拼接方法索引混淆

Question

我知道这听起来很有趣，因为我自己编写了解决方案，但是在一段时间没有看到它之后，我很难理解为什么它会起作用。该算法解决了 freecodecamp 的“库存更新”问题，尽管测试失败（比较并更新存储在 2D 数组中的库存与新交付的第二个 2D 数组。更新当前现有的库存项目数量（在 arr1 中）。如果找不到商品，则将新的商品和数量添加到库存数组中。返回的库存数组应按商品的字母顺序排列。）算法如下：

function updateInventory(arr1, arr2) {
    let newInv = [...arr1, ...arr2]
    //create single list of items only from both arrays
    let temp = newInv.flat().filter(item => typeof item === 'string')
    //create list of the index of all duplicate items
    let duplicates = temp.reduce((acc, item, index) => {
        if (temp.indexOf(item) != index){
            acc.push(index)
        }
        return acc
    }, [])
    //remove duplicate items
    for (let index in duplicates) {
        newInv.splice(index, 1)

    }
    //sort by alphabetical order
    newInv.sort((a,b) => {
        return a[1] > b[1] ? 1 : -1
    })
    return newInv
}
// Example inventory lists
var curInv = [
    [21, "Bowling Ball"],
    [2, "Dirty Sock"],
    [1, "Hair Pin"],
    [5, "Microphone"]
];

var newInv = [
    [2, "Hair Pin"],
    [3, "Half-Eaten Apple"],
    [67, "Bowling Ball"],
    [7, "Toothpaste"]
];

updateInventory(curInv, newInv);

据我了解，预期结果应该是：

[ [ 21, 'Bowling Ball' ],
  [ 2, 'Dirty Sock' ],
  [ 1, 'Hair Pin' ],
  [ 3, 'Half-Eaten Apple' ],
  [ 5, 'Microphone' ],
  [ 7, 'Toothpaste' ] ]

然而，得到的是：

[ [ 67, 'Bowling Ball' ],
  [ 2, 'Dirty Sock' ],
  [ 2, 'Hair Pin' ],
  [ 3, 'Half-Eaten Apple' ],
  [ 5, 'Microphone' ],
  [ 7, 'Toothpaste' ] ]

这是要删除的元素的重复对。我确定我可能遗漏了一些简单的东西，但我就是不明白。

非常感谢您的帮助

Answer 1

有两个问题。第一个是for循环

  let duplicates = [4, 6]

  for (let index in duplicates) {
        newInv.splice(index, 1)
    }

你正在为 in 而不是 of

in 循环遍历索引（in like in index..! :)），for 遍历值
如果您在循环中记录索引，您将看到 0,1 = 重复元素的索引
将其更改为 for 然后它是 4,6 = 重复元素的值 = 要删除的项目的索引。另一种选择是

duplicates.forEach(i => newInv.splice(i, 1))

第二个问题是，当您删除第一项时，第二项的索引会发生变化:-) 所以它不再是索引 6，而是现在 5。这可以通过在循环和拼接之前反转重复项来解决，所以首先从最高索引开始，然后删除“从头到尾”。

所以这应该给出请求的结果

function updateInventory(arr1, arr2) {
    let newInv = [...arr1, ...arr2]
      //create single list of items only from both arrays
    let temp = newInv.flat().filter(item => typeof item === 'string')
      //create list of the index of all duplicate items
    let duplicates = temp.reduce((acc, item, index) => {
        if (temp.indexOf(item) != index) {
          acc.push(index)
        }
        return acc
      }, []).reverse()
      //remove duplicate items
    for (let index of duplicates) {
      newInv.splice(index, 1)
    }
    //sort by alphabetical order
    newInv.sort((a, b) => {
      return a[1] > b[1] ? 1 : -1
    })
    return newInv
  }

  // Example inventory lists
var curInv = [
  [21, "Bowling Ball"],
  [2, "Dirty Sock"],
  [1, "Hair Pin"],
  [5, "Microphone"]
];

var newInv = [
  [2, "Hair Pin"],
  [3, "Half-Eaten Apple"],
  [67, "Bowling Ball"],
  [7, "Toothpaste"]
];


let result = updateInventory(curInv, newInv);

console.log(result);

这将是我解决任务的方法

function updateInventory(curInv, newInv) {
    newInv.forEach(newItem => {
      let newItemName = newItem[1]      
      let inCurrent = curInv.find(currItem => currItem[1] === newItemName)
      if(!inCurrent) curInv.push(newItem)
    })
    return curInv.sort((a,b) => a[1].localeCompare(b[1]))
  }  

  // Example inventory lists
var curInv = [
  [21, "Bowling Ball"],
  [2, "Dirty Sock"],
  [1, "Hair Pin"],
  [5, "Microphone"]
];

var newInv = [
  [2, "Hair Pin"],
  [3, "Half-Eaten Apple"],
  [67, "Bowling Ball"],
  [7, "Toothpaste"]
];


let result = updateInventory(curInv, newInv);

console.log(result);