我有以下对象数组。我需要做的就是从所有连接器阵列中删除匹配的key-val对。
[
{
"connector":[
{
"name":"CC1"
},
{
"name":"App1"
},
{
"name":"CC1"
},
{
"name":"App2"
},
{
"name":"CC1"
},
{
"name":"App3"
}
],
"connections":[
{
"source":"CC1",
"target":"App1"
},
{
"source":"CC1",
"target":"App2"
},
{
"source":"CC1",
"target":"App3"
}
]
},
{
"connector":[
{
"name":"CC1"
},
{
"name":"App1"
},
{
"name":"CC1"
},
{
"name":"App2"
},
{
"name":"CC1"
},
{
"name":"App3"
}
],
"connections":[
{
"source":"CC1",
"target":"App1"
},
{
"source":"CC1",
"target":"App2"
},
{
"source":"CC1",
"target":"App3"
}
]
},
{
"connector":[
{
"name":"CC1"
},
{
"name":"App1"
},
{
"name":"CC1"
},
{
"name":"App2"
},
{
"name":"CC1"
},
{
"name":"App3"
}
],
"connections":[
{
"source":"CC1",
"target":"App1"
},
{
"source":"CC1",
"target":"App2"
},
{
"source":"CC1",
"target":"App3"
}
]
},
{
"connector":[
{
"name":"CC2"
},
{
"name":"App2"
}
],
"connections":[
{
"source":"CC2",
"target":"App2"
}
]
}
]
我尝试在es6中使用filter，map和spreadable运算符的组合，但尚未找到实现此目的的最佳组合。 我想要的输出如下：
[
{
"connector":[
{
"name":"CC1"
},
{
"name":"App1"
},
{
"name":"App2"
},
{
"name":"App3"
}
],
"connections":[
{
"source":"CC1",
"target":"App1"
},
{
"source":"CC1",
"target":"App2"
},
{
"source":"CC1",
"target":"App3"
}
]
},
{
"connector":[
{
"name":"CC1"
},
{
"name":"App1"
},
{
"name":"App2"
},
{
"name":"App3"
}
],
"connections":[
{
"source":"CC1",
"target":"App1"
},
{
"source":"CC1",
"target":"App2"
},
{
"source":"CC1",
"target":"App3"
}
]
},
{
"connector":[
{
"name":"CC1"
},
{
"name":"App1"
},
{
"name":"App2"
},
{
"name":"App3"
}
],
"connections":[
{
"source":"CC1",
"target":"App1"
},
{
"source":"CC1",
"target":"App2"
},
{
"source":"CC1",
"target":"App3"
}
]
},
{
"connector":[
{
"name":"CC2"
},
{
"name":"App2"
}
],
"connections":[
{
"source":"CC2",
"target":"App2"
}
]
}
]
实现这一目标的最佳解决方案是什么？在此先感谢您的帮助..
答案 0 :(得分：1)
setTimeout(doit, 100);
function doit() {
data.forEach(obj => {
obj.connector = obj.connector.filter(({name}, i, arr) =>
arr.findIndex(o => o.name === name) === i
)
});
console.log(data);
}
var data = [{
"connector": [{
"name": "CC1"
},
{
"name": "App1"
},
{
"name": "CC1"
},
{
"name": "App2"
},
{
"name": "CC1"
},
{
"name": "App3"
}
],
"connections": [{
"source": "CC1",
"target": "App1"
},
{
"source": "CC1",
"target": "App2"
},
{
"source": "CC1",
"target": "App3"
}
]
},
{
"connector": [{
"name": "CC1"
},
{
"name": "App1"
},
{
"name": "CC1"
},
{
"name": "App2"
},
{
"name": "CC1"
},
{
"name": "App3"
}
],
"connections": [{
"source": "CC1",
"target": "App1"
},
{
"source": "CC1",
"target": "App2"
},
{
"source": "CC1",
"target": "App3"
}
]
},
{
"connector": [{
"name": "CC1"
},
{
"name": "App1"
},
{
"name": "CC1"
},
{
"name": "App2"
},
{
"name": "CC1"
},
{
"name": "App3"
}
],
"connections": [{
"source": "CC1",
"target": "App1"
},
{
"source": "CC1",
"target": "App2"
},
{
"source": "CC1",
"target": "App3"
}
]
},
{
"connector": [{
"name": "CC2"
},
{
"name": "App2"
}
],
"connections": [{
"source": "CC2",
"target": "App2"
}]
}
];
答案 1 :(得分：1)
如果您只需要普通的操作，例如map，filter，find，那么您不需要lodash：
const data = [{connector:[{name:"CC1"},{name:"App1"},{name:"CC1"},{name:"App2"},{name:"CC1"},{name:"App3"}],connections:[{source:"CC1",target:"App1"},{source:"CC1",target:"App2"},{source:"CC1",target:"App3"}]},{connector:[{name:"CC1"},{name:"App1"},{name:"CC1"},{name:"App2"},{name:"CC1"},{name:"App3"}],connections:[{source:"CC1",target:"App1"},{source:"CC1",target:"App2"},{source:"CC1",target:"App3"}]},{connector:[{name:"CC1"},{name:"App1"},{name:"CC1"},{name:"App2"},{name:"CC1"},{name:"App3"}],connections:[{source:"CC1",target:"App1"},{source:"CC1",target:"App2"},{source:"CC1",target:"App3"}]},{connector:[{name:"CC2"},{name:"App2"}],connections:[{source:"CC2",target:"App2"}]}];
const result = data.map(e => ({ ...e, connector:
e.connector
// Filter the original connector array
// and return only those elements which name is
// the same as a connection source or target
.filter(c => e.connections.find(cn => [cn.source, cn.target].indexOf(c.name) !== -1))
// Remove duplicates by name as @llama notes
.filter((v, i, ary) => ary.findIndex(c => c.name === v.name) === i)
}));
console.log(result)
答案 2 :(得分：-1)
这将过滤掉任何重复的对象，connectors和connections个对象可以包含任意数量的键。
mapper = (input) =>
input.map(x => JSON.stringify(x)) // Stringify
.filter((x, i, a) => a.indexOf(x) === i) // Filter out repeating elements
.map(x => JSON.parse(x)) // Parse stringified object
output = input.map(item =>
({
connector: mapper(item.connector),
connections: mapper(item.connections)
})
)