java 数组去重复

Question

你这个看上去像是json数据，json转成List<javaBean>,在对list进行操作就简单了

Answer 1

{"deviceid":"460001204101325","distance":0.0,"maxspeed":0.0,"runningtime":"0秒"},{"deviceid":"460001204101325","distance":10.8,"maxspeed":42.0,"runningtime":"45分26秒"},
{"deviceid":"460001204101325","distance":15.0,"maxspeed":66.0,"runningtime":"26分30秒"},
{"deviceid":"460001204101325","distance":15.5,"maxspeed":55.0,"runningtime":"29分12秒"},
{"deviceid":"4609320402424","distance":35.2,"maxspeed":74.0,"runningtime":"1时42分27秒"},
{"deviceid":"4609320402424","distance":4.9,"maxspeed":25.0,"runningtime":"30分24秒"},
{"deviceid":"460706130100567","distance":0.0,"maxspeed":0.0,"runningtime":"0秒"},{"deviceid":"460706130100567","distance":88.6,"maxspeed":85.0,"runningtime":"1时22分3秒"},{"deviceid":"460706130100805","distance":26.4,"maxspeed":64.0,"runningtime":"50分59秒"}, 
把上面的数据只要deviceid是一样的都只取一个，后面的参数相加弄成下面这种的{"deviceid":"460001204101325","distance":41.3,"maxspeed":66.0,"runningtime":"1时31分08秒"},
{"deviceid":"4609320402424","distance":40.1,"maxspeed":74.0,"runningtime":"2时12分51秒"},
{"deviceid":"460706130100567","distance":88.6,"maxspeed":85.0,"runningtime":"1时22分3秒"},{"deviceid":"460706130100805","distance":26.4,"maxspeed":64.0,"runningtime":"50分59秒"},
这个怎么弄？？？？

Answer 2

是，但是不是在网页上，而是在service类里用java写出来？？？、

Answer 3

后台自己拼装。List<Map<String,Object>>
map key 放你的deviceid 。

Answer 4

是，但是不是在网页上，而是在service类里用java写出来？？？、简单的写了个，代码实现如下：
import java.util.ArrayList;
import java.util.HashMap;
import java.util.Iterator;
import java.util.List;
import java.util.Map;import net.sf.json.JSONArray;
public class TTT {



public static void main(String[] args) {
String json = "[{\"deviceid\":\"460001204101325\",\"distance\":0.0,\"maxspeed\":0.0,\"runningtime\":\"0秒\"},"+
"{\"deviceid\":\"460001204101325\",\"distance\":15.0,\"maxspeed\":66.0,\"runningtime\":\"26分30秒\"},"+
"{\"deviceid\":\"4609320402424\",\"distance\":35.2,\"maxspeed\":74.0,\"runningtime\":\"1时42分27秒\"},"+
"{\"deviceid\":\"4609320402424\",\"distance\":4.9,\"maxspeed\":25.0,\"runningtime\":\"30分24秒\"}]";

JSONArray ja  = JSONArray.fromObject(json);

List<DeviceInfo> list =  (List<DeviceInfo>) JSONArray.toCollection(ja, DeviceInfo.class);

Map<String,DeviceInfo> map = new HashMap<String,DeviceInfo>();

for(DeviceInfo deviceInfo: list){
 DeviceInfo deviceMap = (DeviceInfo) map
                        .get(deviceInfo.getDeviceid());
                if (deviceMap == null) {
                    map.put(deviceInfo.getDeviceid(), deviceInfo);
                } else {
                 deviceMap.setDistance(deviceMap.getDistance()+deviceInfo.getDistance());
                 deviceMap.setMaxspeed(deviceMap.getMaxspeed()>=deviceInfo.getMaxspeed()?deviceMap.getMaxspeed():deviceInfo.getMaxspeed());
                 //runningtime换个方式，这个你自己搞
                }
 }
java.util.Iterator it = map.entrySet().iterator();

List<DeviceInfo> newList =  new ArrayList<DeviceInfo>(); //最后要的List
while(it.hasNext()){
java.util.Map.Entry entry = (java.util.Map.Entry)it.next();
newList.add((DeviceInfo) entry.getValue());
}

for(DeviceInfo d: newList){
System.out.println(d.getDeviceid()+"--"+d.getDistance()+"--"+d.getMaxspeed());
}
}
}
public class DeviceInfo{

private String deviceid;
private double distance;
private double maxspeed;
private String runningtime;

public String getDeviceid() {
return deviceid;
}
public void setDeviceid(String deviceid) {
this.deviceid = deviceid;
}

public double getDistance() {
return distance;
}
public void setDistance(double distance) {
this.distance = distance;
}

public double getMaxspeed() {
return maxspeed;
}
public void setMaxspeed(double maxspeed) {
this.maxspeed = maxspeed;
}

public String getRunningtime() {
return runningtime;
}
public void setRunningtime(String runningtime) {
this.runningtime = runningtime;
}




