我有一个对象列表，如下所示，
Emp e1 = new Emp(10,"Anitha",1000,"AE");
Emp e2 = new Emp(20,"Chaitanya",2000,"SE");
Emp e3 = new Emp(30,"Chaitanya",3000,"SE");
Emp e4 = new Emp(40,"Deepthi",2100,"AE");
Emp e5 = new Emp(50,"Deepthi",2200,"CE");
Emp e6 = new Emp(60,"Deepthi",2300,"BE");
Emp e7 = new Emp(70,"Anitha",2300,"BE");
Emp e8 = new Emp(80,"Anitha",2400,"ME");
Emp e9 = new Emp(90,"Sita",2200,"CE");
Emp e10 = new Emp(100,"Hari",2200,"CE");
Emp e11 = new Emp(110,"Krishna",2200,"CE");
我想过滤唯一名称上的值，并过滤相同的名称，如
1.on唯一名称：输出应为
(50,"Deepthi",2200,"CE")
(100,"Hari",2200,"CE")
(110,"Krishna",2200,"CE")
并分享相同的名称：
像输出一样(10,"Anitha",1000,"AE")
(70,"Anitha",2300,"BE")
(80,"Anitha",2400,"ME")
(20,"Chaitanya",2000,"SE");
(30,"Chaitanya",3000,"SE");
(40,"Deepthi",2100,"AE");
(50,"Deepthi",2200,"CE");
(60,"Deepthi",2300,"BE");
使用收藏.... 有人能帮助我吗？
先谢谢。 Nithya
答案 0 :(得分：3)
如果您使用的是java 8，请跳到最后！
我可能会创建一个map来做这件事，但看起来你是Java的新手，所以我将描述更基本的方法。
您应首先创建一个列表（arraylist），如下所示：
// create an arraylist (list based on an array)
List<Emp> emps = new ArrayList<Emp>();
然后您可以将对象添加到列表中：
emps.add(new Emp(10,"Anitha",1000,"AE"));
emps.add(new Emp(20,"Chaitanya",2000,"SE"));
.
.
现在你可以开始过滤了！
因此，假设您在类getName()中有Emp方法，您可以编写如下函数：
// this function takes a list and a name, and filters the list by the name
public static List<Emp> filterEmpsByName(List<Emp> emps, String name){
// a place to store the result
List<Emp> result = new ArrayList<Emp>();
// iterate over the list we got
for (Emp emp: emps){
// save only the elements we want
if (emp.getName().equals(name)){
result.add(emp);
}
}
return result;
}
现在，过滤将是调用该函数的一个简单问题：
// print to standard output the result of our function on the "main" list `emp` with name "Anitha"
for (Emp emp : filterEmpsByName(emps, "Anitha")){
System.out.println(emp.toString()); // make sure toString() is overridden in Emp class
}
现在第二部分有点棘手：
// this function takes a list and a name, and filters the list by the name
public static List<Emp> getDistinctlyNamedEmps(List<Emp> emps, String name) {
// this time we use a map which is A LOT faster for this kind of operation
Map<String, Emp> result = new HashMap<String, Emp>();
// iterate over the list we got
for (Emp emp : emps) {
// save only the elements we want
if (result.get(emp.getName()) == null ) {
result.put(emp.getName(), emp);
}
}
// convert map to list - not mandatory if you can use the map as is...
return new ArrayList<Emp>(result.values());
}
请注意，您还可以编写一个comparator来使用名称/任何其他属性来比较对象，但这超出了此评论的范围： - ）
把整个事情放在一起：
主要课程：
import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
import java.util.Map;
public class Main {
public static void main(String[] args) {
// create an [arraylist][4] (list based on an array)
List<Emp> emps = new ArrayList<Emp>();
emps.add(new Emp(10, "Anitha", 1000, "AE"));
emps.add(new Emp(20, "Chaitanya", 2000, "SE"));
// print to standard output the result of our function on the "main"
// list `emp` with name "Anitha"
System.out.println("filterEmpsByName(emps, \"Anitha\") output:");
for (Emp emp : filterEmpsByName(emps, "Anitha")) {
System.out.println(emp.toString()); // make sure toString() is
// overridden in Emp class
}
// print to standard output the result of our second function on the "main"
// list `emp`
System.out.println("getDistinctlyNamedEmps(emps) output:");
for (Emp emp : getDistinctlyNamedEmps(emps)) {
System.out.println(emp.toString()); // make sure toString() is
// overridden in Emp class
}
}
// this function takes a list and a name, and filters the list by the name
public static List<Emp> filterEmpsByName(List<Emp> emps, String name) {
// a place to store the result
List<Emp> result = new ArrayList<Emp>();
// iterate over the list we got
for (Emp emp : emps) {
// save only the elements we want
if (emp.getName().equals(name)) {
result.add(emp);
}
}
return result;
}
// this function takes a list and a name, and filters the list by the name
public static List<Emp> getDistinctlyNamedEmps(List<Emp> emps) {
// this time we use a map which is A LOT faster for this kind of
// operation
Map<String, Emp> result = new HashMap<String, Emp>();
// iterate over the list we got
for (Emp emp : emps) {
// save only the elements we want
if (result.get(emp.getName()) == null) {
result.put(emp.getName(), emp);
}
}
// convert map to list - not necessary
