根据另一个数组和条件筛选对象数组

Question

我很难根据特定的条件从对象数组中检索子集。我有如下格式的对象数组：

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
    },
]

还有一个数组根据摘要方括号中的消息id提供要检索的消息：

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

结果应该是根据messageIds数组中的内容检索最新消息。日期字段是划时代的。

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
    },
]

为了实现上述目标，我尝试将一个过滤器组合在一起，发现它对我不起作用：

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

我希望上面的结果返回数组中的第一个结果，该数组指定了摘要。但是，这总是为我返回完整的数组而不进行过滤。有人能帮我做到这一点吗？

Answer 1

我认为如果我正确理解了你的问题，你只需要在最后插入一个sort。

.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)

​

Answer 2

在查找最新消息时，您可以首先创建一个映射或普通对象，由目标消息ID键控，然后为每个消息分配具有其date属性最大价值的消息：

​

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);

​

Answer 3

这件事很简单。这只是把它们串在一起的问题：

​

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))

​

这个版本只会忽略不匹配的ids。这可能是你想要的，也可能不是你想要的。如果您必须为每个id搜索大量数据，它还会做一些愚蠢的事情:它通过排序和选择第一个找到最大值。对于小数据来说，这很好。对于更大的数据(数亿或数以百万计的数据可能首先是一个问题)，这是O (n log (n))，它比下面的技术O (n)慢

​

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))

​

这使用了一个maxBy函数，它迭代列表，根据提供的函数找到最大值，返回空数组的null。

在没有找到id的情况下，此版本将在输出中包含null。这可能是首选行为，但如果不是，则可以将其包装以筛选非空值的输出。

显然，在这两种解决方案中，您都可以将summary .startsWith ('[' + id)替换为summary .includes (id)，后者更灵活，但可能不太精确。