如何使用javascript或lodash从对象数组中删除不匹配的对象

Question

我从服务器获取两个对象数组,如下所示:

var duplicateTestData = [
    { 
        licenseId: 'xxx',
        batchId: '123',
        reportDate: Fri Dec 11 2015 00:00:00 GMT+0530 (India Standard Time) 
    },
    { 
        licenseId: 'yyy',
        batchId: '124',
        reportDate: Fri Dec 11 2015 00:00:00 GMT+0530 (India Standard Time) 
    },
    { 
        licenseId: 'aaa',
        batchId: '145',
        reportDate: Fri Dec 11 2015 00:00:00 GMT+0530 (India Standard Time) 
    }
];

var finalResult = [
    { 
        reportDate: Fri Dec 11 2015 00:00:00 GMT+0530 (India Standard Time),
        license: {},
        testType: 'P1',
        productType: 'Flower',
        batchId: '123',
        licenseId: 'xxx',
        createType: 'DataUpload' 
    },
    { 
        reportDate: Fri Dec 11 2015 00:00:00 GMT+0530 (India Standard Time),
        testType: 'P1',
        productType: 'Flower',
        batchId: '124',
        licenseId: 'yyy',
        createType: 'DataUpload' 
    },
    { 
        reportDate: Fri Dec 11 2015 00:00:00 GMT+0530 (India Standard Time),
        testType: 'P1',
        productType: 'Flower',
        batchId: '145',
        licenseId: 'aaa',
        createType: 'DataUpload' 
    },
    { 
        reportDate: Fri Dec 14 2015 00:00:00 GMT+0530 (India Standard Time),
        testType: 'P1',
        productType: 'Flower',
        batchId: '145',
        licenseId: 'zzz',
        createType: 'DataUpload' 
    }
]

我试图从finalResult对象中只获取不匹配的对象,最终结果将是这样的:

[
    { 
        reportDate: Fri Dec 14 2015 00:00:00 GMT+0530 (India Standard Time),
        testType: 'P1',
        productType: 'Flower',
        batchId: '145',
        licenseId: 'zzz',
        createType: 'DataUpload' 
    } 
]  

我正在尝试这个,但没有得到正确的结果:

for(var j=0;j < duplicateTestData.length;j++){
    for (var i = 0; i < finalResult.length; i++) {
        if (
            (finalResult[i].licenseId == duplicateTestData[j].licenseId)  && 
            (finalResult[i].reportDate == duplicateTestData[j].reportDate) &&
            (finalResult[i].batchId == duplicateTestData[j].batchId)
        ) {
            finalResult.splice(i, 1);
            break;
        }
    }
}

console.log(finalResult);

Answer 1

简单的出路

finalResult.filter(({batchId:a, licenseId:b, reportDate:c}) =>
  duplicateTestData.find(({batchId:x, licenseId:y, reportDate:z}) =>
    a === x && b === y && c === z) === undefined)

=> [ { reportDate: 'Fri Dec 14 2015 00:00:00 GMT+0530 (India Standard Time)',
    testType: 'P1',
    productType: 'Flower',
    batchId: '145',
    licenseId: 'zzz',
    createType: 'DataUpload' } ]

好的,它有效,但这主要是垃圾.它没有完全准确地描述您正在尝试进行的比较.它太具体了,一旦你的数据发生变化就会中断.

继续阅读,我们可以学到一些有趣的东西

所有对象相等(键和值)...

我将首先制作几个通用程序,以便更好地描述我们问题的解决方案.

与其他解决方案相比,您对此解决方案的注意事项是,它不会对数据内部进行任何假设.此解决方案可能不关心对象中使用的实际键名称.

这意味着我们不会被任何接触batchId,licenseId或reportDate.在这种情况下,通用程序可以解决所有问题,最好的部分是你可以反复使用它们来处理你想要处理的任何数据.

