javascript - 两个包含对象的数组的差分和交集

Question

我有两个数组 list1 和 list2，它们的对象具有某些属性； 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操作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 操作 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

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，并尽可能返回一个列表从函数名来理解。主要逻辑可以放在一个共同的函数 operation 中，这三个函数都依赖。

这里有一些可供选择的操作的实现，您可以在下面找到一个片段:

• 普通的旧 JavaScript for 循环
• 使用filter 和一些 数组方法的箭头函数
• 使用 Set 优化查找

普通的 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 和一些 数组方法的箭头函数

这使用了一些 ES5 和 ES6 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，上述解决方案的时间复杂度为 O(n²) -- some 也表示循环。因此，对于大型数组，您最好在用户 ID 上创建一个(临时)散列。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));