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
private ArrayList<List<String>> chunkArrayList(ArrayList<String> arrayToChunk, int chunkSize) {
ArrayList<List<String>> chunkList = new ArrayList<>();
int guide = arrayToChunk.size();
int index = 0;
int tale = chunkSize;
while (tale < arrayToChunk.size()){
chunkList.add(arrayToChunk.subList(index, tale));
guide = guide - chunkSize;
index = index + chunkSize;
tale = tale + chunkSize;
}
if (guide >0) {
chunkList.add(arrayToChunk.subList(index, index + guide));
}
Log.i("Chunked Array: " , chunkList.toString());
return chunkList;
}
例子
ArrayList<String> test = new ArrayList<>();
for (int i=1; i<=1000; i++){
test.add(String.valueOf(i));
}
chunkArrayList(test,10);
输出
分块:[[1,2,3,4,5,6,7,8,9,10),(11、12、13、14、15、16、17、18、19、20],[21日,22日,23日,24日,25日,26日,27日,28日,29日,30日],[第三十一条、第三十二条、第三十三,34岁,35岁,36岁,37岁,38岁,39岁,40],[41、42、43、44、45、46岁,47岁,48岁,49岁,50],[51岁,52岁,53岁,54岁,55岁,56岁,57岁的58岁的59岁60],[61,62,63,64,65,66,67,68,69,70],[71,72,73,74,75,76,77,78,79,80],[81,82,83,84,85,86,87,88,89,90],[91,92,93,94,95,96,97,98,99,100 ], .........
你会在日志里看到的
其他回答
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]]
**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/
你也可以使用FunctionalJava库- List有分区方法。这个库有自己的集合类型,你可以将它们来回转换为java集合。
import fj.data.List;
java.util.List<String> javaList = Arrays.asList("a", "b", "c", "d" );
List<String> fList = Java.<String>Collection_List().f(javaList);
List<List<String> partitions = fList.partition(2);
让我们假设您想要将列表分割为多个块的类作为库类。
所以让我们说这个类被称为“shared”,in应该是final,以确保它不会被扩展。
import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;
public final class Shared {
List<Integer> input;
int portion;
public Shared(int portion, Integer... input) {
this.setPortion(portion);
this.setInput(input);
}
public List<List<Integer>> listToChunks() {
List<List<Integer>> result = new ArrayList<List<Integer>>();
int size = this.size();
int startAt = 0;
int endAt = this.portion;
while (endAt <= size) {
result.add(this.input.subList(startAt, endAt));
startAt = endAt;
endAt = (size - endAt < this.portion && size - endAt > 0) ? (this.size()) : (endAt + this.portion);
}
return result;
}
public int size() {
return this.input.size();
}
public void setInput(Integer... input) {
if (input != null && input.length > 0)
this.input = Arrays.asList(input);
else
System.out.println("Error 001 : please enter a valid array of integers.");
}
public void setPortion(int portion) {
if (portion > 0)
this.portion = portion;
else
System.out.println("Error 002 : please enter a valid positive number.");
}
}
接下来,让我们尝试从另一个持有公共静态void main(String…args)
public class exercise {
public static void main(String[] args) {
Integer[] numbers = {1, 2, 3, 4, 5, 6, 7};
int portion = 2;
Shared share = new Shared(portion, numbers);
System.out.println(share.listToChunks());
}
}
现在,如果输入一个整数数组[1,2,3,4,5,6,7],分区为2。 结果将是[[1,2],[3,4],[5,6],[7]]
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());
}
}