包含对象的两个数组的差异和交集 [英] Difference and intersection of two arrays containing objects

问题描述

我有两个数组 list1 和 list2，它们有一些属性的对象；userId 是 Id 或唯一属性:

list1 = [{用户ID:1234，用户名:'XYZ'}，{用户ID:1235，用户名:'ABC'}，{用户ID:1236，用户名:'IJKL'}，{用户ID:1237，用户名:'WXYZ'}，{用户ID:1238，用户名:'LMNO'}]列表 2 = [{用户ID:1235，用户名:'ABC'}，{用户ID:1236，用户名:'IJKL'}，{ 用户 ID:1252，用户名:'AAAA' }]

我正在寻找一种简单的方法来执行以下三个操作:

1. list1 操作 list2 应该返回元素的交集:

   <预><代码>[{用户ID:1235，用户名:'ABC'}，{用户ID:1236，用户名:'IJKL'}]

2. list1 operation list2 应该返回 list1 中没有出现在 list2 中的所有元素的列表:

   <预><代码>[{用户ID:1234，用户名:'XYZ'}，{用户ID:1237，用户名:'WXYZ'}，{用户ID:1238，用户名:'LMNO'}]

3. list2 operation list1 应该返回 list2 中没有出现在 list1 中的元素列表:

   <预><代码>[{ 用户 ID:1252，用户名:'AAAA' }]

解决方案

这是对我有用的解决方案.

 var intersect = function (arr1, arr2) {var intersect = [];_.each(arr1, 函数 (a) {_.each(arr2, 函数 (b) {如果(比较(a，b))intersect.push(a);});});返回相交；};var unintersect = function (arr1, arr2) {var unintersect = [];_.each(arr1, 函数 (a) {var 发现 = 假；_.each(arr2, 函数 (b) {如果(比较(a，b)){发现 = 真；}});如果(！找到){unintersect.push(a);}});返回不相交；};函数比较(a，b){如果(a.userId === b.userId)返回真；否则返回假；}

I have two arrays list1 and list2 which have objects with some properties; userId is the Id or unique property:

list1 = [
    { userId: 1234, userName: 'XYZ'  }, 
    { userId: 1235, userName: 'ABC'  }, 
    { userId: 1236, userName: 'IJKL' },
    { userId: 1237, userName: 'WXYZ' }, 
    { userId: 1238, userName: 'LMNO' }
]

list2 = [
    { userId: 1235, userName: 'ABC'  },  
    { userId: 1236, userName: 'IJKL' },
    { userId: 1252, userName: 'AAAA' }
]

I'm looking for an easy way to execute the following three operations:

1. list1 operation list2 should return the intersection of elements:

   [
       { userId: 1235, userName: 'ABC'  },
       { userId: 1236, userName: 'IJKL' }
   ]
   

2. list1 operation list2 should return the list of all elements from list1 which don't occur in list2:

   [
       { userId: 1234, userName: 'XYZ'  },
       { userId: 1237, userName: 'WXYZ' }, 
       { userId: 1238, userName: 'LMNO' }
   ]
   

3. list2 operation list1 should return the list of elements from list2 which don't occur in list1:

   [
       { userId: 1252, userName: 'AAAA' }
   ]
   

解决方案

This is the solution that worked for me.

 var intersect = function (arr1, arr2) {
            var intersect = [];
            _.each(arr1, function (a) {
                _.each(arr2, function (b) {
                    if (compare(a, b))
                        intersect.push(a);
                });
            });

            return intersect;
        };

 var unintersect = function (arr1, arr2) {
            var unintersect = [];
            _.each(arr1, function (a) {
                var found = false;
                _.each(arr2, function (b) {
                    if (compare(a, b)) {
                        found = true;    
                    }
                });

                if (!found) {
                    unintersect.push(a);
                }
            });

            return unintersect;
        };

        function compare(a, b) {
            if (a.userId === b.userId)
                return true;
            else return false;
        }

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