按 id 对唯一数组进行排序并保留最后一个值答案

【问题标题】：Sort unique array by id and keeping last value按 id 对唯一数组进行排序并保留最后一个值
【发布时间】：2021-04-14 18:32:24
【问题描述】：

我有关注

let arr = [
  { id: 1, referenceId: 1, type: "normal", name: "a" },
  { id: 2, referenceId: 1, type: "normal", name: "b" },
  { id: 3, referenceId: 3, type: "chat", name: "c" },
  { id: 4, referenceId: 4, type: "normal", name: "d" },
  { id: 5, referenceId: 5, type: "chat", name: "e" },
  { id: 6, referenceId: 3, type: "chat", name: "f" }
];

我想要如下输出：

[
  { id: 1, referenceId: 1, type: "normal", name: "a" },
  { id: 2, referenceId: 1, type: "normal", name: "b" },
  { id: 4, referenceId: 4, type: "normal", name: "d" },
  { id: 5, referenceId: 5, type: "chat", name: "e" },
  { id: 6, referenceId: 3, type: "chat", name: "f" }
];

我想排序 if type === "chat", 试试这样(type === "normal", no sort) :

arr.filter(item => {
  if (item.type === "normal") {
    return item
  }
  //sort array by referenceId and keeping last item

})

【问题讨论】：

欢迎来到 SO！我对filter 回调中的逻辑有点困惑。 item.type === "normal" 与您的规范的其余部分有什么关系？
只排序 type === "chat", if type === "normal" 不排序

标签： javascript arrays typescript sorting

【解决方案1】：

您可以对referenceId -> lastIndex 进行映射，然后在filter 中查找索引以仅保留与referenceId 的最后一个匹配的索引。进行检查以确保我们仅过滤掉 type: "chat" 项目。时间复杂度是线性的。

const arr = [
  { id: 1, referenceId: 1, type: "normal", name: "a" },
  { id: 2, referenceId: 1, type: "normal", name: "b" },
  { id: 3, referenceId: 3, type: "chat", name: "c" },
  { id: 4, referenceId: 4, type: "normal", name: "d" },
  { id: 5, referenceId: 5, type: "chat", name: "e" },
  { id: 6, referenceId: 3, type: "chat", name: "f" }
];

const lastIdxes = arr.reduce((a, e, i) => {
  a[e.referenceId] = i;
  return a;
}, {});
const result = arr.filter((e, i) => 
  e.type !== "chat" || i === lastIdxes[e.referenceId]
);
console.log(result);

【讨论】：

【解决方案2】：

你可以像这样使用reduce，

let arr = [
  { id: 1, referenceId: 1, type: "normal", name: "a" },
  { id: 2, referenceId: 2, type: "normal", name: "b" },
  { id: 3, referenceId: 3, type: "chat", name: "c" },
  { id: 4, referenceId: 4, type: "normal", name: "d" },
  { id: 5, referenceId: 5, type: "chat", name: "e" },
  { id: 6, referenceId: 3, type: "chat", name: "f" }
];

res = arr.reduce((prev, curr) => {
  index = prev.findIndex(item => item.referenceId === curr.referenceId);
  if(index > -1) {
    prev.splice(index, 1);
  }
  
  prev.push(curr);
  return prev;
}, []);
console.log(res);

【讨论】：

【解决方案3】：

按照此问题的标题按 id 排序，然后跟踪何时删除重复项

// sort by id and drop referenceId duplicates
const arr = [
  { id: 1, referenceId: 1, type: "normal", name: "a" },
  { id: 2, referenceId: 2, type: "normal", name: "b" },
  { id: 3, referenceId: 3, type: "chat", name: "c" },
  { id: 4, referenceId: 4, type: "normal", name: "d" },
  { id: 5, referenceId: 5, type: "chat", name: "e" },
  { id: 6, referenceId: 3, type: "chat", name: "f" }
];

arr.sort((a, b) => a.id > b.id ? 1 : -1);

const toDrop = [];
const mapper = {};
arr.forEach((curr, i) => {
    if (mapper[curr.referenceId] !== undefined) {
    toDrop.unshift(mapper[curr.referenceId]);
  }
  mapper[curr.referenceId] = i;
});

toDrop.forEach(i => {
    arr.splice(i, 1);
});

console.log(arr);

【讨论】：