使用高阶函数合并相等的对象

Question

如果它们的值彼此相等，我想从数组中“合并”对象。

这是一个例子：

"relations_entities": [
  {
    "relation": {
      "label": string,
      "inEntityId": string,
      "outEntityId": string,
      "proof": [
        {
          "text": string,
          "confidence": number,
        }
      ],
    },
    "entity": {
      "entityId": string,
      "label": string,
      "text": string,
    },
  },
  {
    "relation": {
      "label": string,
      "inEntityId": string,
      "outEntityId": string,
      "proof": [
        {
          "text": string,
          "confidence": number,
        }
      ],
    },
    "entity": {
      "entityId": string,
      "label": string,
      "text": string,
    },
  },
]

如果relations_entities[0].relation.label和relations_entities[0].entity.label等于relations_entities[0].relation.label和relations_entities[0].entity.label

然后我需要这个对象：

"relations_entities": [
  {
    "relation": {
      "label": string,
      "inEntityId": string,
      "outEntityId": string,
      "proof": [
        {
          "text": string,
          "confidence": number,
        },
        {
          "text": string,
          "confidence": number,
        }
      ],
    },
    "entity": {
      "entityId": string,
      "label": string,
      "text": string,
    },
  },
]

这两个证明合并在一起。

我尝试使用过滤器实现此行为，但我失去了理智。

也许使用lodash库？

有任何想法吗？

Answer 1

看看它：

 let relations_entities = [ { relation: { label: 'string', inEntityId: 'string', outEntityId: 'string', proof: [ { text: 'string', confidence: 'number', } ], }, entity: { entityId: 'string', label: 'string', text: 'string', }, }, { relation: { label: 'string', inEntityId: 'string', outEntityId: 'string', proof: [ { text: 'string', confidence: 'number', } ], }, entity: { entityId: 'string', label: 'string', text: 'string', }, }, ]; let merged = []; relations_entities.forEach(el => { let found = false; if (merged.length > 0) { merged.forEach(elM => { if (el.relation.label === elM.relation.label && !found) { elM.relation.proof = elM.relation.proof.concat(el.relation.proof); found = true; } }); } if (!found || merged.length === 0) { merged.push(el); } }); console.log(merged); 

Answer 2

这是一个使用es6胖箭头功能的解决方案，这可以通过传入您想要检查的键来改进。 如果您需要此功能，我不介意将其添加到您的答案中以获得完整性。

 var relations_entities = [ { "relation": { "label": 'a', "inEntityId": '123', "outEntityId": '123', "proof": [ { "text": '123', "confidence": '123', } ], }, "entity": { "entityId": '123', "label": '123', "text": '123', }, }, { "relation": { "label": 'b', "inEntityId": '321', "outEntityId": '321', "proof": [ { "text": '321', "confidence": '321', } ], }, "entity": { "entityId": '321', "label": '123', "text": '321', }, }, { "relation": { "label": 'c', "inEntityId": '321', "outEntityId": '321', "proof": [ { "text": '321', "confidence": '321', } ], }, "entity": { "entityId": '321', "label": '123', "text": '321', }, }, { "relation": { "label": 'b', "inEntityId": '321', "outEntityId": '321', "proof": [ { "text": '321', "confidence": '321', } ], }, "entity": { "entityId": '321', "label": '123', "text": '321', }, } ] const compareEntities = (oldEntities) => { // early breakout if there is just one element if (oldEntities.length === 0) { return oldEntities } // here we iterate through each of the passed in entities var newEntities = oldEntities.filter((entity, index, entities) => { // set the internal counter to 1 higher than the filtered counter // (the next object along) var i = index + 1 const relationlabel = entity.relation.label const entityLabel = entity.entity.label // iterate through each remaining entity and remove from array if there is a match while (i < entities.length) { const relationLabelIsEqual = relationlabel === entities[i].relation.label const entityLabelIsEqual = relationlabel === entities[i].relation.label if (relationLabelIsEqual && entityLabelIsEqual) { i = index + 1 return false } // if there is no match we just want to move onto the next entity ++i } // if we have iterated through all entities then there was no match // - return it into the new array return true }) return newEntities } console.log(compareEntities(relations_entities)) 

您也可以使用我设置的repl进行游戏