如何获取集合中每个发件人的最新消息？

Question

我有一个这样的对象数组：

const messages = [ 
  {message: "ghhhhhhhh", receiver: "OX0pReHXfXUTq1XnOnTSX7moiGp2", sender: "14", time: "12:56"},
  {message: "ggggggghjjgcgh", receiver: "OX0pReHXfXUTq1XnOnTSX7moiGp2", sender: "ZCiuWczin3VuibH59MISuEqR3pc2", time: "12:45"},
  {message: "good afternoon", receiver: "OX0pReHXfXUTq1XnOnTSX7moiGp2", sender: "ZCiuWczin3VuibH59MISuEqR3pc2", time: "12:41"},
  {message: "hfdsghfdfhjo", receiver: "OX0pReHXfXUTq1XnOnTSX7moiGp2", sender: "ZCiuWczin3VuibH59MISuEqR3pc2", time: "12:38"},
  {message: "hhhhhhhhhhhhh ", receiver: "OX0pReHXfXUTq1XnOnTSX7moiGp2", sender: "14", time: "11:50"}
];

我想获取每个发件人的最新消息，像这样：

const messages = [
  {message: "ghhhhhhhh", receiver: "OX0pReHXfXUTq1XnOnTSX7moiGp2", sender: "14", time: "12:56"},
  {message: "ggggggghjjgcgh", receiver: "OX0pReHXfXUTq1XnOnTSX7moiGp2", sender: "ZCiuWczin3VuibH59MISuEqR3pc2", time: "12:45"}
];

怎么做？

Answer 1

您可以为此目的使用array.reduce，例如：

const messages = [ {message: "ghhhhhhhh", receiver: "OX0pReHXfXUTq1XnOnTSX7moiGp2", sender: "14", time: "12:56"}
, {message: "ggggggghjjgcgh", receiver: "OX0pReHXfXUTq1XnOnTSX7moiGp2", sender: "ZCiuWczin3VuibH59MISuEqR3pc2", time: "12:45"}
, {message: "good afternoon", receiver: "OX0pReHXfXUTq1XnOnTSX7moiGp2", sender: "ZCiuWczin3VuibH59MISuEqR3pc2", time: "12:41"}
, {message: "hfdsghfdfhjo", receiver: "OX0pReHXfXUTq1XnOnTSX7moiGp2", sender: "ZCiuWczin3VuibH59MISuEqR3pc2", time: "12:38"}
, {message: "hhhhhhhhhhhhh ", receiver: "OX0pReHXfXUTq1XnOnTSX7moiGp2", sender: "14", time: "11:50"}];


function parseTime(timeStr) {
    const fields = timeStr.split(":").map(parseInt);
    return fields[0] * 60 + fields[1];
}

let result = messages.reduce( (map, item) => { 
   if (!map[item.sender] || parseTime(map[item.sender].time) < parseTime(item.time)) {
       map[item.sender] = item;
   } 
   return map;
}, {});

console.log("Latest messages:", Object.values(result));

Answer 2

你可以这样做：-

const messages = [ {message: "ghhhhhhhh", receiver: "OX0pReHXfXUTq1XnOnTSX7moiGp2", sender: "14", time: "12:56"}
, {message: "ggggggghjjgcgh", receiver: "OX0pReHXfXUTq1XnOnTSX7moiGp2", sender: "ZCiuWczin3VuibH59MISuEqR3pc2", time: "12:45"}
, {message: "good afternoon", receiver: "OX0pReHXfXUTq1XnOnTSX7moiGp2", sender: "ZCiuWczin3VuibH59MISuEqR3pc2", time: "12:41"}
, {message: "hfdsghfdfhjo", receiver: "OX0pReHXfXUTq1XnOnTSX7moiGp2", sender: "ZCiuWczin3VuibH59MISuEqR3pc2", time: "12:38"}
, {message: "hhhhhhhhhhhhh ", receiver: "OX0pReHXfXUTq1XnOnTSX7moiGp2", sender: "14", time: "11:50"}]

const getUniqueMessages = (messages) => {
    const msgMap = {};
    const uniqueMsg = [];
    messages.forEach(item => {
        if(!msgMap[item.sender]) {
            msgMap[item.sender] = true;
            uniqueMsg.push(item)
        }
    });

   return uniqueMsg;   
}

// console.log(getUniqueMessages(messages));

Answer 3

const result = [];
const map = new Map();
for (const item of messages) {
    if(!map.has(item.sender)){
        map.set(item.sender, true);
        result.push({
            message: item.message,
    reciever: item.reciever
            sender: item.sender,
            time: item.time
        });
    }
}
console.log(result)