// arrayCompare :: (a -> b -> Bool) -> [a] -> [b] -> Bool
const arrayCompare = f=> ([x,...xs])=> ([y,...ys])=> {
  if (x === undefined && y === undefined)
    return true
  else if (! f (x) (y))
    return false
  else
    return arrayCompare (f) (xs) (ys)
}

// keys :: Object(k:v) -> [k]
const keys = Object.keys

// objectCompare :: (v -> v -> Bool) -> Object(k:v) -> Object(k:v) -> Bool
const objectCompare = f=> a=> b=>
  arrayCompare (x=> y=> f (a[x]) (b[y]) && f (a[y]) (b[y])) (keys(a)) (keys(b))

// objectEqual :: Object -> Object -> Bool
const objectEqual = objectCompare (x=> y=> x === y)

// sample data
let xs = [
  {a:1,b:10},
  {a:2,b:20},
  {a:3,b:30}
]

let ys = [
  {a:1,b:10},
  {a:2,b:20},
  {a:3,b:30},
  {a:4,b:40}
]

// return all ys that are not present in xs
var result = ys.filter(y=> xs.find(objectEqual(y)) === undefined)

console.log(result)
// [{a:4,b:40}]

陷阱!

您必须稍微调整此解决方案,因为您没有比较所有对象键.对象中的对象finalResult比对象多,duplicateTestData因此没有1:1的匹配.

简单来说,如果要与之进行比较,您希望x = {a:1}被视为"匹配" y = {a:1,b:2},只要所有key:值x匹配所有key:值iny

如果我们使用objectEquals上面的比较器,则不会过滤任何内容,finalResult因为没有对象匹配找到的对象duplicateTestData.由于这不是你想要的,让我们定义一个适用于你的情况的比较器......

// subsetObjectEquals :: Object -> Object -> Bool
const subsetObjectEquals = objectCompare (x=> y=> y === undefined || x === y)

// this time use subsetObjectEquals
var result = finalResult.filter(x=>
  duplicateTestData.find(subsetObjectEquals(x)) === undefined)

subsetObjectEquals工作方式有点不同.我真的没想到一个更好的名字,因为这是一个有点奇怪的比较.当y是undefined时,表示该值的密钥不存在在"子集对象",因此并不需要比较

subsetObjectEquals(a,b)
// returns true if all key:value pairs in `a` match all key:value pairs in `b`
// otherwise returns false

完整的工作范例

我附上了一个完整的代码片段,它实际上使用了问题中包含的输入数据.在此处展开​​并运行它以查看它是否有效

// arrayCompare :: (a -> b -> Bool) -> [a] -> [b] -> Bool
const arrayCompare = f=> ([x,...xs])=> ([y,...ys])=> {
  if (x === undefined && y === undefined)
    return true
  else if (! f (x) (y))
    return false
  else
    return arrayCompare (f) (xs) (ys)
}

// keys :: Object(k:v) -> [k]
const keys = Object.keys

// objectCompare :: (v -> v -> Bool) -> Object(k:v) -> Object(k:v) -> Bool
const objectCompare = f=> a=> b=>
  arrayCompare (x=> y=> f (a[x]) (b[x]) && f (a[y]) (b[y])) (keys(a)) (keys(b))

// objectEqual :: Object -> Object -> Bool
const objectEqual = objectCompare (x=> y=> x === y)

// subsetObjectEquals :: Object -> Object -> Bool
const subsetObjectEquals = objectCompare (x=> y=> y === undefined || x === y)

// your data
var duplicateTestData = [{ licenseId: 'xxx',
    batchId: '123',
    reportDate: 'Fri Dec 11 2015 00:00:00 GMT+0530 (India Standard Time)' },
  { licenseId: 'yyy',
    batchId: '124',
    reportDate: 'Fri Dec 11 2015 00:00:00 GMT+0530 (India Standard Time)' },
  { licenseId: 'aaa',
    batchId: '145',
    reportDate: 'Fri Dec 11 2015 00:00:00 GMT+0530 (India Standard Time)' }
  ];

var finalResult = [ { reportDate: 'Fri Dec 11 2015 00:00:00 GMT+0530 (India Standard Time)',
    license: {},
    testType: 'P1',
    productType: 'Flower',
    batchId: '123',
    licenseId: 'xxx',
    createType: 'DataUpload' },
    { reportDate: 'Fri Dec 11 2015 00:00:00 GMT+0530 (India Standard Time)',
    testType: 'P1',
    productType: 'Flower',
    batchId: '124',
    licenseId: 'yyy',
    createType: 'DataUpload' },
    { reportDate: 'Fri Dec 11 2015 00:00:00 GMT+0530 (India Standard Time)',
    testType: 'P1',
    productType: 'Flower',
    batchId: '145',
    licenseId: 'aaa',
    createType: 'DataUpload' },
    { reportDate: 'Fri Dec 14 2015 00:00:00 GMT+0530 (India Standard Time)',
    testType: 'P1',
    productType: 'Flower',
    batchId: '145',
    licenseId: 'zzz',
    createType: 'DataUpload' }               
]

