包含对象的两个数组的差异和交集

Question

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

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' }
]

我正在寻找一种简单的方法来执行以下三个操作：

1. list1 operation list2应该返回元素的交集：

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

2. list1 operation list2应该返回list1中不存在于list2中的所有元素的列表：

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

3. list2 operation list1应返回list2中未在list1出现的元素列表：

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

Answer 1

您可以定义三种功能inBoth ， inFirstOnly和inSecondOnly所有需要两个列表作为参数，并返回一个列表，可以从函数名称可以理解。 主要逻辑可以放在三者都依赖的共同功能operation中。

以下是可供选择的operation的一些实现，您可以在其中找到更远的代码段：

• 简单的旧JavaScript for循环
• 箭头函数使用filter和some数组方法
• 使用Set优化查找

简单旧for循环

 // Generic helper function that can be used for the three operations: function operation(list1, list2, isUnion) { var result = []; for (var i = 0; i < list1.length; i++) { var item1 = list1[i], found = false; for (var j = 0; j < list2.length && !found; j++) { found = item1.userId === list2[j].userId; } if (found === !!isUnion) { // isUnion is coerced to boolean result.push(item1); } } return result; } // Following functions are to be used: function inBoth(list1, list2) { return operation(list1, list2, true); } function inFirstOnly(list1, list2) { return operation(list1, list2); } function inSecondOnly(list1, list2) { return inFirstOnly(list2, list1); } // Sample data var list1 = [ { userId: 1234, userName: 'XYZ' }, { userId: 1235, userName: 'ABC' }, { userId: 1236, userName: 'IJKL' }, { userId: 1237, userName: 'WXYZ' }, { userId: 1238, userName: 'LMNO' } ]; var list2 = [ { userId: 1235, userName: 'ABC' }, { userId: 1236, userName: 'IJKL' }, { userId: 1252, userName: 'AAAA' } ]; console.log('inBoth:', inBoth(list1, list2)); console.log('inFirstOnly:', inFirstOnly(list1, list2)); console.log('inSecondOnly:', inSecondOnly(list1, list2)); 

箭头函数使用filter和some数组方法

这使用了一些ES5和ES6功能：

 // Generic helper function that can be used for the three operations: const operation = (list1, list2, isUnion = false) => list1.filter( a => isUnion === list2.some( b => a.userId === b.userId ) ); // Following functions are to be used: const inBoth = (list1, list2) => operation(list1, list2, true), inFirstOnly = operation, inSecondOnly = (list1, list2) => inFirstOnly(list2, list1); // Sample data const list1 = [ { userId: 1234, userName: 'XYZ' }, { userId: 1235, userName: 'ABC' }, { userId: 1236, userName: 'IJKL' }, { userId: 1237, userName: 'WXYZ' }, { userId: 1238, userName: 'LMNO' } ]; const list2 = [ { userId: 1235, userName: 'ABC' }, { userId: 1236, userName: 'IJKL' }, { userId: 1252, userName: 'AAAA' } ]; console.log('inBoth:', inBoth(list1, list2)); console.log('inFirstOnly:', inFirstOnly(list1, list2)); console.log('inSecondOnly:', inSecondOnly(list1, list2)); 

优化查找

由于嵌套循环，上述解决方案具有O（n²）时间复杂度 - some代表一个循环。 因此，对于大型数组，您最好在user-id上创建（临时）哈希。 这可以通过提供Set （ES6）作为将生成过滤器回调函数的函数的参数来即时完成。 然后，该函数就可以在固定时间内查找与has ：

 // Generic helper function that can be used for the three operations: const operation = (list1, list2, isUnion = false) => list1.filter( (set => a => isUnion === set.has(a.userId))(new Set(list2.map(b => b.userId))) ); // Following functions are to be used: const inBoth = (list1, list2) => operation(list1, list2, true), inFirstOnly = operation, inSecondOnly = (list1, list2) => inFirstOnly(list2, list1); // Sample data const list1 = [ { userId: 1234, userName: 'XYZ' }, { userId: 1235, userName: 'ABC' }, { userId: 1236, userName: 'IJKL' }, { userId: 1237, userName: 'WXYZ' }, { userId: 1238, userName: 'LMNO' } ]; const list2 = [ { userId: 1235, userName: 'ABC' }, { userId: 1236, userName: 'IJKL' }, { userId: 1252, userName: 'AAAA' } ]; console.log('inBoth:', inBoth(list1, list2)); console.log('inFirstOnly:', inFirstOnly(list1, list2)); console.log('inSecondOnly:', inSecondOnly(list1, list2)); 

Answer 2

使用lodash的 _.isEqual方法。 特别：

list1.reduce(function(prev, curr){
  !list2.some(function(obj){
    return _.isEqual(obj, curr)
  }) ? prev.push(curr): false;
  return prev
}, []);

上面给出了相当于A given !B （在SQL术语中， A LEFT OUTER JOIN B ）。 您可以在代码周围移动代码以获得您想要的内容！

Answer 3

简答：

list1.filter(a => list2.some(b => a.userId === b.userId));  
list1.filter(a => !list2.some(b => a.userId === b.userId));  
list2.filter(a => !list1.some(b => a.userId === b.userId));  

更长的答案：
上面的代码将按userId值检查对象，
如果需要复杂的比较规则，可以定义自定义比较器：

comparator = function (a, b) {
    return a.userId === b.userId && a.userName === b.userName
};  
list1.filter(a => list2.some(b => comparator(a, b)));
list1.filter(a => !list2.some(b => comparator(a, b)));
list2.filter(a => !list1.some(b => comparator(a, b)));

还有一种方法可以通过引用比较对象
警告！ 具有相同值的两个对象将被视为不同：

o1 = {"userId":1};
o2 = {"userId":2};
o1_copy = {"userId":1};
o1_ref = o1;
[o1].filter(a => [o2].includes(a)).length; // 0
[o1].filter(a => [o1_copy].includes(a)).length; // 0
[o1].filter(a => [o1_ref].includes(a)).length; // 1

Answer 4

function intersect(first, second) {
    return intersectInternal(first, second, function(e){ return e });
}

function unintersect(first, second){
    return intersectInternal(first, second, function(e){ return !e });  
}

function intersectInternal(first, second, filter) {
    var map = {};

    first.forEach(function(user) { map[user.userId] = user; });

    return second.filter(function(user){ return filter(map[user.userId]); })
}

Answer 5

这是对我有用的解决方案。

 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;
        }

Answer 6

这是一个功能强大的编程解决方案，带有下划线/ lodash来回答你的第一个问题（交集）。

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'}
        ];

_.reduce(list1, function (memo, item) {
        var same = _.findWhere(list2, item);
        if (same && _.keys(same).length === _.keys(item).length) {
            memo.push(item);
        }
        return memo
    }, []);

我会让你改进这个以回答其他问题;-)

Answer 7

只需使用JS的filter和some数组方法，你就可以做到这一点。

let arr1 = list1.filter(e => {
   return !list2.some(item => item.userId === e.userId);
});

这将返回list1但不在list2 。 如果您要查找两个列表中的常用项目。 就这样做吧。

let arr1 = list1.filter(e => {
   return list2.some(item => item.userId === e.userId); // take the ! out and you're done
});