}

Answer 5

感觉把数据封装成对象，在存放在Hashset集合里面 去重，，只是在实体类里面重写一下  equals 方法即可，和hashCode方法（这个其实可以不写，但是建议写）。   在遍历集合得到的就是  去重的，，感觉这样简单

Answer 6

{"deviceid":"460001204101325","distance":0.0,"maxspeed":0.0,"runningtime":"0秒"},{"deviceid":"460001204101325","distance":10.8,"maxspeed":42.0,"runningtime":"45分26秒"},
{"deviceid":"460001204101325","distance":15.0,"maxspeed":66.0,"runningtime":"26分30秒"},
{"deviceid":"460001204101325","distance":15.5,"maxspeed":55.0,"runningtime":"29分12秒"},
{"deviceid":"4609320402424","distance":35.2,"maxspeed":74.0,"runningtime":"1时42分27秒"},
{"deviceid":"4609320402424","distance":4.9,"maxspeed":25.0,"runningtime":"30分24秒"},
{"deviceid":"460706130100567","distance":0.0,"maxspeed":0.0,"runningtime":"0秒"},{"deviceid":"460706130100567","distance":88.6,"maxspeed":85.0,"runningtime":"1时22分3秒"},{"deviceid":"460706130100805","distance":26.4,"maxspeed":64.0,"runningtime":"50分59秒"}, 
把上面的数据只要deviceid是一样的都只取一个，后面的参数相加弄成下面这种的{"deviceid":"460001204101325","distance":41.3,"maxspeed":66.0,"runningtime":"1时31分08秒"},
{"deviceid":"4609320402424","distance":40.1,"maxspeed":74.0,"runningtime":"2时12分51秒"},
{"deviceid":"460706130100567","distance":88.6,"maxspeed":85.0,"runningtime":"1时22分3秒"},{"deviceid":"460706130100805","distance":26.4,"maxspeed":64.0,"runningtime":"50分59秒"},
这个怎么弄？？？？ 

Answer 7

json 转成Bean, 写个equals方法，或者继承排序comparable接口 ，实现comparTo方法，

Answer 8

{"deviceid":"460001204101325","distance":0.0,"maxspeed":0.0,"runningtime":"0秒"},{"deviceid":"460001204101325","distance":10.8,"maxspeed":42.0,"runningtime":"45分26秒"},
{"deviceid":"460001204101325","distance":15.0,"maxspeed":66.0,"runningtime":"26分30秒"},
{"deviceid":"460001204101325","distance":15.5,"maxspeed":55.0,"runningtime":"29分12秒"},
{"deviceid":"4609320402424","distance":35.2,"maxspeed":74.0,"runningtime":"1时42分27秒"},
{"deviceid":"4609320402424","distance":4.9,"maxspeed":25.0,"runningtime":"30分24秒"},
{"deviceid":"460706130100567","distance":0.0,"maxspeed":0.0,"runningtime":"0秒"},{"deviceid":"460706130100567","distance":88.6,"maxspeed":85.0,"runningtime":"1时22分3秒"},{"deviceid":"460706130100805","distance":26.4,"maxspeed":64.0,"runningtime":"50分59秒"}, 
把上面的数据只要deviceid是一样的都只取一个，后面的参数相加弄成下面这种的{"deviceid":"460001204101325","distance":41.3,"maxspeed":66.0,"runningtime":"1时31分08秒"},
{"deviceid":"4609320402424","distance":40.1,"maxspeed":74.0,"runningtime":"2时12分51秒"},
{"deviceid":"460706130100567","distance":88.6,"maxspeed":85.0,"runningtime":"1时22分3秒"},{"deviceid":"460706130100805","distance":26.4,"maxspeed":64.0,"runningtime":"50分59秒"},
这个怎么弄？？？？  

