javascript underscore.js 过滤一个对象数组，基于另一个
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underscore.js filter an array of objects, based on another
提问by bsr
I am trying to filter an array of objects, based on another. The common property id id.
I am not sure filter + each is the best way to do it or map reduce. Anyway, below code doesn't work as outis empty list.
我正在尝试根据另一个对象过滤一组对象。公共属性 id id。我不确定 filter + each 是最好的方法还是 map reduce。无论如何，下面的代码不能像out空列表那样工作。
var aaa = [
{name: "AAA", id: 845},
{name: "BBB", id: 839},
{name: "CCC", id: 854}
];
var bbb = [
{id: 839},
{id: 854}
];
var out = _.filter(aaa, function(val){
return _.each(this, function(val2){
return val['id'] === val2['id']
});
}, bbb);
回答by kay - SE is evil
Just create a "set" of the valid ids and use that "set" to do the filtering:
只需创建一个有效 ID 的“集合”并使用该“集合”进行过滤：
var aaa = [
{name: "AAA", id: 845},
{name: "BBB", id: 839},
{name: "CCC", id: 854}
];
var bbb = [
{id: 839},
{id: 854}
];
var ids = {};
_.each(bbb, function (bb) { ids[bb.id] = true; });
var out = _.filter(aaa, function (val) {
return ids[val.id];
}, bbb);
Filling idsis fast, it's in n * amortizedO(1), i.e O(n). Same holds for the filtering.
填充ids很快，它在 n *摊销O(1) 中，即 O(n)。同样适用于过滤。
If you use each(…)in the inner loop, you will have O(n2). For bigger data sets this would become very slow. Also the additional nesting make the code more difficult to read/understand at first glance.
如果您each(…)在内循环中使用，您将获得 O(n2)。对于更大的数据集，这将变得非常缓慢。此外，额外的嵌套使代码乍一看更难以阅读/理解。
See that code snipped in action: http://jsfiddle.net/SMtX5/
看到在行动中剪断的代码：http: //jsfiddle.net/SMtX5/
回答by anhulife
you can use _.findto filter:
你可以_.find用来过滤：
_.filter(aaa, function(a){
return _.find(bbb, function(b){
return b.id === a.id;
});
});
回答by Tony Broyez
bbb = bbb.map(_ => _.id) && aaa.filter(_ => bbb.indexOf( _.id ) > -1)
You just need pure JS array functions to do that assuming your use case.
假设您的用例，您只需要纯 JS 数组函数就可以做到这一点。
回答by pktangyue
You can use _.some(list, [iterator], [context]).
您可以使用_.some(list, [iterator], [context]).
It returns trueif any of the values in the listpass the iteratortruth test.
如果列表中的任何值通过迭代器真值测试，则返回true。
var out = _.filter(aaa, function(val){
return _.some(this,function(val2){
return val2['id'] === val['id'];
});
}, bbb);
Here is jsfiddle. http://jsfiddle.net/h98ej/
这是jsfiddle。http://jsfiddle.net/h98ej/
