我想根据谓词筛选java.util.Collection。


当前回答

java8之前的简单解决方案:

ArrayList<Item> filtered = new ArrayList<Item>(); 
for (Item item : items) if (condition(item)) filtered.add(item);

不幸的是,这个解决方案不是完全通用的,它输出的是一个列表,而不是给定集合的类型。此外,在我看来,引入库或编写函数来包装这段代码似乎有些过分,除非条件很复杂,但随后可以为该条件编写函数。

其他回答

在Java 8中,您可以直接使用此过滤器方法,然后执行此操作。

 List<String> lines = Arrays.asList("java", "pramod", "example");

 List<String> result = lines.stream()              
         .filter(line -> !"pramod".equals(line))     
         .collect(Collectors.toList());              

 result.forEach(System.out::println); 

您可以使用ForEach DSL编写

import static ch.akuhn.util.query.Query.select;
import static ch.akuhn.util.query.Query.$result;
import ch.akuhn.util.query.Select;

Collection<String> collection = ...

for (Select<String> each : select(collection)) {
    each.yield = each.value.length() > 3;
}

Collection<String> result = $result();

给定一个集合[The, quick, brown, fox, jumping, over, The, lazy, dog],结果是[quick, brown, jumping, over, lazy],即所有字符串都长于三个字符。

ForEach DSL支持的所有迭代样式都是

AllSatisfy AnySatisfy 收集 Counnt CutPieces 检测 GroupedBy IndexOf InjectInto 拒绝 选择

更多详情请参考https://www.iam.unibe.ch/scg/svn_repos/Sources/ForEach

https://code.google.com/p/joquery/

支持不同的可能性,

给定的集合,

Collection<Dto> testList = new ArrayList<>();

的类型,

class Dto
{
    private int id;
    private String text;

    public int getId()
    {
        return id;
    }

    public int getText()
    {
        return text;
    }
}

过滤器

Java 7

Filter<Dto> query = CQ.<Dto>filter(testList)
    .where()
    .property("id").eq().value(1);
Collection<Dto> filtered = query.list();

Java 8

Filter<Dto> query = CQ.<Dto>filter(testList)
    .where()
    .property(Dto::getId)
    .eq().value(1);
Collection<Dto> filtered = query.list();

同时,

Filter<Dto> query = CQ.<Dto>filter()
        .from(testList)
        .where()
        .property(Dto::getId).between().value(1).value(2)
        .and()
        .property(Dto::grtText).in().value(new string[]{"a","b"});

排序(也可用于Java 7)

Filter<Dto> query = CQ.<Dto>filter(testList)
        .orderBy()
        .property(Dto::getId)
        .property(Dto::getName)
    Collection<Dto> sorted = query.list();

分组(也可用于Java 7)

GroupQuery<Integer,Dto> query = CQ.<Dto,Dto>query(testList)
        .group()
        .groupBy(Dto::getId)
    Collection<Grouping<Integer,Dto>> grouped = query.list();

连接(也可用于Java 7)

考虑到,

class LeftDto
{
    private int id;
    private String text;

    public int getId()
    {
        return id;
    }

    public int getText()
    {
        return text;
    }
}

class RightDto
{
    private int id;
    private int leftId;
    private String text;

    public int getId()
    {
        return id;
    }

    public int getLeftId()
        {
            return leftId;
        }

    public int getText()
    {
        return text;
    }
}

class JoinedDto
{
    private int leftId;
    private int rightId;
    private String text;

    public JoinedDto(int leftId,int rightId,String text)
    {
        this.leftId = leftId;
        this.rightId = rightId;
        this.text = text;
    }

    public int getLeftId()
    {
        return leftId;
    }

    public int getRightId()
        {
            return rightId;
        }

    public int getText()
    {
        return text;
    }
}

Collection<LeftDto> leftList = new ArrayList<>();

Collection<RightDto> rightList = new ArrayList<>();

可以像这样连接,

Collection<JoinedDto> results = CQ.<LeftDto, LeftDto>query().from(leftList)
                .<RightDto, JoinedDto>innerJoin(CQ.<RightDto, RightDto>query().from(rightList))
                .on(LeftFyo::getId, RightDto::getLeftId)
                .transformDirect(selection ->  new JoinedDto(selection.getLeft().getText()
                                                     , selection.getLeft().getId()
                                                     , selection.getRight().getId())
                                 )
                .list();

表达式

Filter<Dto> query = CQ.<Dto>filter()
    .from(testList)
    .where()
    .exec(s -> s.getId() + 1).eq().value(2);

这里有一些非常棒的答案。对我来说,我想让事情尽可能简单易懂:

public abstract class AbstractFilter<T> {

    /**
     * Method that returns whether an item is to be included or not.
     * @param item an item from the given collection.
     * @return true if this item is to be included in the collection, false in case it has to be removed.
     */
    protected abstract boolean excludeItem(T item);

    public void filter(Collection<T> collection) {
        if (CollectionUtils.isNotEmpty(collection)) {
            Iterator<T> iterator = collection.iterator();
            while (iterator.hasNext()) {
                if (excludeItem(iterator.next())) {
                    iterator.remove();
                }
            }
        }
    }
}

考虑使用支持泛型的更新的Collections框架谷歌Collections。

更新:谷歌集合库现在已弃用。你应该使用最新发布的番石榴。它仍然具有对集合框架的所有相同扩展,包括基于谓词进行筛选的机制。