如何使用 Lodash _.filter 过滤多条件？

Question

我想从集合中过滤两个obj，conditionArr从Server获取，所以不可预知。

我的代码是：

var users = [{
        user: 'barney',
        age: 36,
        active: true
    },
    {
        user: 'fred',
        age: 40,
        active: false
    },
    {
        user: 'travis',
        age: 37,
        active: true
    }
];

// the Array fetch from Server, so it's unpredictable.
var conditionArr = [{
    user: 'barney'
}, {
    user: 'fred'
}];

// _.filter
result = _.filter(users, conditionArr);

我的预期输出：

 // expect results:
 [{
         user: 'barney',
         age: 36,
         active: true
     },
     {
         user: 'fred',
         age: 40,
         active: false
     }
 ]  

实际结果：

[]

谢谢。

---

我找到了更好的方法：

result =  _.map(conditionArr, (con) => ({
    ...con,
   ...(_.find(users, { user: con.user })),
}))

谢谢...

Answer 1

使用_.intersectionBy():

const users = [{"user":"barney","age":36,"active":true},{"user":"fred","age":40,"active":false},{"user":"travis","age":37,"active":true}];

const conditionArr = [{"user":"barney"},{"user":"fred"}];

const result = _.intersectionBy(users, conditionArr, 'user');

console.log(result);

<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.5/lodash.min.js"></script>

Vanilla JS - 我添加了一个简单的intersectionBy()，它将第二个数组的相关key 值转换为一个集合。然后您可以使用 Set 过滤第一个数组。

const intersectionBy = (arr1, arr2, key) => {
  const keys = new Set(arr2.map(o => o[key]));
  
  return arr1.filter(o => keys.has(o[key]));
};

const users = [{"user":"barney","age":36,"active":true},{"user":"fred","age":40,"active":false},{"user":"travis","age":37,"active":true}];

const conditionArr = [{"user":"barney"},{"user":"fred"}];

const result = intersectionBy(users, conditionArr, 'user');

console.log(result);

Answer 2

与许多其他 lodash 函数一样，_.filter 将函数作为第二个参数（数组中的每个元素作为参数），如果函数返回 true，则该元素将保留在结果中。

var users = [
     { user: 'barney', age: 36, active: true },
     { user: 'fred', age: 40, active: false },
     { user: 'travis', age: 37, active: true }
 ];

// the Array fetch from Server, so it's unpredictable.
var conditionArr = [
  { user: 'barney'},
  { user: 'fred'}
];
 
 // Simplified condition a bit. cond = ['barney', 'fred'];
var cond = _.map(conditionArr, cond => cond.user);
 
 // _.filter
var result = _.filter(users, user => _.indexOf(cond, user.user) !== -1);

console.log('result', result);

<script src="https://cdn.jsdelivr.net/npm/lodash@4.17.5/lodash.min.js"></script>

Answer 3

像使用原生 js -

let userSubset = users.map(function(obj, index) {

   if (obj.user === "barney" || obj.user === "fred") return obj;
}).filter(Boolean);

console.log(userSubset);

// {user: "barney", age: 36, active: true}
// {user: "fred", age: 40, active: false}

Answer 4

在_.filter 方法中你需要像下面的例子一样传递函数

_.filter(users, function(o) { return !o.active; });

var users = [{
    user: 'barney',
    age: 36,
    active: true
  },
  {
    user: 'fred',
    age: 40,
    active: false
  },
  {
    user: 'travis',
    age: 37,
    active: true
  }
];


// _.filter
var result = _.filter(users, function(obj) { return obj.user=='barney'||obj.user=='fred' });

console.log(result)

<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.5/lodash.core.js"></script>

Answer 5

如果您必须使用 lodash 来实现这一点，您可以使用与 array.filter 相同的方式进行操作

_.filter(users, ({user}) => user === 'barney' || user === 'fred');

Answer 6

只需使用array.filter：

var users = [
 { user: 'barney', age: 36, active: true },
 { user: 'fred',  age: 40, active: false },
 { user: 'travis', age: 37, active: true}
];

var result = users.filter(({user}) => user === 'barney' || user === "fred");
console.log(result);