Question

当两个项目具有相同的mapValues时，我正在使用groupBy和keys进行分组并创建typeId。

例如原始数据为

{
    "Name": "One",
    "typeId": 1
},
{
    "Name": "Two",
    "typeId": 2
},
{
    "Name": "One Two",
    "typeId": 1
},
{
    "Name": "Three",
    "typeId": 3
},
{
    "Name": "Three Two",
    "typeId": 3
}

并通过使用groupBy将对象与匹配的typeId和omit的对象typeId中的const GroupedTypes = Object.entries( _.mapValues(_.groupBy(data, 'typeID'), rlist => rlist.map(type => _.omit(type, 'typeID')) ) );分组为...

[
    "1",
    [
      {
        "Name": "One",
      },
      {
        "Name": "One Two",
      }
    ]
  ],
  [
    "2",
    [
      {
        "Name": "Two",
      }
    ]
  ],
  [
    "3",
    [
      {
        "Name": "Three",
      }
    ],
    [
      {
        "Name": "Three Two",
      }
    ]
]

如期返回...

Name

但是，我也想从第一个Name对象中的值0添加[ "1", "One", [ { "Name": "One", }, { "Name": "One Two", } ] ], [ "2", "Two", [ { "Name": "Two", } ] ], [ "3", "Three", [ { "Name": "Three", } ], [ { "Name": "Three Two", } ] ]的对象。因此最终以类似..

lodash

我一直在浏览result = pd.merge(master_df[['Symbol','Strike_Price']],child_df,on=['Symbol','Strike_Price'],indicator=True,how='right')文档，但找不到有效的方法。我该如何实现？

Answer 1

您可以使用reduce根据// if it's the array member.names[0].name // or if it's not an array member.name对项目进行分组：

typeId

您需要创建一个累加器对象，每个const input=[{"Name":"One","typeId":1},{"Name":"Two","typeId":2},{"Name":"One Two","typeId":1},{"Name":"Three","typeId":3},{"Name":"Three Two","typeId":3}] const merged = input.reduce((acc, { Name, typeId }) => { acc[typeId] = acc[typeId] || [ typeId, Name, []]; acc[typeId][2].push({ Name }); return acc; }, {}) console.log(Object.values(merged))作为键，输出中需要的数组作为其值。如果密钥尚不存在，请添加typeId作为值的密钥。这样，第一项的[ typeId, Name, []]将出现在输出中。这就是累加器/ Name的样子：

merged

然后使用Object.values()将该对象的值作为数组获取

Answer 2

只需用{ "1": [1, "One", [{ "Name": "One" }, { "Name": "One Two" }]], "2": [2, "Two", [{ "Name": "Two" }]], "3": [3, "Three", [{ "Name": "Three" }, { "Name": "Three Two" }]] }添加另一个map。

splice

const FilterGuests = [{
  "Name": "One",
  "typeId": 1
}, {
  "Name": "Two",
  "typeId": 2
}, {
  "Name": "One Two",
  "typeId": 1
}, {
  "Name": "Three",
  "typeId": 3
}, {
  "Name": "Three Two",
  "typeId": 3
}];

const GroupedTypes = Object.entries(
  _.mapValues(_.groupBy(FilterGuests, 'typeId'), rlist =>
    rlist.map(roomType => _.omit(roomType, 'typeId'))
  )
).map(e => {
  e.splice(1, 0, e[1][0].Name);
  return e;
});

console.log(GroupedTypes);

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Answer 3

使用_.map()而不是_.mapValues()，它也适用于对象，并返回一个数组。在地图的回调中，从第二个参数获取组键，获取第一个元素Name，然后以请求的格式返回一个数组：

const FilterGuests = [{"Name":"One","typeId":1},{"Name":"Two","typeId":2},{"Name":"One Two","typeId":1},{"Name":"Three","typeId":3},{"Name":"Three Two","typeId":3}];

const GroupedTypes = _.map(
  _.groupBy(FilterGuests, 'typeId'), 
  (rlist, key) =>
  [
    key, 
    _.get(rlist, '[0].Name'), 
    rlist.map(roomType => _.omit(roomType, 'typeId'))
  ]
)

console.log(GroupedTypes);

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<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.11/lodash.js"></script>

使用groupBy

3 个答案: