Javascript 数组 - 根据另一个属性删除重复项

Question

根据另一个属性的值对复杂数组进行重复数据删除的最有效方法是什么？我发现了很多可以对数组或复杂数组进行重复数据删除的示例，但不是这个特定的用例。

我正在尝试查找第 3 列（州）中具有唯一值且第 2 列（许可证）中编号最大的记录

var arrayWithDuplicates = [
    ["Boat", 1, "NV"],
    ["Car", 7, "CA"],
    ["Boat", 3, "NV"],
    ["Boat", 4, "CA"],
    ["Car", 5, "OR"],
    ["Boat", 6, "CA"],
];

期望的结果

var outputArray = [
    ["Car", 7, "CA"],
    ["Boat", 3, "NV"],
    ["Car", 5, "OR"]
];

这可行，但不确定是否适用于大型数据集

var arrayWithDuplicates = [
  ["Boat", 1, "NV"],
  ["Car", 7, "CA"],
  ["Boat", 3, "NV"],
  ["Boat", 4, "CA"],
  ["Car", 5, "OR"],
  ["Boat", 6, "CA"],
];
let arr= arrayWithDuplicates;
let unique = []
for (let i = 0; i < arr.length; i++) {
  let found = false;
  for (let j = 0; j < unique.length; j++) {
    if (arr[i][2] === unique[j][2]) {
      found = true;
      if (arr[i][1] > unique[j][1]) {
        unique[j] = arr[i];
      }
      break;
    }
  }
  if (!found) {
    unique.push(arr[i])
  }
}
console.log(unique);

[["Boat", 3, "NV"], ["Boat", 7, "CA"], ["Car", 5, "OR"]]

您可以看到建议解决方案的性能：https://jsbench.me/eskxxcwnhn/1

Answer 1

使用排序和归约

const arrayWithDuplicates = [ ["Boat", 1, "NV"], ["Car", 7, "CA"], ["Boat", 3, "NV"], ["Boat", 4, "CA"], ["Car", 5, "OR"], ["Boat", 6, "CA"], ]; 

let deduped = [...arrayWithDuplicates]; // take a copy

deduped.sort((a, b) => { // sorts in place
  if (a[0] < b[0]) return 1; // sort on names
  if (a[0] > b[0]) return -1;
  return b[1] - a[1]; // sort on second element
})

// reduce into an object keyed on state
deduped = Object.values(  // take only the values from the object
  deduped.reduce((acc, cur) => {
    const state = cur[2];
    if (!acc[state]) acc[state] = cur;
    return acc;
},{}))
console.log(deduped)

Answer 2

解决问题的一种方法是使用 Map 实例来保存最相关的值。如果遇到另一个，请更换它。然后，当您完成迭代时，您将获取 Map 实例中存在的值。

const arrayWithDuplicates = [
  ["Boat", 1, "NV"],
  ["Car", 7, "CA"],
  ["Boat", 3, "NV"],
  ["Boat", 4, "CA"],
  ["Car", 5, "OR"],
  ["Boat", 6, "CA"],
];

const lookup = new Map();
for (const record of arrayWithDuplicates) {
  const key = record[2];
  
  if (!lookup.has(key)) {
    lookup.set(key, record);
    continue;
  }
  
  const other = lookup.get(key);
  if (record[1] > other[1]) {
    // Iteration order is based on insertion order. By removing the
    // current value first, the new value will be placed at the end.
    // If you don't care about the order, deletion can be omitted.
    lookup.delete(key);
    lookup.set(key, record);
  }
}

const result = Array.from(lookup.values());
console.log(result);

请注意，以下代码序列可能是一个相当繁重的操作：

lookup.delete(key);
lookup.set(key, record);

由于它重新排列了地图内容的迭代顺序。这只会按照您要查找的顺序得到结果。如果结果项的顺序无关紧要，则应删除 lookup.delete(key) 调用以提高执行速度。

尽管使用 Map 实例可能会对小型集合产生一些执行开销。当集合变大时，改进的查找速度真正闪耀。

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