我想根据谓词筛选java.util.Collection。
当前回答
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。
其他回答
我需要根据列表中已经存在的值来筛选列表。例如,删除后面小于当前值的所有值。{2 5 3 4 7 5} ->{2 5 7}。或者例如删除所有重复项{3 5 4 2 3 5 6}->{3 5 4 2 6}。
public class Filter {
public static <T> void List(List<T> list, Chooser<T> chooser) {
List<Integer> toBeRemoved = new ArrayList<>();
leftloop:
for (int right = 1; right < list.size(); ++right) {
for (int left = 0; left < right; ++left) {
if (toBeRemoved.contains(left)) {
continue;
}
Keep keep = chooser.choose(list.get(left), list.get(right));
switch (keep) {
case LEFT:
toBeRemoved.add(right);
continue leftloop;
case RIGHT:
toBeRemoved.add(left);
break;
case NONE:
toBeRemoved.add(left);
toBeRemoved.add(right);
continue leftloop;
}
}
}
Collections.sort(toBeRemoved, new Comparator<Integer>() {
@Override
public int compare(Integer o1, Integer o2) {
return o2 - o1;
}
});
for (int i : toBeRemoved) {
if (i >= 0 && i < list.size()) {
list.remove(i);
}
}
}
public static <T> void List(List<T> list, Keeper<T> keeper) {
Iterator<T> iterator = list.iterator();
while (iterator.hasNext()) {
if (!keeper.keep(iterator.next())) {
iterator.remove();
}
}
}
public interface Keeper<E> {
boolean keep(E obj);
}
public interface Chooser<E> {
Keep choose(E left, E right);
}
public enum Keep {
LEFT, RIGHT, BOTH, NONE;
}
}
这将被这样使用。
List<String> names = new ArrayList<>();
names.add("Anders");
names.add("Stefan");
names.add("Anders");
Filter.List(names, new Filter.Chooser<String>() {
@Override
public Filter.Keep choose(String left, String right) {
return left.equals(right) ? Filter.Keep.LEFT : Filter.Keep.BOTH;
}
});
我将把RxJava加入其中,它在Android上也可用。RxJava可能并不总是最好的选择,但如果您希望在集合上添加更多转换或在过滤时处理错误,RxJava将为您提供更大的灵活性。
Observable.from(Arrays.asList(1, 2, 3, 4, 5))
.filter(new Func1<Integer, Boolean>() {
public Boolean call(Integer i) {
return i % 2 != 0;
}
})
.subscribe(new Action1<Integer>() {
public void call(Integer i) {
System.out.println(i);
}
});
输出:
1
3
5
更多关于RxJava过滤器的细节可以在这里找到。
设置:
public interface Predicate<T> {
public boolean filter(T t);
}
void filterCollection(Collection<T> col, Predicate<T> predicate) {
for (Iterator i = col.iterator(); i.hasNext();) {
T obj = i.next();
if (predicate.filter(obj)) {
i.remove();
}
}
}
的用法:
List<MyObject> myList = ...;
filterCollection(myList, new Predicate<MyObject>() {
public boolean filter(MyObject obj) {
return obj.shouldFilter();
}
});
一些简单明了的Java代码怎么样
List<Customer> list ...;
List<Customer> newList = new ArrayList<>();
for (Customer c : list){
if (c.getName().equals("dd")) newList.add(c);
}
简单、易读、简单(在Android上也适用!) 但如果你使用的是Java 8,你可以用一句简单的话来实现:
List<Customer> newList = list.stream().filter(c -> c.getName().equals("dd")).collect(toList());
注意,toList()是静态导入的
JFilter http://code.google.com/p/jfilter/最适合您的需求。
JFilter是一个简单、高性能的开源库,用于查询Java bean集合。
关键特性
Support of collection (java.util.Collection, java.util.Map and Array) properties. Support of collection inside collection of any depth. Support of inner queries. Support of parameterized queries. Can filter 1 million records in few 100 ms. Filter ( query) is given in simple json format, it is like Mangodb queries. Following are some examples. { "id":{"$le":"10"} where object id property is less than equals to 10. { "id": {"$in":["0", "100"]}} where object id property is 0 or 100. {"lineItems":{"lineAmount":"1"}} where lineItems collection property of parameterized type has lineAmount equals to 1. { "$and":[{"id": "0"}, {"billingAddress":{"city":"DEL"}}]} where id property is 0 and billingAddress.city property is DEL. {"lineItems":{"taxes":{ "key":{"code":"GST"}, "value":{"$gt": "1.01"}}}} where lineItems collection property of parameterized type which has taxes map type property of parameteriszed type has code equals to GST value greater than 1.01. {'$or':[{'code':'10'},{'skus': {'$and':[{'price':{'$in':['20', '40']}}, {'code':'RedApple'}]}}]} Select all products where product code is 10 or sku price in 20 and 40 and sku code is "RedApple".
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