IT技术 - 包含对象的两个数组的差异和交集 - 吾爱随笔录

包含对象的两个数组的差异和交集

IT技术 javascript arrays set-intersection set-difference set-operations

2021-02-05 08:11:18

我有两个数组list1，list2其中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：

list1 operation list2 应该返回元素的交集：

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

list1 operation list2应该返回list1不出现在的所有元素的列表list2：

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

list2 operation list1应该返回list2不出现在的元素列表list1：
```
[
    { userId: 1252, userName: 'AAAA' }
]
```

6个回答

您可以定义三个函数inBoth, inFirstOnly, 并且inSecondOnly它们都以两个列表作为参数，并从函数名称中可以理解返回一个列表。主要逻辑可以放在operation三个都依赖的公共函数中。

以下是一些operation可供选择的实现，您可以在下面找到一个片段：

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

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));

使用`filter`和`some`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) —— 也some代表一个循环。因此，对于大型数组，您最好在用户 ID 上创建一个（临时）散列。这可以通过将(ES6) 作为参数提供给将生成过滤器回调函数的函数来即时完成Set。然后该函数可以在恒定时间内执行查找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));

简短的回答：

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，
如果您需要复杂的比较规则，您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来比较对象
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

惊人的！漂亮干净！

2021-03-19 08:11:18

只需使用filter和someJS的阵列的方法和你能做到这一点。

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

这将返回存在于中list1但不存在于中的项目list2。如果您正在寻找两个列表中的公共项目。就这样做。

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

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
}, []);

以上为您提供了等效的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！

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]); })
}

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普通的旧for循环

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优化查找

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