根据相同的名称分隔数组 [英] Segregate an array based on same name

问题描述

我有一个数组，其中包含多个具有相同名称的对象.如何根据名称分隔数组?

I have an array with multiple objects having the same name. How can I segregate the array based on the name?

我尝试过使用地图，过滤器，但是最终没有用，我也使用了reduce，但是没有得到输出.

I tried with map, filter but didn't work finally I used reduce also but not getting the output.

 const result = Object.values(this.optionsArr.reduce((a, c) => {
        a[c.name] = a[c.name] || {name: c.name};
        a[c.name].options.push(c.options);
        return a;
      }, {})).map(item => ({...item, accessDetails: Object.values(item.accessDetails)}));

输入:

[{name: "audio", options:{name:'true', value: 'T'}},
{name: "audio", options:{name:'false', value: 'F'}},
{name: "audio", options:{name:'yes', value: 'Y'}},
{name: "video", options:{name:'true', value: 'T'}},
{name: "video", options:{name:'yes', value: 'Y'}},
{name: "video", options:{name:'false', value: 'F'}},
{name: "call", options:{name:'true', value: 'T'}},
{name: "call", options:{name:'No', value: 'N'}},
{name: "call", options:{name:'false', value: 'F'}}]

预期产量

 [{name: "audio", options:[{name:'true', value: 'T'},{name:'false', value: 'F'},{name:'yes', value: 'Y'}]},
    {name: "video", options:[{name:'true', value: 'T'},{name:'false', value: 'F'},{name:'yes', value: 'Y'}]},
    {name: "call", options:[{name:'true', value: 'T'},{name:'false', value: 'F'},{name:'yes', value: 'Y'}]}]

推荐答案

您可以使用对象执行归约操作来存储具有相同名称的元素.

You can perform a reduce operation using an object to store elements with the same name.

const arr = [{name: "audio", options:{name:'true', value: 'T'}},
{name: "audio", options:{name:'false', value: 'F'}},
{name: "audio", options:{name:'yes', value: 'Y'}},
{name: "video", options:{name:'true', value: 'T'}},
{name: "video", options:{name:'yes', value: 'Y'}},
{name: "video", options:{name:'false', value: 'F'}},
{name: "call", options:{name:'true', value: 'T'}},
{name: "call", options:{name:'No', value: 'N'}},
{name: "call", options:{name:'false', value: 'F'}}];
const res = Object.values(arr.reduce((acc,{name, options})=>{
  acc[name] = acc[name] || {name, options: []};
  acc[name].options.push(options);
  return acc;
}, {}));
console.log(res);

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