Question

我有一个像下面这样的对象数组

[
{
    "medication": {
        "name": "Turbohaler",
        "time": [
            "2018-09-07"
        ]
    }
},
{
    "medication": {
        "name": "Turbohaler",
        "time": [
            "2018-09-08"
        ]
    }
},
{
    "medication": {
        "name": "Turbohaler",
        "time": [
            "2018-09-09"
        ]
    }
},
{
    "medication": {
        "name": "Turbohaler",
        "time": [
            "2018-09-10"
        ]
    }
},
{
    "medication": {
        "name": "Turbohaler",
        "time": [
            "2018-09-11"
        ]
    }
},
{
    "medication": {
        "name": "Turbohaler",
        "time": [
            "2018-09-07"
        ]
    }
},
{
    "medication": {
        "name": "Turbohaler",
        "time": [
            "2018-09-08"
        ]
    }
},
{
    "medication": {
        "name": "Turbohaler",
        "time": [
            "2018-09-09"
        ]
    }
},
{
    "medication": {
        "name": "Turbohaler",
        "time": [
            "2018-09-10"
        ]
    }
},
{
    "medication": {
        "name": "Turbohaler",
        "time": [
            "2018-09-11"
        ]
    }
},
{
    "medication": {
        "name": "Septron",
        "time": [
            "2018-09-07"
        ]
    }
},
{
    "medication": {
        "name": "Septron",
        "time": [
            "2018-09-08"
        ]
    }
},
{
    "medication": {
        "name": "Septron",
        "time": [
            "2018-09-09"
        ]
    }
},
{
    "medication": {
        "name": "Septron",
        "time": [
            "2018-09-10"
        ]
    }
},
{
    "medication": {
        "name": "Septron",
        "time": [
            "2018-09-11"
        ]
    }
}
]

lodash或下划线提供了使代码最小化的强大功能，但是这些库仍然无法帮助我解决上述问题

我想实现以下目标

[
  {
    "medication": {
    "name": "Turbohaler",
    "time": [
    "2018-09-07",
    "2018-09-08",
    "2018-09-09",
    "2018-09-10",
    "2018-09-11"
  ]
 }
},
{
 "medication": {
 "name": "Septron",
 "time": [
    "2018-09-07",
    "2018-09-08",
    "2018-09-09",
    "2018-09-10",
    "2018-09-11"
  ]
}

} ]

在我们的案例中，通过将具有相同键名（Medication）的对象数组分组，然后将其所有值都提取到一起并将其合并到数组中的单个对象中，形成了以上结果。我尝试了不同的方法，但失败了，如果有人可以尝试获得超出预期的输出，将不胜感激

预先感谢

Answer 1

这是您面临的问题的O(n)解决方案。另外，对数组使用 spread语法，这样，如果time数组有多个项目，则可以正确处理。

var arr = [{
    "medication": {
      "name": "Turbohaler",
      "time": [
        "2018-09-07"
      ]
    }
  },
  {
    "medication": {
      "name": "Turbohaler",
      "time": [
        "2018-09-08"
      ]
    }
  },
  {
    "medication": {
      "name": "Turbohaler",
      "time": [
        "2018-09-09"
      ]
    }
  },
  {
    "medication": {
      "name": "Turbohaler",
      "time": [
        "2018-09-10"
      ]
    }
  },
  {
    "medication": {
      "name": "Turbohaler",
      "time": [
        "2018-09-11"
      ]
    }
  },
  {
    "medication": {
      "name": "Turbohaler",
      "time": [
        "2018-09-07"
      ]
    }
  },
  {
    "medication": {
      "name": "Turbohaler",
      "time": [
        "2018-09-08"
      ]
    }
  },
  {
    "medication": {
      "name": "Turbohaler",
      "time": [
        "2018-09-09"
      ]
    }
  },
  {
    "medication": {
      "name": "Turbohaler",
      "time": [
        "2018-09-10"
      ]
    }
  },
  {
    "medication": {
      "name": "Turbohaler",
      "time": [
        "2018-09-11"
      ]
    }
  },
  {
    "medication": {
      "name": "Septron",
      "time": [
        "2018-09-07"
      ]
    }
  },
  {
    "medication": {
      "name": "Septron",
      "time": [
        "2018-09-08"
      ]
    }
  },
  {
    "medication": {
      "name": "Septron",
      "time": [
        "2018-09-09"
      ]
    }
  },
  {
    "medication": {
      "name": "Septron",
      "time": [
        "2018-09-10"
      ]
    }
  },
  {
    "medication": {
      "name": "Septron",
      "time": [
        "2018-09-11"
      ]
    }
  }
];
var tempObj = {};
arr.forEach((item)=>{
  if(!tempObj[item.medication.name]){
    tempObj[item.medication.name] = item;
  } else {
    tempObj[item.medication.name].medication.time.push(...item.medication.time);
  }
});
var res = Object.values(tempObj);
console.log(res);

Answer 2

您可以使用Map，然后收集给定名称的所有值，然后呈现新数组。

var data = [{ medication: { name: "Turbohaler", time: ["2018-09-07"] } }, { medication: { name: "Turbohaler", time: ["2018-09-08"] } }, { medication: { name: "Turbohaler", time: ["2018-09-09"] } }, { medication: { name: "Turbohaler", time: ["2018-09-10"] } }, { medication: { name: "Turbohaler", time: ["2018-09-11"] } }, { medication: { name: "Turbohaler", time: ["2018-09-07"] } }, { medication: { name: "Turbohaler", time: ["2018-09-08"] } }, { medication: { name: "Turbohaler", time: ["2018-09-09"] } }, { medication: { name: "Turbohaler", time: ["2018-09-10"] } }, { medication: { name: "Turbohaler", time: ["2018-09-11"] } }, { medication: { name: "Septron", time: ["2018-09-07"] } }, { medication: { name: "Septron", time: ["2018-09-08"] } }, { medication: { name: "Septron", time: ["2018-09-09"] } }, { medication: { name: "Septron", time: ["2018-09-10"] } }, { medication: { name: "Septron", time: ["2018-09-11"] } }],
    grouped = Array.from(
        data.reduce((m, { medication: { name, time } }) => m.set(name, (m.get(name) || []).concat(time)), new Map),
        ([name, time]) => ({ medication: { name, time } })
    );
    
console.log(grouped);

.as-console-wrapper { max-height: 100% !important; top: 0; }

当存在某些特定名称的键时，数组内部的数组推入对象

2 个答案: