javascript - 在数组中查找丢失的组成员

我有四个列表-两个用于用户，两个用于组。我必须检查第一用户列表中的用户是否是第二列表中的用户。

我的代码：

importPackage(java.io);
importPackage(java.lang);

var list1 = "Han SOLO, Luke SKYWALKER, Darth VADER, Boba FETT".split(', ')
,list2 = "Luke SKYWALKER, Han SOLO, Lando CALRISSIAN, Boba FETT".split(', ')
// group lists
,group1 = "Group1, Group1, Group1, Group2".split(', ')
,group2 = "Group1, Group1, Group2, Group2".split(', ')
,j = 0
,msg = "";

   for(var j = 0; j < list2.length; j++) {

      if (list1.indexOf(list2[j]) == -1)
         msg = msg + list1[j] + " has been added to " + group1[j] + "<br />";

   }

print(msg);

该代码的结果是：


  Darth VADER已添加到组1


这个结果对我来说是好的，但是当我进行更改时：

var list1 = "Han SOLO, Luke SKYWALKER, Darth VADER, Boba FETT".split(', ')
,list2 = "Luke SKYWALKER, Han SOLO, Darth VADER, Boba FETT".split(', ')
// group lists
,group1 = "Group1, Group1, Group1, Group2".split(', ')
,group2 = "Group1, Group1, Group2, Group2".split(', ')

结果为空，但应为：


  Darth VADER已添加到组1


我想知道应该更改什么以更正第二个脚本（两个用户列表相同）。

最佳答案

您可以在检查之前做一些准备工作：

1）按喜欢的用户分组

function group(users, groups) {
  var result = {};
  for (var i = 0; i < users.length; i++) {
    var user = users[i];
    if (!result[user]) result[user] = {};
    result[users[i]][groups[i]] = true;
  }
  return result;
}

2）检查不同的组

function comp(one, two) {
  function getGroups(u) {
    var groups = [];
    for (var i in u) groups.push(i);

    return groups;
  }

  var res = {};
  for (var i in one) {
    if (!two[i]) {
      res[i] = getGroups(one[i]);
    } else {
      var g = [],
        g1 = one[i],
        g2 = two[i];
      for (var j in g1) {
        if (!g2[j]) g.push(j);
      }
      res[i] = g;
    }
  }
  return res;
}

结果我们有像

{
    "Han SOLO": [],
    "Luke SKYWALKER": [],
    "Darth VADER": [
        "Group1"
    ],
    "Boba FETT": []
}

因此，通过简单的循环，我们可以获得有关不同的所有信息，例如

var msg = '';

for (var i in res) {
  if (res[i].length) {
    msg += 'user ' + i + ' added in group: ' + res[i].join() + '<br>';
  }
}

var list1 = "Han SOLO, Luke SKYWALKER, Darth VADER, Boba FETT".split(', '),
  list2 = "Luke SKYWALKER, Han SOLO, Darth VADER, Boba FETT".split(', ')
  // group lists
  ,
  group1 = "Group1, Group1, Group1, Group2".split(', '),
  group2 = "Group1, Group1, Group2, Group2".split(', ');

function group(users, groups) {
  var result = {};
  for (var i = 0; i < users.length; i++) {
    var user = users[i];
    if (!result[user]) result[user] = {};
    result[users[i]][groups[i]] = true;
  }
  return result;
}

function comp(one, two) {
  function getGroups(u) {
    var groups = [];
    for (var i in u) groups.push(i);

    return groups;
  }

  var res = {};
  for (var i in one) {
    if (!two[i]) {
      res[i] = getGroups(one[i]);
    } else {
      var g = [],
        g1 = one[i],
        g2 = two[i];
      for (var j in g1) {
        if (!g2[j]) g.push(j);
      }
      res[i] = g;
    }
  }
  return res;
}

var res = comp(group(list1, group1), group(list2, group2));

var msg = '';

for (var i in res) {
  if (res[i].length) {
    msg += 'user ' + i + ' added in group: ' + res[i].join() + '<br>';
  }
}

document.getElementById('res').innerHTML = msg;

<p id="res"></p>