return new ArrayList<Emp>(result.values());
}
}
部分Emp类：
public class Emp {
private String name;
public Emp(int stubi, String name, int j, String stubs) {
this.name = name;
}
public String getName() {
return this.name;
}
public String toString() {
return "[" + this.name + "]";
}
}
Java 8：
Java 8具有lambda expressions（匿名函数），它是许多其他语言中用于过滤和其他操作的简洁工具。
您可以详细了解如何使用它们here。
答案 1 :(得分：0)
这不是集合，虽然它是传统意义上的“列表”，但它不是java.util.ArrayList。
为此，您可以尝试：
ArrayList<Emp> employeeList = new ArrayList<Emp>();
employeeList.add(new Emp(10,"Anitha",1000,"AE"));
// repeat
答案 2 :(得分：0)
据我所知，答案直到现在才假设任务是搜索查找特定名称，或者查找具有唯一名称的元素 - 我认为这不是问的问题对于。
为了按照原始问题中描述的方式过滤列表，可以创建一个从“密钥”（即本例中的“名称”）到共享此元素的元素列表的映射键。使用此地图，可以轻松找到
这些任务非常相似，可能有进一步的概括，但这里有一种方法可以实现：
import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
import java.util.Map;
import java.util.Map.Entry;
public class CriterionFilter
{
public static void main(String[] args)
{
List<Emp> list = new ArrayList<Emp>();
list.add(new Emp(10,"Anitha",1000,"AE"));
list.add(new Emp(20,"Chaitanya",2000,"SE"));
list.add(new Emp(30,"Chaitanya",3000,"SE"));
list.add(new Emp(40,"Deepthi",2100,"AE"));
list.add(new Emp(50,"Deepthi",2200,"CE"));
list.add(new Emp(60,"Deepthi",2300,"BE"));
list.add(new Emp(70,"Anitha",2300,"BE"));
list.add(new Emp(80,"Anitha",2400,"ME"));
list.add(new Emp(90,"Sita",2200,"CE"));
list.add(new Emp(100,"Hari",2200,"CE"));
list.add(new Emp(110,"Krishna",2200,"CE"));
Function<Emp, String> keyFunction = new Function<Emp, String>()
{
@Override
public String apply(Emp s)
{
return s.getName();
}
};
List<Emp> fiteredOnUnique = filterOnUnique(list, keyFunction);
System.out.println("Filtered on unique:");
print(fiteredOnUnique);
List<Emp> filteredOnSame = filterOnSame(list, keyFunction);
System.out.println("Filtered on same:");
print(filteredOnSame);
}
private static void print(Iterable<?> elements)
{
for (Object element : elements)
{
System.out.println(element);
}
}
/**
* Create a map that maps the keys that are provided for the given
* elements to the list of elements that have this key
*
* @param elements The input elements
* @param keyFunction The key function
* @return The map
*/
private static <T, K> Map<K, List<T>> map(
Iterable<? extends T> elements, Function<? super T, K> keyFunction)
{
Map<K, List<T>> map = new HashMap<K, List<T>>();
for (T t : elements)
{
K key = keyFunction.apply(t);
List<T> list = map.get(key);
if (list == null)
{
list = new ArrayList<T>();
map.put(key, list);
}
list.add(t);
}
return map;
}
/**
* Uses the given key function to compute the keys associated with the
* given elements, and returns a list containing the element of
* the given sequence that have unique keys
*
* @param elements The input elements
* @param keyFunction The key function
* @return The filtered list
*/
private static <T, K> List<T> filterOnUnique(
Iterable<? extends T> elements, Function<? super T, K> keyFunction)
{
List<T> result = new ArrayList<T>();
Map<K, List<T>> map = map(elements, keyFunction);
for (Entry<K, List<T>> entry : map.entrySet())
{
List<T> list = entry.getValue();
if (list.size() == 1)
{
result.add(list.get(0));
}
}
return result;
}
/**
* Uses the given key function to compute the keys associated with the
* given elements, and returns a list containing all elements of
* the given sequence that have a key that occurs multiple times.
*
* @param elements The input elements
* @param keyFunction The key function
* @return The filtered list
*/
private static <T, K> List<T> filterOnSame(
Iterable<? extends T> elements, Function<? super T, K> keyFunction)
{
List<T> result = new ArrayList<T>();
Map<K, List<T>> map = map(elements, keyFunction);
for (Entry<K, List<T>> entry : map.entrySet())
{
List<T> list = entry.getValue();
if (list.size() > 1)
{
result.addAll(list);
}
}
return result;
}
/**
* Interface for a generic function
*/
static interface Function<S, T>
{
T apply(S s);
}
}
class Emp
{
private int i;
private String name;
private int j;
private String whatever;
public Emp(int i, String name, int j, String whatever)
{
this.i = i;
this.name = name;
this.j = j;
this.whatever = whatever;
}
@Override
public String toString()
{
return "Emp [i=" + i + ", name=" + name + ", j=" + j + ", whatever=" + whatever + "]";
}
String getName()
{
return name;
}
}
（编辑：调整第一个案例，见原始问题的评论）