从数组中删除重复条目，但保留最完整的条目

Question

我有一个多维数组，其值如下：

let unfilteredArray = [
  ["1234", "no email entered", "other value", null, 7, "another value"],
  ["3333", "b@example.com", "another value", 2, 10, "something else"],
  ["1234", "a@example.com", "random value", 2, null, "something else"],
  ["4444", "c@example.com", "another value", 29, 3, "xxx"],
  ["5555", "abcd", "another value", 3, 41, "yyy"],
  ["1234", "another random text", "another value", 4, 8, "zzz"],
  ["5555", "efgh", "another value", null, 0, null]
];

我想提取唯一的用户 ID，但在列索引 1 中保留带有电子邮件的条目。如果给定的用户 ID 没有电子邮件，则使用给定用户 ID 的最后一行。所选行的所有其他列应保留在输出中。

上述值的结果应如下所示：

let unfilteredArray = [
  ["1234", "a@example.com", "random value", 2, null, "something else"],
  ["3333", "b@example.com", "another value", 2, 10, "something else"],
  ["4444", "c@example.com", "another value", 29, 3, "xxx"],
  ["5555", "efgh", "another value", null, 0, null] // no email for User ID 5555 => last 5555 row used
];

到目前为止，我有一个删除重复的脚本：

// index = index of the column with User ID
function removeDuplicates(array, index) {
  // Gets only the first entry - the rest are removed
  let filteredArray = [];
  for (let i = 1; i < array.length; i++) {
    let isDuplicate = false;
    for (let j = i-1; j >= 0; j--) {
      if (array[i][index] == array[j][index]) {
        isDuplicate = true;
        break;
      }
    }
    if (!isDuplicate) {
      filteredArray.push(array[i]);
    }
  }
  return filteredArray;
}

如何确保返回带有电子邮件的行？

Answer 1

将数组简化为 Map，并在 Map 中仅包含 ID (key) 但 Map 上不存在的行，或者其值为电子邮件的行。使用数组扩展将 Map 的 .values() 转换回数组，或 Array.from()。

注意：validateEmail(email) 函数取自 answer。

function validateEmail(email) {
  const re = /^(([^<>()[\]\\.,;:\s@"]+(\.[^<>()[\]\\.,;:\s@"]+)*)|(".+"))@((\[[0-9]{1,3}\.[0-9]{1,3}\.[0-9]{1,3}\.[0-9]{1,3}\])|(([a-zA-Z\-0-9]+\.)+[a-zA-Z]{2,}))$/;
  return re.test(String(email).toLowerCase());
}

function removeDuplicates(array) {
  return Array.from(
    array.reduce((acc, row) => {
      const [key, value] = row;
      
      if(!acc.has(key) || validateEmail(value)) {
        acc.set(key, row);
      }
     
      return acc;
    }, new Map()).values()
  )
}

const unfilteredArray = [["1234","no email entered","other value",null,7,"another value"],["3333","b@example.com","another value",2,10,"something else"],["1234","a@example.com","random value",2,null,"something else"],["4444","c@example.com","another value",29,3,"xxx"],["5555","abcd","another value",3,41,"yyy"],["1234","another random text","another value",4,8,"zzz"],["5555","efgh","another value",null,0,null]];

const result = removeDuplicates(unfilteredArray);

console.log(result);

Answer 2

我在 Stackoverflow 上结合了一些答案，得到了这段代码。我将您的数组更改为对象。如果不重要，这里是我的结果代码

var unfilteredArray = [
  {id:1234, text: "no email entered"},
  {id:3333, text: "b@example.com"},
  {id:1234, text: "a@example.com"},
  {id:4444, text: "c@example.com"},
  {id:5555, text: "abcd"},
  {id:1234, text: "another random text"},
  {id:5555,text: "efgh"}
];

function isValidEmail(email) {
    const re = /^(([^<>()[\]\\.,;:\s@"]+(\.[^<>()[\]\\.,;:\s@"]+)*)|(".+"))@((\[[0-9]{1,3}\.[0-9]{1,3}\.[0-9]{1,3}\.[0-9]{1,3}\])|(([a-zA-Z\-0-9]+\.)+[a-zA-Z]{2,}))$/;
  return re.test(String(email).toLowerCase());
}

var arr2 = unfilteredArray.reduce( (a,b) => {
    var i = a.findIndex( x => x.id === b.id);
    return i === -1 ? a.push({ id : b.id, text : b.text }) : a[i].text = 
    this.isValidEmail(a[i].text) ? a[i].text: b.text, a;
}, []);

console.log(arr2)

*验证邮件功能链接 => here