如何根據某些特定值對地圖中的值進行分組？

Question

我的取值從地圖來像下面

Key = 1_1, Value = 02/04/2012 
Key = 1_2, Value = 03/04/2012 
Key = 1_3, Value = 04/04/2012 
Key = 1_4, Value = 05/04/2012 
Key = 1_5, Value = 06/04/2012 
Key = 1_6, Value = 09/04/2012 
Key = 1_7, Value = 10/04/2012 
Key = 1_8, Value = 11/04/2012 
Key = 1_9, Value = 12/04/2012 
Key = 1_10, Value = 13/04/2012 
Key = 1_11, Value = 18/04/2012 
Key = 1_12, Value = 19/04/2012 
Key = 1_13, Value = 20/04/2012 
Key = 1_14, Value = 23/04/2012 
Key = 1_15, Value = 24/04/2012 
Key = 1_16, Value = 25/04/2012 
Key = 1_17, Value = 26/04/2012 
Key = 1_18, Value = 27/04/2012 
Key = 1_19, Value = 30/04/2012 
Key = 10_20, Value = 02/04/2012 
Key = 10_21, Value = 03/04/2012 
Key = 10_22, Value = 04/04/2012 
Key = 10_23, Value = 05/04/2012 
Key = 10_24, Value = 06/04/2012 
Key = 10_25, Value = 09/04/2012 
Key = 10_26, Value = 10/04/2012 
Key = 10_27, Value = 11/04/2012 
Key = 10_28, Value = 12/04/2012 
Key = 10_29, Value = 13/04/2012 
Key = 10_30, Value = 16/04/2012 
Key = 10_31, Value = 17/04/2012 
Key = 10_32, Value = 18/04/2012 
Key = 10_33, Value = 19/04/2012 
Key = 10_34, Value = 23/04/2012 
Key = 10_35, Value = 24/04/2012 
Key = 10_36, Value = 26/04/2012 
Key = 10_37, Value = 27/04/2012 

我真的很努力分開這些值，並把它們放在單獨的地圖。

我想分組如下。

1_1到1_19這個我想根據「_」進行分割並單獨獲得第一個值並將它們分組到一個單獨的地圖中。

像1是關鍵，值將是日期。

編輯：

employeeMap = showExelData(sheetData); 
     String previousEemployeeID = "",employeeID[]; 
     Iterator<Map.Entry> entries = employeeMap.entrySet().iterator(); 
     while (entries.hasNext()) { 
      Map.Entry entry = entries.next(); 
      employeeID = entry.getKey().toString().split("_"); 

       // this is the place where i want to check the values if 1 than group the values it can be even Key = 1_0, Value = 25/04/2012 to If Key = 1_18, Value = 30/04/2012 
    but when the other one comes ex : Key = 10_0, Value = 25/04/2012 to If Key = 10_17, Value = 30/04/2012it has to go to new Map 

我這是在哪裏欠缺的地方。 }

Answer 1

如果您想將1_1，1_2拆分爲1_19 ..請使用String類的split（）函數。

Ex。

String x = 1_19; 

String[] y = x.split("_"); 

Y [0]將是等於1且y [1]將19

至於使用所述第一值用於在地圖的密鑰，這將是不可能的，因爲它需要一地圖中每個條目的唯一鍵，就像npinti在您的帖子中發表的評論一樣。

public class Mapping { 

    Map<String, String> coMap; 
    List<String> coList; 

    public Mapping() 
    { 
     init(); 
    } 

    public static void main(String[] args) 
    { 
     Mapping oMapping = new Mapping(); 

     Map<String, Map<String, String>> oMap = oMapping.classifyMapEntries(); 

     for (String sParentKey : oMapping.coList) 
     { 
      Map<String, String> oChildMap = oMap.get(sParentKey); 
      Iterator<String> oIterator = oChildMap.keySet().iterator(); 

      System.out.println("Map"); 
      while(oIterator.hasNext()) 
      { 
       String sChildKey = oIterator.next(); 
       System.out.print("Key: " + sChildKey + ", Value: " 
            + oChildMap.get(sChildKey) + "\n"); 
      } 
     } 
    } 

    private void init() 
    { 
     coMap = new HashMap<String, String>(); 
     coList = new ArrayList<String>(); 

     coMap.put("1_1", "a"); 
     coMap.put("1_19", "a"); 
     coMap.put("10_1", "b"); 
     coMap.put("10_19", "b"); 
    } 

    private Map<String, Map<String, String>> classifyMapEntries() 
    { 
     Map<String, Map<String, String>> oClassified = 
      new HashMap<String, Map<String,String>>(); 

