.sort() 不适用于数组（javascript）

Question

我正在尝试通过首先创建对象来将“新货”数组与当前库存数组合并，以使数据更易于管理。因此，任何相同的项目都会添加到库存中的任何现有项目中。

.sort 运行但 flat 似乎没有做任何事情。我怀疑存在一些与我如何制作数组和弄乱索引有关的问题？

function updateInventory(arr1, arr2) {
    let invObj = {}
    let updateObj = {}
    let result = []

    arr1.forEach( x => invObj[x[1]] = x[0])
    arr2.forEach( x => updateObj[x[1]] = x[0])

    for(let key in updateObj) {
        if (invObj[key]) {
            invObj[key] += updateObj[key]
        } else {
            invObj[key] = updateObj[key]
        }
    }

    result =  Object.keys(invObj).map(key=>[invObj[key],key])
    .sort((a,b)=>{
    // attempting to sort inventory alphabetically here as required by my course's test
        return a[1] - b[1]
    })

    return result
}
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"]
];

console.log(updateInventory(curInv, newInv));

Answer 1

我会创建一个查找对象并保留对数组项的引用。在我遍历当前项目之后，我将遍历新项目。检查它是否存在并更新计数。如果它不存在，则将该项目添加到库存中。

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"]
];

// make a look up object to reference by the key
var lookup = curInv.reduce( (obj, item) => ({ ...obj, [item[1]]: item }), {})

// loop over the new inventory and add it on
newInv.forEach((item) => {
  // check to see if we have the item
  var key = item[1]
  var exisiting = lookup[key]
  // if exists add it
  if (exisiting) {
    exisiting[0] += item[0]
  } else {
    // new item
    // add to our look up table in case it repeats
    lookup[key] = item
    // add it to the inventory list
    curInv.push(item)
  }
})

console.log(curInv)

Answer 2

对字符串进行排序时会稍微复杂一些，因为您还必须考虑大小写。这是我用js制作的表格中的sn-p，希望对您有所帮助。你可以像 taplar 所说的那样使用 localecompare。

            .sort(
                function(a,b) {
                    a = a[1];
                    b = b[1];
                    if (a < b) {
                        return -1; 
                    } else if (a > b) {
                        return 1;
                    } else {
                        return 0; // Equal
                    }
                });

Answer 3

您是否希望数组与第一个实例中的数据格式相同？

function updateInventory(arr1, arr2) {
    let invObj = {}
    let updateObj = {}
    let result = []

    arr1.forEach( x => invObj[x[1]] = x[0])
    arr2.forEach( x => updateObj[x[1]] = x[0])

    for(let key in updateObj) {
        if (invObj[key]) {
            invObj[key] += updateObj[key]
        } else {
            invObj[key] = updateObj[key]
        }
    }
    
    return invObj;
}
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"]
];

console.log(updateInventory(curInv, newInv));

Answer 4

在对字符串值进行排序时，您应该使用localeCompare 方法。

function updateInventory (arr1, arr2) {
  let invObj = {};
  let updateObj = {};
  let result = [];

  arr1.forEach(x => invObj[x[1]] = x[0]);
  arr2.forEach(x => updateObj[x[1]] = x[0]);

  for (let key in updateObj) {
    if (invObj[key]) {
      invObj[key] += updateObj[key];
    } else {
      invObj[key] = updateObj[key];
    }
  }

  result = Object.keys(invObj)
    .sort((a, b) => a.localeCompare(b))
    .map(key => [invObj[key], key]);

  return result;
}

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']
];

console.log(
  updateInventory(curInv, newInv)
);