从第二个数组中存在的第一个数组中删除值

Question

I have this variable:

var first = [{name: "john", key: "1"}, 
             {name: "george", key: "3"}, 
             {name: "paul", key: "2"},
             {name: "ringo", key: "4"}];

var second = [{key: "2"}, 
              {key: "4"}, 
              {key: "3"}]

I am trying to create a new object that omits value that is missing at second , in this case, name: "john" :

var third = [{name: "george", key: "3"}, 
             {name: "paul", key: "2"},
             {name: "ringo", key: "4"}];

Here's the code I tried:

var third = first.map(obj => {
    var retVal = {};
    retVal["name"] = obj.name;
    retVal["key"] = obj.key;
    return retVal;
});

This code adds all and becomes:

var third = [{name: "john", key: "1"}, 
             {name: "george", key: "3"}, 
             {name: "paul", key: "2"},
             {name: "ringo", key: "4"}];

, but can't figure out the restriction that wouldn't add the one I want to omit.

Please note, that regular loop is not a deal, it has to be with .map .

Answer 1

You can use the functions filter , map and includes as follow:

 var first = [{name: "john", key: "1"}, {name: "george", key: "3"}, {name: "paul", key: "2"},{name: "ringo", key: "4"}], second = [{key: "2"}, {key: "4"}, {key: "3"}], keys = second.map(o => o.key), result = first.filter(o => keys.includes(o.key)); console.log(result); 

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