Answer 9

用TreeSet去重，要写排序方法吧？  排序相同的会去掉？

Answer 10

{"deviceid":"460001204101325","distance":0.0,"maxspeed":0.0,"runningtime":"0秒"},{"deviceid":"460001204101325","distance":10.8,"maxspeed":42.0,"runningtime":"45分26秒"},
{"deviceid":"460001204101325","distance":15.0,"maxspeed":66.0,"runningtime":"26分30秒"},
{"deviceid":"460001204101325","distance":15.5,"maxspeed":55.0,"runningtime":"29分12秒"},
{"deviceid":"4609320402424","distance":35.2,"maxspeed":74.0,"runningtime":"1时42分27秒"},
{"deviceid":"4609320402424","distance":4.9,"maxspeed":25.0,"runningtime":"30分24秒"},
{"deviceid":"460706130100567","distance":0.0,"maxspeed":0.0,"runningtime":"0秒"},{"deviceid":"460706130100567","distance":88.6,"maxspeed":85.0,"runningtime":"1时22分3秒"},{"deviceid":"460706130100805","distance":26.4,"maxspeed":64.0,"runningtime":"50分59秒"}, 
把上面的数据只要deviceid是一样的都只取一个，后面的参数相加弄成下面这种的{"deviceid":"460001204101325","distance":41.3,"maxspeed":66.0,"runningtime":"1时31分08秒"},
{"deviceid":"4609320402424","distance":40.1,"maxspeed":74.0,"runningtime":"2时12分51秒"},
{"deviceid":"460706130100567","distance":88.6,"maxspeed":85.0,"runningtime":"1时22分3秒"},{"deviceid":"460706130100805","distance":26.4,"maxspeed":64.0,"runningtime":"50分59秒"},
这个怎么弄？？？？  json 转成Bean 的方法：import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collection;
import java.util.List;import net.sf.json.JSONArray;
import net.sf.json.JSONObject;public class JsonUtils {
  
  public static String encodeList(Object[] objs) {
    return encodeList(Arrays.asList(objs));
  }  @SuppressWarnings("unchecked")
  public static String encodeList(Collection objs) {
    JSONArray list = new JSONArray();
    if (objs == null || objs.size() == 0)
      return list.toString();
    for (Object ae : objs) {
      list.add(ae);
    }
    return list.toString();
  }
  
  @SuppressWarnings("unchecked")
  public static String encodeObject(Object obj){
    if(obj instanceof Collection)
      return encodeList((Collection)obj);
    JSONObject jo = JSONObject.fromObject(obj);
    return jo.toString();
  }
  
  @SuppressWarnings("unchecked")
  public static <T>List<T> decodeList(String str, Class<T> clazz) {
    if (str == null || "".equals(str))
      return null;
    JSONArray obj = JSONArray.fromObject(str);
    Object[] rts = obj.toArray();
    List<T> result = new ArrayList<T>(rts.length);
    for (int i = 0; i < rts.length; i++) {
      Object jo = rts[i];
      T ele = (T)JSONObject.toBean((JSONObject) jo, clazz);
      result.add( ele );
    }
    return result;
  }
  
  @SuppressWarnings("unchecked")
  public static <T>T decodeObject(String json,Class<T> clz){
    JSONObject jsonObject = JSONObject.fromObject(json);
    T bean = (T) JSONObject.toBean(jsonObject, clz);
    return bean;
  }
} 有适合你的  ，可以转成list 里放的javaBean的型式。BEAN重写equals（）方法，只要deviceid字段相同就认为是同一个对象。之后就可以遍历这个集合a(原始集合)，放到b（去重后的集合），每次放的时候都检查b集合中是否已经有这个对像。b.contains(o) 用这个判断是否已经存在。存在就不放进去。2，直接遍历，放到hashset中也可以。会把重复的去掉。
3，treeSet是按照排序去重的，如果用就实现排序接口的排序方法。排序相同会去掉。

Answer 11

这种数据类型要是在JS中做去重要简单更合适，jquery中有响应的方法。

Answer 12

用TreeSet去重，要写排序方法吧？  排序相同的会去掉？  恩呢 我说错了  实现comparable或者  comparator    谢谢

Answer 13

大哥不对啊，，，HashSet是重写equals和HashCOde啊，，Treeset是实现接口啊    没看清楚。

Answer 14

大哥不对啊，，，HashSet是重写equals和HashCOde啊，，Treeset是实现接口啊    没看清楚。

Answer 15

整成MAP KEY为deviceid，如果MAP中有KEY就进行累加操作，如果没有就直接PUT。
循环咯