Groovy：如何按 2 个不同的标准对地图值进行分组？

发布于 2024-12-28 22:52:18 字数 1424 浏览 3 评论 0原文

如何按 2 个不同的标准对地图值进行分组以获得以下输出？

def listOfMaps = [
  [name:'Clark', city:'London', hobby: 'chess'], [name:'Sharma', city:'London', hobby: 'design'],
  [name:'Maradona', city:'LA', hobby: 'poker'], [name:'Zhang', city:'HK', hobby: 'chess'],
  [name:'Ali', city: 'HK', hobby: 'poker'], [name:'Liu', city:'HK', hobby: 'poker'],
  [name:'Doe', city:'LA', hobby: 'design'], [name:'Smith', city:'London', hobby: 'poker'],
  [name:'Johnson', city: 'London', hobby: 'poker'], [name:'Waters', city:'LA', hobby: 'poker'],
  [name:'Hammond', city:'LA', hobby: 'design'], [name:'Rogers', city:'LA', hobby: 'chess'],
]

团体顺序：爱好、城市

扑克
    伦敦
        史密斯
        约翰逊
    洛杉矶
        马拉多纳
        沃特斯
    香港
        阿里
        刘
设计
    伦敦
        夏尔马
    洛杉矶
        美国能源部
        哈蒙德
    香港
棋
    伦敦
        克拉克   
    洛杉矶
        罗杰斯
    香港
        张

团体顺序：城市、爱好

伦敦
    扑克
        史密斯
        约翰逊
    设计
        夏尔马
    棋
        克拉克
洛杉矶
    扑克
        马拉多纳
        沃特斯
    设计
        美国能源部
        哈蒙德
    棋
        罗杰斯

香港
    扑克
        阿里
        刘
    设计
    棋
        张

编辑：

我真正需要的是迭代的方法，以有效地循环组结构，并能够构造结果（组/子组/名称）。

类似于：

对于每个组，打印/输出组名称；
对于组内的每个子组，打印/输出
每个子组内的子组名称，打印名称。

它将产生上面概述的结果。

顺便说一句，我想对整个数据结构（组和名称）进行排序。

原文

How can I group map values by 2 different criteria to get the outputs below ?

def listOfMaps = [
  [name:'Clark', city:'London', hobby: 'chess'], [name:'Sharma', city:'London', hobby: 'design'],
  [name:'Maradona', city:'LA', hobby: 'poker'], [name:'Zhang', city:'HK', hobby: 'chess'],
  [name:'Ali', city: 'HK', hobby: 'poker'], [name:'Liu', city:'HK', hobby: 'poker'],
  [name:'Doe', city:'LA', hobby: 'design'], [name:'Smith', city:'London', hobby: 'poker'],
  [name:'Johnson', city: 'London', hobby: 'poker'], [name:'Waters', city:'LA', hobby: 'poker'],
  [name:'Hammond', city:'LA', hobby: 'design'], [name:'Rogers', city:'LA', hobby: 'chess'],
]

group order : hobby, city

poker
    London
        Smith
        Johnson
    LA
        Maradona
        Waters
    HK
        Ali
        Liu
design
    London
        Sharma
    LA
        Doe
        Hammond
    HK
chess
    London
        Clark   
    LA
        Rogers
    HK
        Zhang

group order : city, hobby

London
    poker
        Smith
        Johnson
    design
        Sharma
    chess
        Clark
LA
    poker
        Maradona
        Waters
    design
        Doe
        Hammond
    chess
        Rogers

HK
    poker
        Ali
        Liu
    design
    chess
        Zhang

Edit :

What I really need is the way to iterate to effectively loop through the group structure, and be able to construct the result (group / subgroup / name ).

Something like :

for each group, print/output the group name;
for each subgroup inside a group, print/output the subgroup name
inside each subgroup, print the names.

It would yield the result outlined above.

As a nice aside, I would like to sort the whole data structure (groups and names).

