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
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]]
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]]
List<List<Integer>> allChunkLists = new ArrayList<List<Integer>>();
List<Integer> chunkList = null;
int fromIndex = 0;
int toIndex = CHUNK_SIZE;
while (fromIndex < origList.size()) {
chunkList = origList.subList(fromIndex, (toIndex > origList.size() ? origList.size() : toIndex));
allChunkLists.add(chunkList);
fromIndex = toIndex;
toIndex += CHUNK_SIZE;
}
没有库,只有Java的subList()。toIndex需要适当地有界,以避免在subList()中出现越界错误。
Apache Commons Collections 4在ListUtils类中有一个分区方法。下面是它的工作原理:
import org.apache.commons.collections4.ListUtils;
...
int targetSize = 100;
List<Integer> largeList = ...
List<List<Integer>> output = ListUtils.partition(largeList, targetSize);
我猜你遇到的问题是命名100个数组列表并填充它们。您可以创建一个数组列表数组,并使用循环填充每个数组列表。
最简单(也是最愚蠢的)的方法是这样的:
ArrayList results = new ArrayList(1000);
// populate results here
for (int i = 0; i < 1000; i++) {
results.add(i);
}
ArrayList[] resultGroups = new ArrayList[100];
// initialize all your small ArrayList groups
for (int i = 0; i < 100; i++) {
resultGroups[i] = new ArrayList();
}
// put your results into those arrays
for (int i = 0; i < 1000; i++) {
resultGroups[i/10].add(results.get(i));
}