如何使用lodash在对象的嵌套数组中查找对象？

Question

我有这个数据集

var records = [{
    gc: '2',
    time: 1231232423,
    cards: [{
        cardCode: '12121',
        rssi: 23
      }, {
        cardCode: '12122',
        rssi: 33
      }, {
        cardCode: '12421',
        rssi: 43
      }

    ]
  },
  {
    gc: '3',
    time: 4331232233,
    cards: [{
        cardCode: '6423',
        rssi: 23
      }, {
        cardCode: '12421',
        rssi: 13
      }

    ]
  }, , {
    gc: '4',
    time: 4331232233,
    cards: [{
        cardCode: '8524',
        rssi: 03
      },
      {
        cardCode: '6423',
        rssi: 23
      }, {
        cardCode: '12421',
        rssi: 67
      }
    ]
  }
]

我有一个对象的数组(records) ，并且在每个对象中都有另一个数组(cards) 。 这意味着如果我将始终在“记录”中存在的对象内放置cards阵列。 因此，从一开始我就想遍历整个列表records ，并将所有对象的cards数组相互比较，然后找到匹配的对象的rssi值，然后将具有最小rssi值的对象推入另一个新数组。 最后，我希望以同样的方式在数组中包含所有具有最小rssi值的匹配对象。 我正在使用lodash并尝试过

 matchedRecords =   records.forEach(record=>{
 record.cards.forEach(record=>{
  _.filter(records, _.flow(
    _.property('cards'),
     _.partialRight(_.some, { cardCode: record.cardCode })
         ));
      })
   })

我想要的结果是

 [
   {
    gc : 3, 
    cards : [{
     cardCode : '12421',
     rssi : 13
    }]
   }
 ]

注意： 应在cardCode key的重低音上比较纸牌数组对象

Answer 1

更新

该版本基于对问题的评论中的长期需求讨论之后达成的理解。 它不使用lodash。 它已经变得足够复杂，可能会分解成较小的部分，而lodash可能会对此有所帮助。

它不仅包括gc ，还包括原始记录中的time ，实际上，它将包括那里的其他任何内容。 如果只需要gc ，只需从代码中删除...rest即可。

 const sharedCards = records => Object.values( Object .entries (records .reduce ( (a, {cards, ...rest}) => cards .reduce ( (a, {cardCode, rssi}) => ({ ...a, [cardCode]: [...(a[cardCode] || []), {...rest, rssi}] }), a, ), {} )) .filter ( ([code, rs]) => rs.length == records.length ) .reduce((a, [code, rs]) => { const {gc, rssi, ...rest} = rs.reduce( (r1, r2) => r2.rssi < r1.rssi ? r2 : r1, {rssi: Infinity} ) return { ...a, [gc]: { ...(a[gc] || {...rest, gc}), cards: [...((a[gc] || {gc}).cards || []), {cardCode: code, rssi}] } } }, {}) ) const records = [{gc: "2", time: 1231232423, cards: [{cardCode: "12121", rssi: 23}, {cardCode: "12122", rssi:33}, {cardCode: "12421", rssi: 43}]}, {gc: "3", time: 4331232233, cards: [{cardCode: "6423", rssi: 23}, {cardCode: "12421", rssi: 13}]}, {gc: "4", time: 4331232233, cards: [{cardCode: "8524", rssi: 3}, {cardCode: "6423", rssi: 23}, {cardCode: "12421", rssi: 67}]}]; console .log ( sharedCards (records) ) // Now we add `{cardCode: "6423", rssi: 7}` to the first record const records2 = [{gc: "2", time: 1231232423, cards: [{cardCode: "6423", rssi: 7}, {cardCode: "12121", rssi: 23}, {cardCode: "12122", rssi:33}, {cardCode: "12421", rssi: 43}]}, {gc: "3", time: 4331232233, cards: [{cardCode: "6423", rssi: 23}, {cardCode: "12421", rssi: 13}]}, {gc: "4", time: 4331232233, cards: [{cardCode: "8524", rssi: 3}, {cardCode: "6423", rssi: 23}, {cardCode: "12421", rssi: 67}]}]; console .log ( sharedCards (records2) ) 

原始答案

这只是部分解决方案，如果我们可以解决评论中的讨论，请予以更新。 它获取出现在每条记录中的所有卡，并选择rssi最低的版本。 这可能接近要求，也可能不接近。

它不使用lodash。 使用lodash可能会简化一点，但可能不会很多。

 const sharedCards = records => Object .entries (records .reduce ( (a, {cards}) => cards .reduce ( (a, {cardCode, rssi}) => ({...a, [cardCode]: [...(a[cardCode] || []), rssi]}), a, ), {} )) .filter ( ([code, rssis]) => rssis.length == records.length ) .map ( ([code, rssis]) => ({ cardCode: code, rssi: Math .min (...rssis) }) ) const records = [{gc: "2", time: 1231232423, cards: [{cardCode: "12121", rssi: 23}, {cardCode: "12122", rssi:33}, {cardCode: "12421", rssi: 43}]}, {gc: "3", time: 4331232233, cards: [{cardCode: "6423", rssi: 23}, {cardCode: "12421", rssi: 13}]}, {gc: "4", time: 4331232233, cards: [{cardCode: "8524", rssi: 3}, {cardCode: "6423", rssi: 23}, {cardCode: "12421", rssi: 67}]}]; console .log ( sharedCards (records) ) // Now we add `{cardCode: "6423", rssi: 7}` to the first record const records2 = [{gc: "2", time: 1231232423, cards: [{cardCode: "6423", rssi: 7}, {cardCode: "12121", rssi: 23}, {cardCode: "12122", rssi:33}, {cardCode: "12421", rssi: 43}]}, {gc: "3", time: 4331232233, cards: [{cardCode: "6423", rssi: 23}, {cardCode: "12421", rssi: 13}]}, {gc: "4", time: 4331232233, cards: [{cardCode: "8524", rssi: 3}, {cardCode: "6423", rssi: 23}, {cardCode: "12421", rssi: 67}]}]; console .log ( sharedCards (records2) ) 

Answer 2

与洛达什

 let records = [{"gc":"2","time":1231232423,"cards":[{"cardCode":"12121","rssi":23},{"cardCode":"12122","rssi":33},{"cardCode":"12421","rssi":43}]},{"gc":"3","time":4331232233,"cards":[{"cardCode":"6423","rssi":23},{"cardCode":"12421","rssi":13}]},{"gc":"4","time":4331232233,"cards":[{"cardCode":"8524","rssi":3},{"cardCode":"6423","rssi":23},{"cardCode":"12421","rssi":67}]}] let mapped = _.reduce(records, (acc, rec) => { _.forEach(rec.cards, (card) => { card = _.assign({}, card, { gc: rec.gc, multiple: false }) if (card.cardCode in acc) { let lc = acc[card.cardCode].rssi < card.rssi acc[card.cardCode] = lc ? acc[card.cardCode] : card acc[card.cardCode].multiple = lc } else { acc[card.cardCode] = card; } }) return acc }, {}) let result = _.reduce(records, (acc, rec) => { rec.cards = _.filter(rec.cards, (card) => mapped[card.cardCode].multiple && mapped[card.cardCode].gc === rec.gc ) rec.cards.length && acc.push(rec) return acc }, []); console.log(result) 

 .as-console-wrapper { max-height: 100%!important } 

 <script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.11/lodash.js"></script>