如何将一个数组列表(size=1000)拆分为多个相同大小(=10)的数组列表?
ArrayList<Integer> results;
如何将一个数组列表(size=1000)拆分为多个相同大小(=10)的数组列表?
ArrayList<Integer> results;
当前回答
您需要知道您划分列表的块大小。假设您有一个包含108个条目的列表,您需要25个块大小。因此,你最终会得到5个列表:
4项各有25项; 有8个元素的。
代码:
public static void main(String[] args) {
List<Integer> list = new ArrayList<Integer>();
for (int i=0; i<108; i++){
list.add(i);
}
int size= list.size();
int j=0;
List< List<Integer> > splittedList = new ArrayList<List<Integer>>() ;
List<Integer> tempList = new ArrayList<Integer>();
for(j=0;j<size;j++){
tempList.add(list.get(j));
if((j+1)%25==0){
// chunk of 25 created and clearing tempList
splittedList.add(tempList);
tempList = null;
//intializing it again for new chunk
tempList = new ArrayList<Integer>();
}
}
if(size%25!=0){
//adding the remaining enteries
splittedList.add(tempList);
}
for (int k=0;k<splittedList.size(); k++){
//(k+1) because we started from k=0
System.out.println("Chunk number: "+(k+1)+" has elements = "+splittedList.get(k).size());
}
}
其他回答
这对我很有用
/**
* Returns List of the List argument passed to this function with size = chunkSize
*
* @param largeList input list to be portioned
* @param chunkSize maximum size of each partition
* @param <T> Generic type of the List
* @return A list of Lists which is portioned from the original list
*/
public static <T> List<List<T>> chunkList(List<T> list, int chunkSize) {
if (chunkSize <= 0) {
throw new IllegalArgumentException("Invalid chunk size: " + chunkSize);
}
List<List<T>> chunkList = new ArrayList<>(list.size() / chunkSize);
for (int i = 0; i < list.size(); i += chunkSize) {
chunkList.add(list.subList(i, i + chunkSize >= list.size() ? list.size()-1 : i + chunkSize));
}
return chunkList;
}
例如:
List<Integer> stringList = new ArrayList<>();
stringList.add(0);
stringList.add(1);
stringList.add(2);
stringList.add(3);
stringList.add(4);
stringList.add(5);
stringList.add(6);
stringList.add(7);
stringList.add(8);
stringList.add(9);
List<List<Integer>> chunkList = getChunkList1(stringList, 2);
您可以将Guava库添加到项目中并使用列表。划分方法,例如:
List<Integer> bigList = ...
List<List<Integer>> smallerLists = Lists.partition(bigList, 10);
您需要知道您划分列表的块大小。假设您有一个包含108个条目的列表,您需要25个块大小。因此,你最终会得到5个列表:
4项各有25项; 有8个元素的。
代码:
public static void main(String[] args) {
List<Integer> list = new ArrayList<Integer>();
for (int i=0; i<108; i++){
list.add(i);
}
int size= list.size();
int j=0;
List< List<Integer> > splittedList = new ArrayList<List<Integer>>() ;
List<Integer> tempList = new ArrayList<Integer>();
for(j=0;j<size;j++){
tempList.add(list.get(j));
if((j+1)%25==0){
// chunk of 25 created and clearing tempList
splittedList.add(tempList);
tempList = null;
//intializing it again for new chunk
tempList = new ArrayList<Integer>();
}
}
if(size%25!=0){
//adding the remaining enteries
splittedList.add(tempList);
}
for (int k=0;k<splittedList.size(); k++){
//(k+1) because we started from k=0
System.out.println("Chunk number: "+(k+1)+" has elements = "+splittedList.get(k).size());
}
}
**Divide a list to lists of n size**
import java.util.AbstractList;
import java.util.ArrayList;
import java.util.List;
public final class PartitionUtil<T> extends AbstractList<List<T>> {
private final List<T> list;
private final int chunkSize;
private PartitionUtil(List<T> list, int chunkSize) {
this.list = new ArrayList<>(list);
this.chunkSize = chunkSize;
}
public static <T> PartitionUtil<T> ofSize(List<T> list, int chunkSize) {
return new PartitionUtil<>(list, chunkSize);
}
@Override
public List<T> get(int index) {
int start = index * chunkSize;
int end = Math.min(start + chunkSize, list.size());
if (start > end) {
throw new IndexOutOfBoundsException("Index " + index + " is out of the list range <0," + (size() - 1) + ">");
}
return new ArrayList<>(list.subList(start, end));
}
@Override
public int size() {
return (int) Math.ceil((double) list.size() / (double) chunkSize);
}
}
Function call :
List<List<String>> containerNumChunks = PartitionUtil.ofSize(list, 999)
详情:https://e.printstacktrace.blog/divide-a-list-to-lists-of-n-size-in-Java-8/
Java 8
我们可以根据大小或条件拆分列表。
static Collection<List<Integer>> partitionIntegerListBasedOnSize(List<Integer> inputList, int size) {
return inputList.stream()
.collect(Collectors.groupingBy(s -> (s-1)/size))
.values();
}
static <T> Collection<List<T>> partitionBasedOnSize(List<T> inputList, int size) {
final AtomicInteger counter = new AtomicInteger(0);
return inputList.stream()
.collect(Collectors.groupingBy(s -> counter.getAndIncrement()/size))
.values();
}
static <T> Collection<List<T>> partitionBasedOnCondition(List<T> inputList, Predicate<T> condition) {
return inputList.stream().collect(Collectors.partitioningBy(s-> (condition.test(s)))).values();
}
然后我们可以把它们用作:
final List<Integer> list = Arrays.asList(1,2,3,4,5,6,7,8,9,10);
System.out.println(partitionIntegerListBasedOnSize(list, 4)); // [[1, 2, 3, 4], [5, 6, 7, 8], [9, 10]]
System.out.println(partitionBasedOnSize(list, 4)); // [[1, 2, 3, 4], [5, 6, 7, 8], [9, 10]]
System.out.println(partitionBasedOnSize(list, 3)); // [[1, 2, 3], [4, 5, 6], [7, 8, 9], [10]]
System.out.println(partitionBasedOnCondition(list, i -> i<6)); // [[6, 7, 8, 9, 10], [1, 2, 3, 4, 5]]