// get all finalResult items that do not subsetObjectEqual items in duplicateTestData
var result = finalResult.filter(x=>
  duplicateTestData.find(subsetObjectEquals(x)) === undefined)

console.log(result)

Answer 2

var res = _.filter(finalResult, function(item) {
  return !_.find(duplicateTestData, {
    batchId: item.batchId,
    licenseId: item.licenseId,
    reportDate: item.reportDate
  });
});
console.log(res);

的jsfiddle

Answer 3

您可以使用哈希表并返回一个新的结果集,而不会在迭代时拼接数组.

function getKey(o) {
    return o.licenseId + '|' + o.reportDate + '|' + o.batchId;
}

var duplicateTestData = [{ licenseId: 'xxx', batchId: '123', reportDate: 'Fri Dec 11 2015 00:00:00 GMT+0530 (India Standard Time)' }, { licenseId: 'yyy', batchId: '124', reportDate: 'Fri Dec 11 2015 00:00:00 GMT+0530 (India Standard Time)' }, { licenseId: 'aaa', batchId: '145', reportDate: 'Fri Dec 11 2015 00:00:00 GMT+0530 (India Standard Time)' }],
    finalResult = [{ reportDate: 'Fri Dec 11 2015 00:00:00 GMT+0530 (India Standard Time)', license: {}, testType: 'P1', productType: 'Flower', batchId: '123', licenseId: 'xxx', createType: 'DataUpload' }, { reportDate: 'Fri Dec 11 2015 00:00:00 GMT+0530 (India Standard Time)', testType: 'P1', productType: 'Flower', batchId: '124', licenseId: 'yyy', createType: 'DataUpload' }, { reportDate: 'Fri Dec 11 2015 00:00:00 GMT+0530 (India Standard Time)', testType: 'P1', productType: 'Flower', batchId: '145', licenseId: 'aaa', createType: 'DataUpload' }, { reportDate: 'Fri Dec 14 2015 00:00:00 GMT+0530 (India Standard Time)', testType: 'P1', productType: 'Flower', batchId: '145', licenseId: 'zzz', createType: 'DataUpload' }],
    hash = Object.create(null),
    result = [];

duplicateTestData.forEach(function (a) {            
    hash[getKey(a)] = true;
});

result = finalResult.filter(function (a) {
    return !hash[getKey(a)];
});

console.log(result);

ES6

function getKey(o) {
    return o.licenseId + '|' + o.reportDate + '|' + o.batchId;
}

var duplicateTestData = [{ licenseId: 'xxx', batchId: '123', reportDate: 'Fri Dec 11 2015 00:00:00 GMT+0530 (India Standard Time)' }, { licenseId: 'yyy', batchId: '124', reportDate: 'Fri Dec 11 2015 00:00:00 GMT+0530 (India Standard Time)' }, { licenseId: 'aaa', batchId: '145', reportDate: 'Fri Dec 11 2015 00:00:00 GMT+0530 (India Standard Time)' }],
    finalResult = [{ reportDate: 'Fri Dec 11 2015 00:00:00 GMT+0530 (India Standard Time)', license: {}, testType: 'P1', productType: 'Flower', batchId: '123', licenseId: 'xxx', createType: 'DataUpload' }, { reportDate: 'Fri Dec 11 2015 00:00:00 GMT+0530 (India Standard Time)', testType: 'P1', productType: 'Flower', batchId: '124', licenseId: 'yyy', createType: 'DataUpload' }, { reportDate: 'Fri Dec 11 2015 00:00:00 GMT+0530 (India Standard Time)', testType: 'P1', productType: 'Flower', batchId: '145', licenseId: 'aaa', createType: 'DataUpload' }, { reportDate: 'Fri Dec 14 2015 00:00:00 GMT+0530 (India Standard Time)', testType: 'P1', productType: 'Flower', batchId: '145', licenseId: 'zzz', createType: 'DataUpload' }],
    map = duplicateTestData.reduce((r, a) => r.set(getKey(a)), new Map),
    result = finalResult.filter(a => !map.has(getKey(a)));

console.log(result);