我来自Python背景,目前正在将我的Python程序移植到Java。我需要关于解决问题的最佳方法的建议。
最初,我用Python创建了一个元组列表:
loft = [('india',1),('accepts',1),('narendra',1), ('modi',1),('manmohan',1),('singh',1),('sonia gandhi',1),('rajkot',1),('sharma',1),('raja',1),('india',2),('manmohan',2),('singh',2),('nepal',2),('prime minister',2),('meeting',2),('economy',2),('manmohan',3),('narendra',3),('modi',3),('gupta',3),('rajkot',3),('patel',3),('singh',3),('rajiv',3),('aajtak',3),('manmohan',4),('nepal',4),('bahadur',4),('king',4),('meeting',4),('economy',4),('wife',4),('plane',4)](其中,印度,accepts是关键字,数字是从数据库中获取的id。)现在,应用:
di = {}
for x,y in ll:
di.setdefault(x,[]).append(y)
newdi = {}我的列表变成了一本字典:
di = {'manmohan': [1, 2, 3, 4], 'sonia gandhi': [1], 'raja': [1], 'india': [1, 2], 'narendra': [1, 3], 'patel': [3], 'sharma': [1], 'nepal': [2, 4], 'gupta': [3], 'singh': [1, 2, 3], 'meeting': [2, 4], 'economy': [2, 4], 'rajkot': [1, 3], 'prime minister': [2], 'plane': [4], 'bahadur': [4], 'king': [4], 'wife': [4], 'accepts': [1], 'modi': [1, 3], 'aajtak': [3], 'rajiv': [3]}Java部分:
public void step1() throws SQLException{
Connection con= new Clustering().connect();
Statement st = con.createStatement();
Statement st1 = con.createStatement();
ResultSet rs = st.executeQuery("select uid from url where artorcat=1");
ArrayList<Tuples> allkeyword = new ArrayList<Tuples>();
long starttime = System.currentTimeMillis();
while (rs.next()) {
int id = rs.getInt("uid");
String query = "select tags.tagname from tags left join tag_url_relation on tags.tid=tag_url_relation.tid where tag_url_relation.uid="+id;
ResultSet rs1 = st1.executeQuery(query);
while (rs1.next()){
String tag = rs1.getString(1);
//Creating an object t of type Tuples
//and pass values to constructor
Tuples t = new Tuples(id,tag);
//adding the above tuple to arraylist allkeyword
allkeyword.add(t);
}//job done, now lets test by iterating
}
Iterator<Tuples> it = allkeyword.iterator();
while(it.hasNext()){
Tuples t = it.next();
System.out.println(t.getId());
System.out.println(t.getKeyword());
}
long endtime = System.currentTimeMillis();
long totaltime = endtime-starttime;
System.out.println("Total time:" + totaltime);
}
And here is Tuples class :
/**
*
*
* Tuple class is created to create a multiple data type tuple. We are using this tuples object to retrieve keyword and
* id in step1 in Clustering.java.
* @author akshayy
*
*/
public class Tuples {
int i;
String s;
public Tuples(int i, String s) {
this.i= i;
this.s=s;
}
public int getId(){
return this.i;
}
public String getKeyword(){
return this.s;
}
}到目前一切尚好。我创建了一个包含关键字和id元组类的数组列表。现在,下一步如何查找id中关键字的匹配项。像'manmohan‘可以在id 1,2,3,4中找到,等等。
di = {'manmohan': [1, 2, 3, 4], 'sonia gandhi': [1], 'raja': [1], 'india': [1, 2], 'narendra': [1, 3], 'patel': [3], 'sharma': [1], 'nepal': [2, 4], 'gupta': [3], 'singh': [1, 2, 3], 'meeting': [2, 4], 'economy': [2, 4], 'rajkot': [1, 3], 'prime minister': [2], 'plane': [4], 'bahadur': [4], 'king': [4], 'wife': [4], 'accepts': [1], 'modi': [1, 3], 'aajtak': [3], 'rajiv': [3]}请建议我下一步应该如何在arraylist中找到类似的项目,并像上面那样对它们进行排序。或者我需要一个完全不同的东西?
发布于 2013-04-29 21:31:15
您需要创建一个具有列表或设置值的Map。根据您的需要,您可以保留Tuples类,也可以单独使用String和Integer。
下面是一个示例:
// construct a map with string key (tag) and list of integers (ids) as the value
Map<String, List<Integer>> keywords = new HashMap<String, List<Integer>>();
while (rs.next()) {
int id = rs.getInt("uid");
String query = "select tags.tagname from tags left join tag_url_relation on tags.tid=tag_url_relation.tid where tag_url_relation.uid="+id;
ResultSet rs1 = st1.executeQuery(query);
while (rs1.next()){
String tag = rs1.getString(1);
// construct the List for this keyword
if (!keywords.containsKey(tag)) {
keywords.put(tag, new ArrayList<Integer>());
}
keywords.get(tag).add(id);
}
}keywords将是一个与您的Python实现中的数据结构类似的数据结构:
List<Integer> manmohanList = keywords.get("manmohan"); // will get you a list containing the numbers 1,2,3,4
for (Integer id: manmohanList) {
System.out.println(id); // prints 1,2,3,4
}发布于 2013-04-29 21:32:51
看一下java.lang.Map接口。您实际上是在构建一个
Map<String,List<Integer>> 使用纯集合类,您可以使用contains和Collections.sort等方法(关注性能,如果需要,您可以考虑自己的排序算法)
对于新的Java开发人员来说,迭代Map并非易事,但是您可以迭代KeySet,在每个迭代点对map执行get,然后对值执行contains,在本例中是一个列表。
Integer bar = whatever you are evaluating
Map<String, List<Integer>> fooMap = new HashMap<String, List<Integer>>();
... build your map ...
for(String key:fooMap.keySet()){
if(fooMap.get(key).contains(bar)){
...logic when found...
}
}发布于 2013-04-29 21:49:35
与其为元组创建一个类,不如声明一个HashMap来存储字典、关键字和位置。比如
Map<String, ArrayList<Integer>> dictionary = new HashMap<String, ArrayList<Integer>>();
//Now before adding any new keyword to the map just check if it contains it or not.
while (rs1.next()){
//Your
//Old
//Code
if(dictionary.contains(tag)){
id_list = dictionary.get(tag);
id_list.add(id);
dictionary.put(tag, id_list);
}else{
dictionary.put(tag, id);
}
}还没有测试它的打字错误。但我想你应该有个想法。希望这能有所帮助。
https://stackoverflow.com/questions/16279342
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