根據另一個陣列和條件過濾一組物件

我正在努力根據特定條件從一組物件中檢索一個子集。我有以下格式的物件陣列：

const messages = [
    {
        summary: '[x1fg66pwq-qft] Second reminder to submit supporting documents',
        date: '1624652200',
        type: 1
    },
    {
        summary: '[x1fg66pwq-fgh] Final reminder to submit supporting documents',
        date: '1629058600',
        type: 4
    },
    {
        summary: '[zy9l89ptb-yuw] Please submit your proof of address',
        date: '1631708200',
        type: 2
    },
    {
        summary: '[ggk9nygh8-pmn] Your application has been successful.',
        date: '1634300200',
        type: 1
    },
]

還有另一個陣列，它根據摘要方括號中的訊息 ID 提供要檢索的訊息：

const messageIds = ['x1fg66pwq', 'zy9l89ptb'];

結果應該是根據messageIds陣列中的內容檢索最新訊息。日期欄位在紀元中。

const result = [
    {
        summary: '[x1fg66pwq] Final reminder to submit supporting documents',
        date: '1629058600',
        type: 4
    },
    {
        summary: '[zy9l89ptb] Please submit your proof of address',
        date: '1631708200',
        type: 2
    },
]

為了實作上述目標，我嘗試組合一個過濾器并找到對我不起作用的過濾器：

const result = messages.filter((message) =>
        messageIds.find(id => message.summary.includes(testEvent))
    );

我希望上面的內容回傳具有指定摘要的陣列中的第一個結果。但是，這總是為我回傳完整的陣列而無需過濾。有人可以幫我實作這一目標嗎？

uj5u.com熱心網友回復：

在查找latest 時，您可以首先創建一個映射或普通物件，以目標訊息 ID 為鍵，然后將其date屬性具有最大值的訊息分配給每個訊息：

const messages = [{summary: '[x1fg66pwq-qft] Second reminder to submit supporting documents',date: '1624652200',type: 1},{ summary: '[x1fg66pwq-fgh] Final reminder to submit supporting documents',date: '1629058600', type: 4},{ summary: '[zy9l89ptb-yuw] Please submit your proof of address',date: '1631708200',type: 2},{  summary: '[ggk9nygh8-pmn] Your application has been successful.', date: '1634300200',type: 1},];

const messageIds = ['x1fg66pwq', 'zy9l89ptb'];

// Create object keyed by the message ids:
let obj = Object.fromEntries(messageIds.map(id => [id, null]));
// Link the message to each id with minimal date
for (let msg of messages) {
    let id = msg.summary.match(/^\[(.*?)-/)?.[1];
    if (obj[id] === null ||  msg.date >  obj[id]?.date) obj[id] = msg;
}

let result = Object.values(obj);

console.log(result);

uj5u.com熱心網友回復：

這部分很容易。這只是將它們串在一起的問題：

const matches = (messages, ids) => ids .flatMap (
  id => messages .filter (({summary}) => summary .startsWith ('['   id))
                 .sort (({date: a}, {date: b}) => b - a)
                 .slice (0, 1)
)

const messages = [{summary: '[x1fg66pwq-qft] Second reminder to submit supporting documents', date: '1624652200', type: 1}, {summary: '[x1fg66pwq-fgh] Final reminder to submit supporting documents', date: '1629058600', type: 4}, {summary: '[zy9l89ptb-yuw] Please submit your proof of address', date: '1631708200', type: 2}, {summary: '[ggk9nygh8-pmn] Your application has been successful.', date: '1634300200', type: 1}]

const messageIds = ['x1fg66pwq', 'zy9l89ptb']

console .log (matches (messages, messageIds))

此版本將簡單地忽略不匹配的 ID。這可能是也可能不是您想要的。如果您必須為每個 id 搜索大量資料，它也會做一些愚蠢的事情：它通過對它們進行排序并選擇第一個來找到最大值。對于小資料，這很好。對于較大的資料（數十萬或數百萬可能首先成為問題的地方），這是O (n log (n))，這比以下技術慢，這是O (n)：

const maxBy = (fn) => (xs) => xs .reduce (
  ({curr, max}, x) => fn(x) > max ? {curr: x, max: fn (x)} : {curr, max}, 
  {curr: null, max: -Infinity}
) .curr

const matches = (messages, ids) => ids .flatMap (
  id => maxBy (({date}) => Number(date)) 
              (messages .filter (({summary}) => summary .startsWith ('['   id)))
)

const messages = [{summary: '[x1fg66pwq-qft] Second reminder to submit supporting documents', date: '1624652200', type: 1}, {summary: '[x1fg66pwq-fgh] Final reminder to submit supporting documents', date: '1629058600', type: 4}, {summary: '[zy9l89ptb-yuw] Please submit your proof of address', date: '1631708200', type: 2}, {summary: '[ggk9nygh8-pmn] Your application has been successful.', date: '1634300200', type: 1}]

const messageIds = ['x1fg66pwq', 'zy9l89ptb']

console .log (matches (messages, messageIds))

這使用一個maxBy函式遍歷串列以根據提供的函式找到最大值，回傳null一個空陣列。

如果null未找到 id，此版本將包含在輸出中。這可能是首選行為，但如果不是，您可以包裝它以過濾非空值的輸出。

顯然，在這些解決方案中的任何一個中，您都可以替換summary .startsWith ('[' id)為summary .includes (id)，這更靈活，但可能不太準確。

uj5u.com熱心網友回復：

sort如果我正確理解你的問題，我認為你只需要在最后加上一個

.sort((b,a) => a.date - b.date)

const messages = [{
    summary: '[x1fg66pwq-qft] Second reminder to submit supporting documents',
    date: '1624652200',
    type: 1
  },
  {
    summary: '[x1fg66pwq-fgh] Final reminder to submit supporting documents',
    date: '1629058600',
    type: 4
  },
  {
    summary: '[zy9l89ptb-yuw] Please submit your proof of address',
    date: '1631708200',
    type: 2
  },
  {
    summary: '[ggk9nygh8-pmn] Your application has been successful.',
    date: '1634300200',
    type: 1
  },
]

const messageIds = ['x1fg66pwq', 'zy9l89ptb'];

const result = messages.filter((message) =>
  messageIds.some(id => message.summary.includes(id))
).sort((b, a) =>  a.date -  b.date);

console.log(result)

標籤：javascript 数组 算法 ecmascript-6 过滤

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