Question

这是我的数据

[{name:'james',grade:'A'},
{name:'john',grade:'B'},
{name:'iris',,grade:'A'},
{name:'ivan',,grade:'C'}]

我想保留具有A级和C级的对象，它很容易我可以像

一样进行过滤

person.filter(obj => obj.grade === 'A' || obj.grade === 'C')

但现在我有一个对象数组。

[{grade:'A'},{grade:'C'}]

任何线索如何立即进行过滤？我需要嵌套循环吗？

Answer 1

使用Array.prototype.some：

＆＃13;

let person = [{name:'james', grade:'A'},
              {name:'john', grade:'B'},
              {name:'iris', grade:'A'},
              {name:'ivan', grade:'C'}];

let predicate = [{grade:'A'},{grade:'C'}];
        
let result = person.filter(obj => predicate.some(p => p.grade == obj.grade))

console.log('result:', result)

＆＃13;

如果您的谓词比这更具动态，请比较所有对象属性，而不仅仅是p.grade。

person.filter(obj => predicate.some(p => {
    return Object.keys(p).every(k => obj[k] == p[k]);
}));

Answer 2

使用下划线库例如 -

    var bbb = [
        {id: 839},
        {id: 854}
    ];
     var ids = {};
    _.each(bbb, function (bb) { ids[bb.id] = true; });

var data = [{grade:'A'},{grade:'C'}];
var value = {};
_.each(data , function (d) { value[data.garde] === 'A' | value[data.garde] === 'C' ; });

Answer 3

成绩为select AnyCommunicationProblem, count(*) cnt from PROTOS_PersonalInformation GROUP BY AnyCommunicationProblem和A的对象应该被过滤为（经典的js语法），

Answer 4

如果你有很多成绩要检查（我不知道所有的成绩;））。您可以先将数组转换为Set

const grades = new Set([{grade:'A'},{grade:'C'}].map(({grade}) => grade))

const persons = [{name:'james',grade:'A'},
  {name:'john',grade:'B'},
  {name:'iris',grade:'A'},
  {name:'ivan',grade:'C'}]

然后使用has

过滤人物数组

const filtered = persons.filter(({grade}) => grades.has(grade))

Answer 5

您可以使用Set作为谓词并相应地过滤

let person = [{ name: 'james', grade: 'A' }, { name: 'john', grade: 'B' }, { name: 'iris', grade: 'A' }, { name: 'ivan', grade: 'C' }],
    predicate = [{ grade: 'A' }, { grade: 'C' }],
    result = person.filter((s => p => s.has(p.grade))(new Set(predicate.map(p => p.grade))));

console.log(result)

.as-console-wrapper { max-height: 100% !important; top: 0; }

过滤对象数组的对象数组

5 个答案: