过滤数组：获取最高价格和唯一对象？

Question

 store = [{ "item": "shirt", "price": 20 }, { "item": "shirt", "price": 50 }, { "item": "pants", "price": 10 }, { "item": "pants", "price": 20 } ] //im filtering the array to get objects without duplication here console.log(store.filter((v, i, a) => a.findIndex(v2 => ['item'].every(k => v2[k] === v[k])) === i))

我也想在同一个过滤器中获得最高价格，所以我如何在执行后获得这个 output？

预计 output：

[{
    "item": "shirt",
    "price": 50
 },
 {
    "item": "pants",
    "price": 20
}]

Answer 1

您可以在过滤之前按价格对它们进行sort() 。

 store = [{ "item": "shirt", "price": 20 }, { "item": "shirt", "price": 50 }, { "item": "pants", "price": 10 }, { "item": "pants", "price": 20 } ] const result = store.sort((a, b) => b.price - a.price).filter((v, i, a) => a.findIndex(v2 => ['item'].every(k => v2[k] === v[k])) === i); console.log(result);

Answer 2

我喜欢@axtck 的回答。 ....过滤对象后，现在使用.map()通过排序从原始数据中找到最高价格：

.map(
    ({item,price}) => 
    ({item,price:store.filter(p => p.item === item).sort((a,b) => b.price - a.price)[0].price})
)

 store = [{ "item": "shirt", "price": 20 }, { "item": "shirt", "price": 50 }, { "item": "pants", "price": 10 }, { "item": "pants", "price": 20 } ] //im filtering the array to get objects without duplication here console.log( store.filter((v, i, a) => a.findIndex(v2 => ['item'].every(k => v2[k] === v[k])) === i).map( ({item,price}) => ({item,price:store.filter(p => p.item === item).sort((a,b) => b.price - a.price)[0].price}) ) )

Answer 3

您可以使用Array.reduce()来获取每件商品的最高价格。

我们将枚举每个值，创建 map object，如果该项目不存在条目或现有条目的价格较低，我们将更新：

 const store = [{ "item": "shirt", "price": 20 }, { "item": "shirt", "price": 50 }, { "item": "pants", "price": 10 }, { "item": "pants", "price": 20 } ] const result = Object.values(store.reduce((acc, { item, price }) => { // Either item does not exist or has a lower price... if (.acc[item] || acc[item],price < price) { acc[item] = { item; price }; } return acc, }. {})) console.log(result)

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