Difference and intersection of two arrays containing objects

后端 未结 7 894
予麋鹿
予麋鹿 2020-11-28 06:10

I have two arrays list1 and list2 which have objects with some properties; userId is the Id or unique property:

list1          


        
7条回答
  •  予麋鹿
    予麋鹿 (楼主)
    2020-11-28 06:45

    You could define three functions inBoth, inFirstOnly, and inSecondOnly which all take two lists as arguments, and return a list as can be understood from the function name. The main logic could be put in a common function operation that all three rely on.

    Here are a few implementations for that operation to choose from, for which you can find a snippet further down:

    • Plain old JavaScript for loops
    • Arrow functions using filter and some array methods
    • Optimised lookup with a Set

    Plain old for loops

    // 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)); 

    Arrow functions using filter and some array methods

    This uses some ES5 and ES6 features:

    // 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));

    Optimising lookup

    The above solutions have a O(n²) time complexity because of the nested loop -- some represents a loop as well. So for large arrays you'd better create a (temporary) hash on user-id. This can be done on-the-fly by providing a Set (ES6) as argument to a function that will generate the filter callback function. That function can then perform the look-up in constant time with 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));

提交回复
热议问题