其他内容。
英文:
We can use filter to remove the duplicates, at the same time incrementing the count.
<!-- begin snippet: js hide: false console: true babel: false -->
<!-- language: lang-js -->
const input = [{"id": "fishing_rod","emoji": "🎣","description": "Used to go fishing.","category": "fishing",},{"id": "fishing_rod","emoji": "🎣","description": "Used to go fishing.","category": "fishing",},{"id": "pickaxe","emoji": "⛏️","description": "Used to go mining.","category": "mining",}]const output = input.filter((val, i, arr) => {const first = arr.findIndex(({ id }) => val.id === id)if(first === i)return val.amount = 1 // the current value is the first occurenceelse arr[first].amount++ // the current value is not the first occurence})console.log(output)
<!-- end snippet -->
答案3
得分: 0
const source = [{"id": "fishing_rod","emoji": "🎣","description": "用于钓鱼。","category": "钓鱼",},{"id": "fishing_rod","emoji": "🎣","description": "用于钓鱼。","category": "钓鱼",},{"id": "pickaxe","emoji": "⛏️","description": "用于采矿。","category": "采矿",},];const result = {};source.forEach((item) => result[item.id] ? result[item.id].amount ++ : result[item.id] = { ...item, amount: 1});console.log(Object.values(result));
请注意，我已将表情符号和描述部分翻译成中文。
英文:
Technically a one-liner 
<!-- begin snippet: js hide: false console: true babel: false -->
<!-- language: lang-js -->
const source = [{"id": "fishing_rod","emoji": "🎣","description": "Used to go fishing.","category": "fishing",},{"id": "fishing_rod","emoji": "🎣","description": "Used to go fishing.","category": "fishing",},{"id": "pickaxe","emoji": "⛏️","description": "Used to go mining.","category": "mining",},];const result = {};source.forEach((item) => result[item.id] ? result[item.id].amount ++ : result[item.id] = { ...item, amount: 1});console.log(Object.values(result));
<!-- end snippet -->
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