包含对象

时间：2015-10-26 22:08:14

标签： javascript arrays set-intersection set-difference set-operations

我有两个数组list1和list2，它们包含具有某些属性的对象; 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list2未出现的所有元素的列表：

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

  3. list2 operation list1应该返回list2中未出现的list1元素列表：

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

7 个答案:

答案 0 :(得分：19)

您可以定义三个函数inBoth，inFirstOnly和inSecondOnly，它们都将两个列表作为参数，并返回一个列表，可以从函数名称中理解。主逻辑可以放在一个共同的函数operation中，这三个函数都依赖于它。

以下是operation可供选择的一些实现，您可以在其中找到更远的代码段：

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

普通的旧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也表示循环。因此，对于大型数组，您最好在user-id上创建（临时）哈希。这可以通过为将生成过滤器回调函数的函数提供Set（ES6）参数来 on-the-fly 完成。然后，该函数可以使用has：

在恒定时间内执行查找 

// 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));

答案 1 :(得分：4)

使用lodash's _.isEqual方法。具体做法是：

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

Above为您提供相当于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！

答案 2 :(得分：1)

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]); })
}

答案 3 :(得分：1)

这是适合我的解决方案。 

 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;
        }

答案 4 :(得分：1)

简短答案： 

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，
如果您需要复杂的比较规则，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

答案 5 :(得分：0)

这是一个功能编程解决方案，带有下划线/ lodash来回答你的第一个问题（十字路口）。

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; - ）

答案 6 :(得分：0)

只需使用JS的filter和some数组方法，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中不存在的项目。如果您要查找两个列表中的常见项目。就是这样

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