如何从数组中获取不匹配的元素？

Question

所以我有从缓存中获取的具有rxInfos和specialMembers的请求，因此如果rxInfos.indexID匹配，则进行其他处理，如果rxInfos indexID存在，但由于某种原因与缓存indexID不同，我想将该indexID推送到mismatchedIndexIDs数组。 就像第二个元素一样，rxInfos.indexID与任何specialMembers不匹配，因此应推送到mismatchedIndexIDs。

下面的代码将所有specialMembers id推送到不匹配的数组。

main.ts

 for (const member of specialMembers) {
        for (const rxInfo of this.rxInfos) {
            if (member.indexID === rxInfo.indexID) {
                this.indexIDs.push(rxInfo.indexID);
                proxyMember = member;
                if (!member.dateOfBirth) {
                    statusDesc = "member dateOfbirth not found";
                    return Promise.reject(this.errorHandler(request, statusDesc));
                }
                const requestBody: any = this.buildSingleRequestBody(proxyMember, rxInfo);
                const requestObject = this.specialtyQuestionRequest(requestBody);
                this.requestArray.push(requestObject);
            } else {
                this.mismatchIndexIDS.push(rxInfo.indexID);
                this.indexIdMismatchCounter++;
            }
        }
    }

数据

"rxInfos": [
      {
            "drugNdc": "10101",
            "rxNumber": "14556459709",
            "firstFillIndicator": "N",
            "sourceSystem": "TBS",
            "indexID": "RPT0ifQ"
        },
      {
            "drugNdc": "101",
            "rxNumber": "145945000709",
            "firstFillIndicator": "N",
            "sourceSystem": "TBS",
            "indexID": "GJhQ1MrQnZkTFRR"
        }
    ]

    "specialyMembers":[
      {
        "dob":"12-12-1970"
        "firstName": "jimmy",
        "lasteName": "shew",
         "indexID": "RPT0ifQ"
      },
      {
         "dob":"18-10-1970"
        "firstName": "Timmy",
        "lasteName": "Doug",
         "indexID": "GJhQ1MrQ"
      },
        {
         "dob":"17-06-1981"
        "firstName": "John",
        "lasteName": "owascar",
         "indexID": "GJhQ1MrTGDSRQ"
      }
      ]

Answer 1

const memberMatched = member => {
  return rxInfos.find(rxInfo => rxInfo.indexID === member.indexID)
}

const mismatchIndexIDS = specialMembers.reduce((res, member) => {
  return memberMatched(member) ? res : res.concat(member.indexID)
}, [])

const indexIdMismatchCounter = mismatchIndexIDS.length

const matchedMembers = specialMembers.reduce((res, member) => {
  const rxRecord = memberMatched(member)
  return rxRecord ? res.concat({rxRecord, member}) : res
}, [])

// DO YOUR STUFF WITH MATCHED MEMBERS

Answer 2

有一个更好的方法与做这个find ，将直接获得成员specialMembers具有一定indexID 。 而且，您只需要处理一个for循环（另一个则在find是隐式的）：

specialMembers.find(member => member.indexID === rxInfo.indexID)

for循环的简化版本如下：

 var rxInfos = [{ "drugNdc": "10101", "rxNumber": "14556459709", "firstFillIndicator": "N", "sourceSystem": "TBS", "indexID": "RPT0ifQ" }, { "drugNdc": "101", "rxNumber": "145945000709", "firstFillIndicator": "N", "sourceSystem": "TBS", "indexID": "GJhQ1MrQnZkTFRR" } ]; var specialMembers = [{ "dob": "12-12-1970", "firstName": "jimmy", "lasteName": "shew", "indexID": "RPT0ifQ" }, { "dob": "18-10-1970", "firstName": "Timmy", "lasteName": "Doug", "indexID": "GJhQ1MrQ" }, { "dob": "17-06-1981", "firstName": "John", "lasteName": "owascar", "indexID": "GJhQ1MrTGDSRQ" } ]; var indexIDs = [], mismatchIndexIDS = [], indexIdMismatchCounter = 0; for (const rxInfo of rxInfos) { const member = specialMembers.find(member => member.indexID === rxInfo.indexID); if (member) { indexIDs.push(rxInfo.indexID); console.log('process', member); } else { mismatchIndexIDS.push(rxInfo.indexID); indexIdMismatchCounter++; } } console.log('index ids', indexIDs); console.log('mismatches', mismatchIndexIDS); console.log('counter', indexIdMismatchCounter);