数组过滤出多重复项时生成可展示错误详情数组的优化方案咨询

我有一个从Excel表格提取的对象数组，目前使用.filter()方法对比主对象的特定键来返回新数组，处理逻辑如下：

• 如果返回数组长度为0，就不把额外数据合并到原对象中
• 如果返回数组长度为1，就将其键展开到当前对象，之后按需替换内容
• 但如果数组长度大于1，我想要给用户展示找到的重复项详情，让他们可以修正表格后重试

现有代码示例 #

/* Current user I am getting from for looping the firstSheet */
let currentUser = {'First Name': 'Billy', 'Last Name': 'Bob', 'Member Number': 3333,'Home Phone': 9764582917, 'Cell Phone': 2345677123, 'Day Phone': 9928844574, 'Email': 'seeBill@gmail.com', 'plan price': '10.10' }

/* this is found from checking further in the first sheet by a small for loop that runs as it finds duplicates, ussaly there is only 1 object in this. There is a line per plan*/
let dupeArray = [{'First Name': 'Billy', 'Last Name': 'Bob', 'Member Number': 1111,'Home Phone': 9764582917, 'Cell Phone': 2345677123, 'Day Phone': 9928844574, 'Email': 'seeBill@gmail.com', 'plan Price': '10.10'  }]

let secondSheet = [
    {'First Name': 'Billy', 'Last Name': 'Bob', 'plan1': 1111, 'plan2': 2222, 'plan3': 3333, 'Home Phone': 9764582917, 'Cell Phone': 2345677123, 'Day Phone': 9928844574, 'Email': 'seeBill@gmail.com', 'plan1 Price': '10.10', 'plan2 Price': '10.10', 'plan3 Price': '10.10'  },
    {'First Name': 'Benny', 'Last Name': 'Socks', 'plan1': 5315, 'plan2': 5562, 'plan3': 3333, 'Home Phone': 1122331467, 'Cell Phone': 9867485926, 'Day Phone': 1759603752, 'Email': 'seeBennySocks@gmail.com', 'plan1 Price': '10.10', 'plan2 Price': '10.10', 'plan3 Price': '10.10'  },
    {'First Name': 'Sara', 'Last Name': 'Jones', 'plan1': 5645, 'plan2': 5522, 'plan3': 6678, 'Home Phone': 6672854673, 'Cell Phone': 7766552897, 'Day Phone': 9856301788, 'Email': 'seeSara@gmail.com', 'plan1 Price': '10.10', 'plan2 Price': '10.10', 'plan3 Price': '10.10'  },
    {'First Name': 'Jones', 'Last Name': 'Smith', 'plan1': 2344, 'plan2': 7766, 'plan3': 8763, 'Home Phone': 2221113323, 'Cell Phone': 7766552897, 'Day Phone': 1115566779, 'Email': 'seeJones@gmail.com', 'plan1 Price': '10.10', 'plan2 Price': '10.10', 'plan3 Price': '10.10'  },
]

let existingContacts;
function handleUpdate(){
    existingContacts = secondSheet.filter((currContact) => {
        if( [currContact['plan1'], currContact['plan2'], currContact['plan3']].indexOf(currentUser['memberNumber']) > -1 || currContact['Email'] === currentUser['Email'] || currContact['Cell Phone'] === currentUser['Cell Phone'] ) {
            return true;
        } else {
            for (let a = 0; a < dupeArray.length; a++) {
                if([currContact['plan1'], currContact['plan2'], currContact['plan3'],].indexOf(dupeArray[a]['Member Number']) > -1) {
                    return true
                }
            }
            return false;
        }
    })
    if (existingContacts.length === 1) {return true;}
    else if (existingContacts.length > 1) { /* Wanna push was found in the .filter() here */ return false; }
    else { return false; }
}

console.log(handleUpdate())
/* This is just to explain why the function itself returns a boolean */
// filteredSheet.push({
//     ...(handleUpdate() && existingContacts[0]),
//     /* then I replace what i need to from currentUser here */
// })

我已经处理了一些细节，比如转换值避免空邮箱之类的错误，现在只需要实现：当找到多个重复项时，生成一个可以传给弹窗组件的详情数组，示例错误日志如下：

Error Log:
Billy Bob: duplicate with Benny Socks at Member Number
...etc

我的初步思路 #

我想到的方法是在函数开头声明modalArray，然后在existingContacts.length > 1的分支里遍历数组，逐一判断重复原因并添加到数组中，但感觉这是 brute force（暴力解法），不够优雅：

else if (existingContacts.length > 1) { 
        let modalArray = []; // 需要在函数顶部声明
        for(let b=0; b < existingContacts.length; b++) {
            if([existingContacts[b]['plan1'], existingContacts[b]['plan2'], existingContacts[b]['plan3']].indexOf(currentUser['memberNumber']) > -1) {
                modalArray.push({ currentUserFirstName: currentUser['First Name'], currentUserlastName: currentUser['Last Name'],  issueFound: 'Member Number', currContactFirstName: existingContacts[b]['First Name'], currContactLastName: existingContacts[b]['Last Name']})
            
            } else if (existingContacts[b]['Email'] === currentUser['Email']) { 
                modalArray.push({ currentUserFirstName: currentUser['First Name'], currentUserlastName: currentUser['Last Name'],  issueFound: 'Email', currContactFirstName: existingContacts[b]['First Name'], currContactLastName: existingContacts[b]['Last Name']}) 
            
            } else if (existingContacts[b]['Cell Phone'] === currentUser['Cell Phone']) {
                modalArray.push({ currentUserFirstName: currentUser['First Name'], currentUserlastName: currentUser['Last Name'],  issueFound: 'Cell Phone', currContactFirstName: existingContacts[b]['First Name'], currContactLastName: existingContacts[b]['Last Name']}) 
            
            } else {
                for (let c = 0; c < dupeArray.length; c++) { // lol
                    if([existingContacts[b]['plan1'], existingContacts[b]['plan2'], existingContacts[b]['plan3'],].indexOf(dupeArray[c]['Member Number']) > -1) {
                        modalArray.push({ currentUserFirstName: currentUser['First Name'], currentUserlastName: currentUser['Last Name'],  issueFound: 'Member Number', currContactFirstName: existingContacts[b]['First Name'], currContactLastName: existingContacts[b]['Last Name']}) 
                    }
                }
            }
        }
        return false;
    }

我希望能找到更简洁、可维护的方法来生成这个可以传给弹窗组件的数组。

备注：内容来源于stack exchange，提问作者Zu0s