     Iterator<String> oIterator = coMap.keySet().iterator(); 
     while(oIterator.hasNext()) 
     { 
      String sKey = oIterator.next(); 

      String sFirst = sKey.substring(0,sKey.indexOf("_")); 
      if (!coList.contains(sFirst)) 
      { 
       coList.add(sFirst); 
      } 
     } 

     for (String sKey : coList) 
     { 
      Map<String, String> oChildMap = new HashMap<String, String>(); 

      Iterator<String> oIterator2 = coMap.keySet().iterator(); 
      while(oIterator2.hasNext()) 
      { 
       String sChildKey = oIterator2.next(); 
       String sParentKey = sChildKey.substring(0,sChildKey.indexOf("_")); 

       if (sKey.equals(sParentKey)) 
       { 
        oChildMap.put(sChildKey, coMap.get(sChildKey)); 
       } 
      } 

      oClassified.put(sKey, oChildMap); 
     } 

     return oClassified; 
    } 

} 

Answer 2

從澄清評論我假設你想過濾你的鑰匙：只包含開始1_，另一個與10_等

使用普通的Java鍵一個地圖，可以通過使用Map<String, Map<String, String>>並遍歷原始地圖做到這一點：

Map<String, Map<String, String>> filtered = new HashMap<String, Map<String, String>>(); 
for (Entry<String, String> sourceEntry : source.entrySet()) { 
    String keyPart = sourceEntry.getKey().split("_")[0]; 
    Map<String, String> filteredTarget = filtered.get(keyPart); 
    if (filteredTarget == null) { 
    filteredTarget = new HashMap<String, String>(); 
    filtered.put(keyPart, filteredTarget); 
    } 
    filteredTarget.put(sourceEntry.getKey(), sourceEntry.getValue()); 
} 

Map<String, String> oneMap = filtered.get("1"); 
assert oneMap.get("1_19").equals("30/04/2012"); 
assert filtered.get("10").get("10_37").equals("27/04/2012"); 

請注意在源地圖或地圖過濾的後續變化不會更新其他地圖。如果你想有這樣的事情，比Guava可以幫助：

Map<String, Map<String, String>> filtered = Maps.newHashMap(); 
for (Entry<String, String> sourceEntry : source.entrySet()) { 
    final String keyPart = sourceEntry.getKey().split("_")[0]; 
    Map<String, String> filteredTarget = filtered.get(keyPart); 
    if (filteredTarget == null) { 
    filteredTarget = Maps.filterKeys(source, new Predicate<String>() { 

     @Override 
     public boolean apply(String input) { 
     return input.startsWith(keyPart + "_"); 
     } 

    }); 
    filtered.put(keyPart, filteredTarget); 
    } 
} 

Map<String, String> oneMap = filtered.get("1"); 
assert oneMap.get("1_19").equals("30/04/2012"); 
assert filtered.get("10").get("10_37").equals("27/04/2012"); 

oneMap.put("1_50", "Test"); 
assert source.get("1_50").equals("Test"); 

Answer 3

你想創建地圖，其中對外部映射的關鍵是給定的按鍵的第一數量的Map地圖。以下代碼未經測試，但您可以將其作爲提示...

Map<String, Date> givenMap; // this is the given Map 
Map<Integer, Map<Integer, Date> > newMap = new HashMap<Integer, Map<Integer, Date> >(); 
for (Map.Entry<String, Date> entry : givenMap.entrySet()) { 
    String givenKey = entry.getKey(); 
    Date givenDate = entry.getValue(); 
    String[] splittedKey = givenKey.split("_"); 
    int newOuterKey = Integer.parseInt(splittedKey[0]); 
    int newInnerKey = Integer.parseInt(splittedKey[1]); 
    if (!newMap.containsKey(newOuterKey)) { 
     newMap.put(newOuterKey, new HashMap<Integer, Date>(); 
    } 
    newMap.get(newOuterKey).put(newInnerKey, givenDate); 
}