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强者自强 2025-01-04 22:52:18

对于第一种情况：

result = map.groupBy( { it.hobby }, { it.city } )

对于第二种情况：

result = map.groupBy( { it.city }, { it.hobby } )

您最终会得到地图中的原始值而不仅仅是名称，但您将能够执行以下操作：

result[ 'poker' ][ 'HK' ].name

获得结果

["Ali", "Liu"]

顺便说一句：此 groupBy 的形式有自 Groovy 以来才可用1.8.1，所以如果你坚持使用早期版本，这将不起作用

编辑2

根据你下面的评论，你可以这样做：

result.each { a, b ->
  println "$a"
  b.each { c, d ->
    println "  $c"
    d.each {
      println "    $it.name"
    }
  }
}

这与GVdP在他的回答中的逻辑相同，但我觉得使用像我这里一样具有多个参数的 groupBy 使您的代码更具可读性并且其意图更明显

For the first case:

result = map.groupBy( { it.hobby }, { it.city } )

and for the second:

result = map.groupBy( { it.city }, { it.hobby } )

You will end up with the original values in the map rather than just the name, but you will be able to do:

result[ 'poker' ][ 'HK' ].name

to get the result

["Ali", "Liu"]

btw: This form of groupBy has only been available since Groovy 1.8.1, so if you're stuck on an earlier version, this won't work

edit 2

Based on your comment below, you can then do:

result.each { a, b ->
  println "$a"
  b.each { c, d ->
    println "  $c"
    d.each {
      println "    $it.name"
    }
  }
}

This is the same logic as GVdP had in his answer, but I feel using the groupBy with multiple parameters like I have here makes your code more readable and obvious as to its intent

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等往事风中吹 2025-01-04 22:52:18

如果您希望结果是几个嵌套的映射，请尝试以下操作：

def byHobbyCity = map.groupBy{it.hobby}.collectEntries{k, v -> [k, v.groupBy{it.city}]}
def byCityHobby = map.groupBy{it.city}.collectEntries{k, v -> [k, v.groupBy{it.hobby}]}

println byHobbyCity['chess']['London']*.name
==> [Clark]

println byCityHobby['London']['chess']*.name
==> [Clark]

If you want the result to be a couple of nested maps, try something like this:

def byHobbyCity = map.groupBy{it.hobby}.collectEntries{k, v -> [k, v.groupBy{it.city}]}
def byCityHobby = map.groupBy{it.city}.collectEntries{k, v -> [k, v.groupBy{it.hobby}]}

println byHobbyCity['chess']['London']*.name
==> [Clark]

println byCityHobby['London']['chess']*.name
==> [Clark]

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谎言月老 2025-01-04 22:52:18

根据您的评论回答，这将按照您的要求进行：

def listOfMaps = [
  [name:'Clark', city:'London', hobby: 'chess'], [name:'Sharma', city:'London', hobby: 'design'],
  [name:'Maradona', city:'LA', hobby: 'poker'], [name:'Zhang', city:'HK', hobby: 'chess'],
  [name:'Ali', city: 'HK', hobby: 'poker'], [name:'Liu', city:'HK', hobby: 'poker'],
  [name:'Doe', city:'LA', hobby: 'design'], [name:'Smith', city:'London', hobby: 'poker'],
  [name:'Johnson', city: 'London', hobby: 'poker'], [name:'Waters', city:'LA', hobby: 'poker'],
  [name:'Hammond', city:'LA', hobby: 'design'], [name:'Rogers', city:'LA', hobby: 'chess'],
]

listOfMaps.groupBy {it.hobby}.each { k,v ->
 println k
 v.groupBy {it.city}.each {k1, v1->
  println "  $k1"
  v1.each {
   println "    $it.name"
  }
 }
}

用 Groovy 1.7.10 测试

Answering based on your comments, this will do as you ask:

def listOfMaps = [
  [name:'Clark', city:'London', hobby: 'chess'], [name:'Sharma', city:'London', hobby: 'design'],
  [name:'Maradona', city:'LA', hobby: 'poker'], [name:'Zhang', city:'HK', hobby: 'chess'],
  [name:'Ali', city: 'HK', hobby: 'poker'], [name:'Liu', city:'HK', hobby: 'poker'],
  [name:'Doe', city:'LA', hobby: 'design'], [name:'Smith', city:'London', hobby: 'poker'],
  [name:'Johnson', city: 'London', hobby: 'poker'], [name:'Waters', city:'LA', hobby: 'poker'],
  [name:'Hammond', city:'LA', hobby: 'design'], [name:'Rogers', city:'LA', hobby: 'chess'],
]

listOfMaps.groupBy {it.hobby}.each { k,v ->
 println k
 v.groupBy {it.city}.each {k1, v1->
  println "  $k1"
  v1.each {
   println "    $it.name"
  }
 }
}

Tested with Groovy 1.7.10

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