使用 jQuery 根据字符串内容和重复计数过滤数组

Question

我会解释需要什么。希望您能提供帮助。 我有一个数组。首先，两个数字是无关紧要的权重，然后是空格和一个电子邮件。

var arr= new Array("10 firstemail@email.com","15 secondemail@email.com","25 secondemail@email.com","35 secondemail@email.com","15 firstemail@email.com"); // huge array with many emails not just 2

我想要一个新数组，按电子邮件过滤，重复次数最少。

应该是：

["10 firstemail@email.com","15 firstemail@email.com"];

因为 firstemail@email.com 只重复 2 次，而 secondemail@email.com 重复 3 次。

使用的浏览器是带有 jQ​​uery 的 IE8（必须使用 jQuery，因为原生 .filter() 不起作用）

我知道这可能可以通过几个 foreach/fors 来完成，但我正在寻找优化的解决方案。

例如：

new_arr = $(arr).filter(function (str) { return arr[str].indexOf("firstemail@email.com") >=0; });

如果您知道 firstemail@email.com 在数组中重复次数最少，则此方法有效，但这不是解决方案，因为我不知道哪个电子邮件重复次数最少.. 可能 Math.min 可以以某种方式使用，但我会很高兴得到帮助。

编辑：请看一下这段代码

var arr= new Array("10 firstemail@email.com","15 secondemail@email.com","25 secondemail@email.com","35 secondemail@email.com","15 firstemail@email.com"); // huge array with many emails not just 2

function extractEmails (text) { return text.match(/([a-zA-Z0-9._-]+@[a-zA-Z0-9._-]+\.[a-zA-Z0-9._-]+)/gi); } // Please edit the regEx to get least repeated email instead all

var get_least_repeated=extractEmails(arr.toString());

new_arr = $(arr).filter(function (str) { return arr[str].indexOf(get_least_repeated) >=0; });

console.log(new_arr);

这似乎效果最好，但我根本不擅长 regEx 来更改 extractEmails 函数以获取最少重复的电子邮件（目前得到所有）

另一个想法是以某种方式使用 jQuery.grep()。

编辑 2：我想出了我自己的优化代码，以 @arielb 的想法将电子邮件作为键和重复计数作为值的对象。

最终代码

var arr = new Array(
  "15 secondemail@email.com",
  "10 firstemail@email.com",
  "25 secondemail@email.com",
  "35 secondemail@email.com",
  "15 firstemail@email.com",
 "12 third@email.com",
 "35 secondemail@email.com"
); // huge array with many emails nut just 2


    var new_emails={}; //init an object
$.each(arr,function(k,v){
  var justmail=v.split(" ")[1];
  if(typeof(new_emails[justmail])=='undefined') new_emails[justmail]=0;
  new_emails[justmail]+=1; 
  });

var selectedemail=''; var eminvalue=Number.MAX_VALUE;
$.each(new_emails,function(k,v){
if(v<eminvalue) {eminvalue=v;selectedemail=k;}
});

var new_arr=$(arr).filter(function (str) { return arr[str].indexOf(selectedemail) >=0; });

console.log(new_arr);

Answer 1

选项 1

var arr= new Array(
  "10 firstemail@email.com",
  "15 secondemail@email.com",
  "25 secondemail@email.com",
  "35 secondemail@email.com",
  "15 firstemail@email.com"); // huge array with many emails nut just 2

// Collect each unique email with its count
// and an array of each occurrence of the email
// in the original array
var emails = {};
$.each(arr, function(index, str) {
  var email = str.split(' ')[1];
  if (!emails[email]) {
    emails[email] = {values: [], count: 0}
  }
  emails[email].values.push(str);
  emails[email].count++;
});

var emailCounts = [];
// Create an array with unique emails and their counts
$.each(emails, function(email, data) {
  emailCounts.push([email, data.count])
})

// Sort the unique emails by count
emailCounts.sort(function(a, b) {
    return a[1] - b[1];
});

// First element in the sorted array is the least frequent
var leastFrequentEmail = emailCounts[0][0];

console.log(emails[leastFrequentEmail].values)

https://jsbin.com/milemugota/edit?js,console

它在整个数组上迭代一次，然后在唯一数组上迭代两次（一次用于收集计数，一次用于排序）。

选项 2

遵循https://www.sneppets.com/java/find-least-common-element-in-unsorted-array/的算法1

function emailPart(str) {
  return str.split(' ')[1];
}

var arr = new Array(
  "10 firstemail@email.com",
  "15 secondemail@email.com",
  "25 secondemail@email.com",
  "35 secondemail@email.com",
  "15 firstemail@email.com"); // huge array with many emails nut just 2

// sort by email
arr.sort(function(a, b) {
  var emailA = emailPart(a);
  var emailB = emailPart(b)
  if (emailA > emailB) {
    return 1;
  } else if (emailA < emailB) {
    return -1;
  } else {
    return 0;
  }
})

var currentEmailCount = 1;
var minEmailCount = arr.length + 1;
var leastCommonEmail = emailPart(arr[0]);

for (var i = 1; i < arr.length; i++) {
  if (emailPart(arr[i]) == emailPart(arr[i - 1])) {
    currentEmailCount++;
  } else {
    if (currentEmailCount < minEmailCount) {
      minEmailCount = currentEmailCount;
      leastCommonEmail = emailPart(arr[i - 1])
    }
    currentEmailCount = 1;
  }
}

if (currentEmailCount < minEmailCount) {
  leastCommonEmail = emailPart(arr[arr.length - 1]);
}

console.log(leastCommonEmail)

https://jsbin.com/gitacoqipa/edit?js,console

Answer 2

我会使用一个对象\字典来计算每封电子邮件的出现次数，而不需要任何过滤器。 也许是这样的：

    function check(arr){
    var emails={}; //init an object
    for(var i=0;i<arr.length;i++){
     emails[arr[i]]=++emails[arr[i]] || 1;
    }
    //now we are going to check which email is the least.
    var arrOccurences = [];
    var minValue = arr.length;
    var selectedEmail = '';
    jQuery.each(emails,function(key,value){
     if(value<minValue){
       selectedEmail=key;
       minValue = value;
    }

    });
    console.log(selectedEmail);
  }