如何筛选每个匹配项只存储一次的对象数组

Question

我正在使用Javascript，并具有以下对象数组：

const offersStep = [
     {a: 1, price: 67.10},
     {a: 3, price: 88.20},
     {a: 5, price: 88.20}, 
     {a: 7, price: 57.10},
     {a: 13, price: 57.10},
     {a: 15, price: 57.10},
     {a: 29, price: 57.10},
     {a: 30, price: 57.10},
]

为了使存储在一个新的数组中，这些对象，忽略它--那些已经存储在新数组上的支柱price具有一个值的对象--我必须做什么和如何做呢？

预期产出：

offers = [
  {a: 1, price: 67.10},
  {a: 3, price: 88.20},
  {a: 7, price: 57.10}
]

Answer 1

使用迭代来确定先前节省的价格是否与正在迭代的当前对象相匹配。

const offersStep = [
     {a: 1, price: 67.10},
     {a: 3, price: 88.20},
     {a: 5, price: 88.20}, 
     {a: 7, price: 57.10},
     {a: 13, price: 57.10},
     {a: 15, price: 57.10},
     {a: 29, price: 57.10},
     {a: 30, price: 57.10},
]

// Vars to store previously recorded prices.
var storedPrices = [];

// Var to store unique offersStep.
var uniqueOffersStep = [];

// Iterates through each of the offersStep array objects.
offersStep.forEach (function(value) {
  // Sets a boolean to false .. used later to determine if the current object is added to the unique array.
  var isSkip = false;
  
  // Iterates through each of the storedPrices array.
  storedPrices.forEach (function(spvalue) {
    // If the current objects price matches one of the storedPrices then it sets the skip flag to true.
    if (value.price == spvalue) {
      isSkip = true;
    }
  })
  // Adds the current objects price to the storedPrices array.
  storedPrices.push(value.price);
  
  // Check if the skip flag is still false after checking stored prices and adds the object to the uniqueOffers variable.
  if (isSkip == false) {
    uniqueOffersStep.push(value);
  }
})

// Outputs the final var to console.
console.log(uniqueOffersStep);

由于依赖于多个迭代层，因此对于较大的数据集来说，它的资源相当多。

希望这能有所帮助。

Answer 2

您可以将对象存储在地图键中，该键从它的price中删除。在存储了唯一的价格之后，将地图的值()迭代器扩展到一个新的数组中，以获得结果。

​

const offersStep = [
     {a: 1, price: 67.10},
     {a: 3, price: 88.20},
     {a: 5, price: 88.20}, 
     {a: 7, price: 57.10},
     {a: 13, price: 57.10},
     {a: 15, price: 57.10},
     {a: 29, price: 57.10},
     {a: 30, price: 57.10},
];

function filterOffers(offers) {
  let offersByPrice = new Map();
  for (let offer of offers) {
    if (!offersByPrice.has(offer.price)) {
      // Store the offer
      offersByPrice.set(offer.price, offer);
    }
  }
  
  return [...offersByPrice.values()];
}

console.log(filterOffers(offersStep));

​

Answer 3

Idea

过滤原始数组，通过一个集合来跟踪到目前为止遇到的价格。

码

​

const offersStep = [
     {a: 1, price: 67.10},
     {a: 3, price: 88.20},
     {a: 5, price: 88.20}, 
     {a: 7, price: 57.10},
     {a: 13, price: 57.10},
     {a: 15, price: 57.10},
     {a: 29, price: 57.10},
     {a: 30, price: 57.10}
];

let offer
  , nset_prices = new Set() 
  ;

offer = offersStep.filter ( po_item => {
            let b_ok = !nset_prices.has(po_item.price)
              ;

            if (b_ok) {
                nset_prices.add(po_item.price)
            }
            return b_ok;
        });

console.log(offer);

​