Javascript 如何基于不同索引合并两个对象

我在一个API端点返回的数组中有两个对象，该端点根据区域设置公开当选官员。一个是

办公室

，另一个是

官员

。我需要了解如何将这两个对象合并为一个更有意义的对象

{
  "offices": [
    {
      "name": "President of the United States",
      "officialIndices": [0]
    }, {
      "name": "Vice-President of the United States",
      "officialIndices": [1]
    }, {
      "name": "United States Senate",
      "officialIndices": [2, 3]
    }, {
      "name": "Governor",
      "officialIndices": [4]
    }
  ],
  "officials": [
    {
      "name": "Donald J. Trump"
    }, {
      "name": "Mike Pence"
    }, {
      "name": "Dianne Feinstein"
    }, {
      "name": "Kamala D. Harris"
    }, {
      "name": "Edmund G. Brown Jr.",
    }
  ]
}

我的最终目标如下：

[
  "officials": [
    {
      "name": "Donald J. Trump",
      "office": "President of the United States"
    }, {
      "name": "Mike Pence",
      "office": "Vice-President of the United States"
    }, {
      "name": "Dianne Feinstein",
      "office": "United States Senate"
    }, {
      "name": "Kamala D. Harris",
      "office": "United States Senate",
    }, {
      "name": "Edmund G. Brown Jr.",
      "office": "Governor"
    }
  ]
}

obj.officials.map(function(item, i){
  var j = 0;
  for (j in obj.offices)
  {
    if (obj.offices[j].officialIndices.includes(i)) {
        break;
    }
  }
  return {
    "name": item.name,
    "office": obj.offices[j].name
  }
})

所以这是我失败的尝试

var representatives = data;

function addOffice(name, indices){
  _.forEach(indices, function(value, key) {
    representatives.officials[key].office = name;
  });
}

_.forEach(representatives.offices, function(value, key) {
  // console.log("Office: '" + value.name + "' should be applied to officials " + value.officialIndices);
  addOffice(value.name, value.officialIndices);
});

// remove `offices` as it's now redundant
delete representatives.offices;

console.log(representatives);

使用此功能：

function addOffice(obj) {
  return {
    officials : obj.officials.map(function(official, idx){
      return {
        name : official.name,
        office : obj.offices.filter(function(office) {
          return office.officialIndices.indexOf(idx) > -1
        })[0].name
      }
    })
  }
}

---

您可以这样做：

[
  "officials": [
    {
      "name": "Donald J. Trump",
      "office": "President of the United States"
    }, {
      "name": "Mike Pence",
      "office": "Vice-President of the United States"
    }, {
      "name": "Dianne Feinstein",
      "office": "United States Senate"
    }, {
      "name": "Kamala D. Harris",
      "office": "United States Senate",
    }, {
      "name": "Edmund G. Brown Jr.",
      "office": "Governor"
    }
  ]
}

obj.officials.map(function(item, i){
  var j = 0;
  for (j in obj.offices)
  {
    if (obj.offices[j].officialIndices.includes(i)) {
        break;
    }
  }
  return {
    "name": item.name,
    "office": obj.offices[j].name
  }
})

---

这里有一个解决方案，用于将

官员

在办公室的

官方骰子

中使用的结果展平。最后，我们使用转换所有

官员

及其关联的

办公室

的结果值

var result = _.flatMap(data.offices, function(office) {
  return _(data.officials)
    .pick(office.officialIndices)
    .map(function(official) {
      return _(official)
        .pick('name')
        .set('office', office.name)
        .value();
    }).value();
});

var数据={
“办事处”：[{
“姓名”：“美国总统”，
“officialIndices”：[0]
}, {
“姓名”：“美国副总统”，
“药剂师”：[1]
}, {
“名称”：“美国参议院”，
“药剂师”：[2,3]
}, {
“名称”：“总督”，
“药剂师”：[4]
}],
“官员”：[{
“姓名”：“唐纳德·J·特朗普”
}, {
“姓名”：“迈克·彭斯”
}, {
“姓名”：“Dianne Feinstein”
}, {
“姓名”：“卡马拉·D·哈里斯”
}, {
“姓名”：“Edmund G.Brown Jr.”，
}]
};
var结果=uu.flatMap（data.offices，函数（office））{
返回(数据.官员)
.pick（office.officialIndices）
.map（功能（官方）{
返回(官方)
.pick（'名称'）
.set（'office'，office.name）
.value（）；
}).value（）；
});
控制台日志（结果）
body>div{min height:100%；top:0；}