比较两个对象数组,并将匹配值的元素排除在JS中的新数组中

Question

这是我在JavaScript中的用例:

我有两个对象数组,它们具有匹配的属性(id和name).

var result1 = [
    {id:1, name:'Sandra', type:'user', username:'sandra'},
    {id:2, name:'John', type:'admin', username:'johnny2'},
    {id:3, name:'Peter', type:'user', username:'pete'},
    {id:4, name:'Bobby', type:'user', username:'be_bob'}
];

var result2 = [
    {id:2, name:'John', email:'johnny@example.com'},
    {id:4, name:'Bobby', email:'bobby@example.com'}
];

var props = ['id', 'name'];

我的目标是让另一个对象数组只包含不匹配的元素.像这样:

var result = [
    {id:1, name:'Sandra'},
    {id:3, name:'Peter'}
];

我知道有一种方法可以通过从result1比较每个对象和result2的对象,然后比较它们的键,如果没有匹配,将值放在另一个对象然后将它推入新数组,但我奇怪是否有更优雅的方式,如使用低破折号或下划线或类似的东西.

谢谢!

Answer 1

只需使用JS内置的Array迭代方法就可以了:

var result1 = [
    {id:1, name:'Sandra', type:'user', username:'sandra'},
    {id:2, name:'John', type:'admin', username:'johnny2'},
    {id:3, name:'Peter', type:'user', username:'pete'},
    {id:4, name:'Bobby', type:'user', username:'be_bob'}
];

var result2 = [
    {id:2, name:'John', email:'johnny@example.com'},
    {id:4, name:'Bobby', email:'bobby@example.com'}
];

var props = ['id', 'name'];

var result = result1.filter(function(o1){
    // filter out (!) items in result2
    return !result2.some(function(o2){
        return o1.id === o2.id;          // assumes unique id
    });
}).map(function(o){
    // use reduce to make objects with only the required properties
    // and map to apply this to the filtered array as a whole
    return props.reduce(function(newo, name){
        newo[name] = o[name];
        return newo;
    }, {});
});

document.body.innerHTML = '<pre>' + JSON.stringify(result, null, 4) +
        '</pre>';

如果您正在做这么多,那么请务必查看外部库来帮助您,但是首先要学习基础知识,并且基础知识将为您提供良好的服务.

Answer 2

好吧,这使用lodash或香草javascript它取决于情况.

但是对于只返回包含重复项的数组,可以通过以下方式实现,当然它是从@ 1983获得的

var result = result1.filter(function (o1) {
    return result2.some(function (o2) {
        return o1.id === o2.id; // return the ones with equal id
   });
});
// if you want to be more clever...
let result = result1.filter(o1 => result2.some(o2 => o1.id === o2.id));

Answer 3

使用Lodash可以获得相同的结果.

var result1 = [
    {id:1, name:'Sandra', type:'user', username:'sandra'},
    {id:2, name:'John', type:'admin', username:'johnny2'},
    {id:3, name:'Peter', type:'user', username:'pete'},
    {id:4, name:'Bobby', type:'user', username:'be_bob'}
];

var result2 = [
    {id:2, name:'John', email:'johnny@example.com'},
    {id:4, name:'Bobby', email:'bobby@example.com'}
];

var result3 = _(result1) 
        .differenceBy(result2, 'id', 'name')
        .map(_.partial(_.pick, _, 'id', 'name'))
        .value();

console.log(result3);

<script src="https://cdn.jsdelivr.net/lodash/4.16.4/lodash.min.js"></script>

您可以使用属性"id"和"name"作为在它们之间"链接"元素的方法,使用两个数组之间的差异来获得所需的结果.如果这些属性中的任何一个不同,则元素被认为是不同的(在您的情况下不太可能,因为id似乎是唯一的).

最后,您必须映射结果以"省略"对象的不需要的属性.

希望能帮助到你.

Answer 4

我已经搜索了很多解决方案,我可以在其中比较具有不同属性名称的两个对象数组(类似于左外连接).我想出了这个解决方案.我在这里使用了Lodash.我希望这能帮到您.

var Obj1 = [
    {id:1, name:'Sandra'},
    {id:2, name:'John'},   
];

var Obj2 = [
    {_id:2, name:'John'},
    {_id:4, name:'Bobby'}
];

var Obj3 = lodash.differenceWith(Obj1, Obj2, function (o1, o2) {
    return o1['id'] === o2['_id']
});

console.log(Obj3);
//  {id:1, name:'Sandra'}

Answer 5

这是使用 Lodash 的另一个解决方案：

var _ = require('lodash');

var result1 = [
    {id:1, name:'Sandra', type:'user', username:'sandra'},
    {id:2, name:'John', type:'admin', username:'johnny2'},
    {id:3, name:'Peter', type:'user', username:'pete'},
    {id:4, name:'Bobby', type:'user', username:'be_bob'}
];

var result2 = [
    {id:2, name:'John', email:'johnny@example.com'},
    {id:4, name:'Bobby', email:'bobby@example.com'}
];

// filter all those that do not match
var result = types1.filter(function(o1){
    // if match found return false
    return _.findIndex(types2, {'id': o1.id, 'name': o1.name}) !== -1 ? false : true;
});

console.log(result);