的Groovy - ArrayList的一個遞增項目數量，而不是重新添加

Question

我有一個看起來像這兩數組列表，（這我要簡化這個問題）：

def originalList = [[name: "Ben", age: 21, weight: 80], [name: "Martin", age: 36, weight: 99], [name: "Sammy", age: 18, weight: 65], [name: "Ben", age: 28, weight: 75], [name: "Ben", age: 28, weight: 120]] 

另一位我會請撥打newList並開始爲空，即def newList = []，但會添加屬性name,age和occurrences。

現在我想做的是循環遍歷originalList，並添加其項目newList但如果newList已經包含具有相同名稱和年齡屬性的項目我想遞增occurrences屬性的項目，如：

def newList = [[name: "Ben", age: 21, occurrences: 1], [name: "Martin", age: 36, occurrences: 1], [name: "Sammy", age: 18, occurrences: 1], [name: "Ben", age: 28, occurrences: 2]] 

我該如何着手完成這項工作？這是我嘗試過的。

originalList.each { 
    newList.eachWithIndex { nl, i -> 
     if(nl.name.equals(it.name) && nl.age == it.age) { 
      nl.occurrances++ 
     } else { 
      newList.add([name: it.name, age: it.age, occurrances: 0]) 
     } 
    } 
} 

它看起來像它不工作怎麼把newList.eachWithIndex永遠循環怎麼把它開始是空的，怎麼把它永遠循環可以永遠遞增occurrences財產或項目添加到列表中。

Answer 1

你可以這種方式例如：

def originalList = [ 
    [name: "Ben", age: 21, weight: 80], 
    [name: "Martin", age: 36, weight: 99], 
    [name: "Sammy", age: 18, weight: 65], 
    [name: "Ben", age: 28, weight: 75], 
    [name: "Ben", age: 28, weight: 120] 
] 

originalList 
    .groupBy({ it.name }, { it.age }) 
    .collect { k1, v1 -> v1.collect { k2, v2 -> [name: k1, age: k2, occurrences: v2.size()] } } 
    .flatten() 

它的第一步驟的輸入被兩個name分組並age這導致Map，然後values（v1）每個鍵​​被處理獲取統計信息。由於上述操作導致嵌套List，所以在最後調用flatten()。

Answer 2

我不是一個時髦的開發人員，所以我會在java中編寫代碼片段。請翻譯你自己。

Map<String,Integer> map=new HashMap<String,Integer>();//this map stores name and occurances 

for(Employee e: originaleList){ 
    int oldCount=map.get(e.getName()) == null ? 0 : map.get(e.getName()); 
    map.put(e.getName(), oldCount + 1); 

    newList.add(new Employee(e.getName(), e.getAge(), map.get(e.getName()))); 
    } 

Answer 3

只是在你原來的算法修復的bug（使用地圖的方式更有效，雖然！）：

originalList.each { 
    def found=false 
    newList.eachWithIndex { nl, i -> 
     if(nl.name.equals(it.name) && nl.age == it.age) { 
      nl.occurrances++ 
      found = true 
     } 
    } 
    if (!found) { 
     newList.add([name: it.name, age: it.age, occurrances: 1]) 
    } 
} 

UPDATE：版本地圖，根據要求...

def map = [:] 

originalList.each { 
    def key = it.name+"@"+it.age // or something better, keeping it simple here 

    if (map[key]) { 
     map[key].occurances++ 
    } 
    else { 
     def entry = [name: it.name, age: it.age, occurrances: 1] 
     map[key] = entry 
     newList.add(entry) 
    } 
}