两个包含对象的数组的差和交集

Question

我有两个数组list1和list2有一些属性的对象;userId为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中没有出现在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应该返回list2中没有出现在list1中的元素列表:

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

Answer 1

可以定义三个函数inBoth、inFirstOnly和inSecondOnly，它们都接受两个列表作为参数，并返回一个可以从函数名中理解的列表。主逻辑可以放在三者都依赖的通用函数operation中。

这里有几个operation的实现可供选择，您可以在下面找到一个片段:

• 普通的旧JavaScript for循环
• 使用filter和some数组方法的箭头函数
• Set
优化查找

普通老式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)); 

使用filter和some数组方法的箭头函数

这使用了一些ES5和ES6的特性:

// 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，上述解具有O(n²)时间复杂度——some也表示一个循环。因此，对于大型数组，您最好在user-id上创建一个(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实时完成。然后，该函数可以使用has: 在常量时间内执行查找。

// 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

短答:

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));  

长答:
上面的代码将根据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)));

还有一种通过引用比较对象的方法
警告!两个具有相同值的对象将被认为不同:

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 3

只要使用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中存在但list2中不存在的项。如果您正在寻找两个列表中的共同项目。就这么做。

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

Answer 4

使用lodash的_.isEqual方法。具体来说:

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

上面的

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

Answer 5

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 6

这是一个带有下划线/lodash的函数式编程解决方案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
    }, []);

我会让你改进这个来回答其他问题;-)

Answer 7

这就是我的解决方案。

 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;
        }