Answer 4

一种快速 (O(n)) 且可读性强的方法

实际上，这是一个非常好的问题。我觉得我可以提供一个比以前的答案更高效的答案（如果您要一遍又一遍地运行此操作，这很重要），同时也非常易于阅读和理解。

您基本上有一个集合，并希望减少按某个属性对它们进行分组的对象。

案例1：集合排序

如果我们可以假设集合首先按最新消息排序（看起来很像），我们可以只保留我们看到的第一个消息：

const messages = [ 
  {message: "ghhhhhhhh", receiver: "OX0pReHXfXUTq1XnOnTSX7moiGp2", sender: "14", time: "12:56"},
  {message: "ggggggghjjgcgh", receiver: "OX0pReHXfXUTq1XnOnTSX7moiGp2", sender: "ZCiuWczin3VuibH59MISuEqR3pc2", time: "12:45"},
  {message: "good afternoon", receiver: "OX0pReHXfXUTq1XnOnTSX7moiGp2", sender: "ZCiuWczin3VuibH59MISuEqR3pc2", time: "12:41"},
  {message: "hfdsghfdfhjo", receiver: "OX0pReHXfXUTq1XnOnTSX7moiGp2", sender: "ZCiuWczin3VuibH59MISuEqR3pc2", time: "12:38"},
  {message: "hhhhhhhhhhhhh ", receiver: "OX0pReHXfXUTq1XnOnTSX7moiGp2", sender: "14", time: "11:50"}
];

/** @type {Map<String, Object>} */
const latestMessageBySender = new Map();

// retain the most recent message from every unique sender
for (const message of messages) {
    if (!latestMessageBySender.has(message.sender)) {
        latestMessageBySender.set(message.sender, message);
    }
}

// collect and show the resulting messages
for (const message of latestMessageBySender.values()) {
    console.info(message);
}

我们在这里使用Map，这是正确的现代方式（我们的意图比使用旧的、hacky 的{} 更清晰）。

由于最新消息最先出现，我们只需要检查我们的地图是否已经包含该键。如果是，什么也不做；否则，将当前消息添加到地图中。

案例2：集合未排序

另一方面，如果不能保证集合已排序，则不必像其他答案建议的那样解析时间字段，我们可以通过按字典顺序比较字符串来更快地做到这一点：

if (message1.time > message2.time) {
    // message 1 is more recent than message 2
}

我们在这里考虑时间戳的格式正确（2 位代表小时，2 位代表分钟，即/\d\d:\d\d/）。换句话说，我们期望04:07 而不是4:7。我们还假设我们只有同一天的消息，因为没有给出日期。

因此，在这种情况下，最终代码将是：

const messages = [ 
  {message: "ghhhhhhhh", receiver: "OX0pReHXfXUTq1XnOnTSX7moiGp2", sender: "14", time: "12:56"},
  {message: "ggggggghjjgcgh", receiver: "OX0pReHXfXUTq1XnOnTSX7moiGp2", sender: "ZCiuWczin3VuibH59MISuEqR3pc2", time: "12:45"},
  {message: "good afternoon", receiver: "OX0pReHXfXUTq1XnOnTSX7moiGp2", sender: "ZCiuWczin3VuibH59MISuEqR3pc2", time: "12:41"},
  {message: "hfdsghfdfhjo", receiver: "OX0pReHXfXUTq1XnOnTSX7moiGp2", sender: "ZCiuWczin3VuibH59MISuEqR3pc2", time: "12:38"},
  {message: "hhhhhhhhhhhhh ", receiver: "OX0pReHXfXUTq1XnOnTSX7moiGp2", sender: "14", time: "11:50"}
];

/** @type {Map<String, Object>} */
const latestMessageBySender = new Map();

// retain the most recent message from every unique sender
for (const message of messages) {
    const previousMessage = latestMessageBySender.get(message.sender);
    if (!previousMessage || message.time > previousMessage.time) {
        latestMessageBySender.set(message.sender, message);
    }
}

// collect and show the resulting messages
for (const message of latestMessageBySender.values()) {
    console.info(message);
}

在向映射添加消息之前，我们首先检查是否已经存在相同键的消息。如果没有，只需添加新的；如果有，按字典顺序比较 time 字段并替换现有字段，以防新字段更新。

这里需要注意的重要一点是，我们实际上并不希望首先对数组进行排序。那将是O(n log n)，但我们可以通过在进行时进行比较来避免O(n)。