无论如何将ArrayList拆分成不同的部分而不知道它的大小直到运行时？我知道有一种叫做的方法：
list.subList(a,b);
但我们需要明确提到盯着和结束范围的清单。我的问题是，我们得到一个包含帐号的arraylist，其中包含2000,4000个帐号的数据（编码时间内不会知道数字），我需要将此符号传递给PL / SQL的IN查询，如IN不支持超过1000个值，我试图分成多个块并将其发送到查询
注意：我不能使用像番石榴等任何外部库.. :(在这方面的任何指南表示赞赏。
这应该给你所有的部分：
int partitionSize = 1000;
List<List<Integer>> partitions = new LinkedList<List<Integer>>();
for (int i = 0; i < originalList.size(); i += partitionSize) {
partitions.add(originalList.subList(i,
Math.min(i + partitionSize, originalList.size())));
}
通用功能：
public static <T> ArrayList<T[]> chunks(ArrayList<T> bigList,int n){
ArrayList<T[]> chunks = new ArrayList<T[]>();
for (int i = 0; i < bigList.size(); i += n) {
T[] chunk = (T[])bigList.subList(i, Math.min(bigList.size(), i + n)).toArray();
chunks.add(chunk);
}
return chunks;
}
享受吧〜:)
Java 8（不是它有优势）：
List<String> list = new ArrayList<>();
Collections.addAll(list, "a","b","c","b","c","a","c","a","b");
分组大小：
final int G = 3;
final int NG = (list.size() + G - 1) / G;
旧式：
List<List<String>> result = new ArrayList(NG);
IntStream.range(0, list.size())
.forEach(i -> {
if (i % G == 0) {
result.add(i/G, new ArrayList<>());
}
result.get(i/G).add(list.get(i));
});
新风格：
List<List<String>> result = IntStream.range(0, NG)
.mapToObj(i -> list.subList(3 * i, Math.min(3 * i + 3, list.size())))
.collect(Collectors.toList());
感谢@StuartMarks为忘记了toList。
如果您受到PL / SQL in限制的约束，那么您想知道如何将列表拆分为大小<= n的块，其中n是限制。这是一个更简单的问题，因为它不需要事先知道列表的大小。
伪代码：
for (int n=0; n<list.size(); n+=limit)
{
chunkSize = min(list.size,n+limit);
chunk = list.sublist(n,chunkSize);
// do something with chunk
}
如果您已经或不介意添加Guava库，则无需重新发明轮子。
简单地说：final List<List<String>> splittedList = Lists.partition(bigList, 10);
其中bigList实现List接口，10是每个子列表的所需大小（最后可能更小）
listSize = oldlist.size();
chunksize =1000;
chunks = list.size()/chunksize;
ArrayList subLists;
ArrayList finalList;
int count = -1;
for(int i=0;i<chunks;i++){
subLists = new ArrayList();
int j=0;
while(j<chunksize && count<listSize){
subList.add(oldList.get(++count))
j++;
}
finalList.add(subLists)
}
您可以使用此决赛入围者，因为它包含oldList的块列表。
我也在做关键：索引值的值映射。
public static void partitionOfList(List<Object> l1, List<Object> l2, int partitionSize){
Map<String, List<Object>> mapListData = new LinkedHashMap<String, List<Object>>();
List<Object> partitions = new LinkedList<Object>();
for (int i = 0; i < l1.size(); i += partitionSize) {
partitions.add(l1.subList(i,Math.min(i + partitionSize, l1.size())));
l2=new ArrayList(partitions);
}
int l2size = l2.size();
System.out.println("Partitioned List: "+l2);
int j=1;
for(int k=0;k<l2size;k++){
l2=(List<Object>) partitions.get(k);
// System.out.println(l2.size());
if(l2.size()>=partitionSize && l2.size()!=1){
mapListData.put("val"+j+"-val"+(j+partitionSize-1), l2);
j=j+partitionSize;
}
else if(l2.size()<=partitionSize && l2.size()!=1){
// System.out.println("::::@@::"+ l2.size());
int s = l2.size();
mapListData.put("val"+j+"-val"+(j+s-1), l2);
//k++;
j=j+partitionSize;
}
else if(l2.size()==1){
// System.out.println("::::::"+ l2.size());
//int s = l2.size();
mapListData.put("val"+j, l2);
//k++;
j=j+partitionSize;
}
}
System.out.println("Map: " +mapListData);
}
public static void main(String[] args) {
List l1 = new LinkedList();
l1.add(1);
l1.add(2);
l1.add(7);
l1.add(4);
l1.add(0);
l1.add(77);
l1.add(34);
partitionOfList(l1,l2,2);
}
输出：
分区列表：[[1,2]，[7,4]，[0,77]，[34]]
地图：{val1-val2 = [1,2]，val3-val4 = [7,4]，val5-val6 = [0,77]，val7 = [34]}
以下代码：
private static List<List<Object>> createBatch(List<Object> originalList, int
batch_size) {
int Length = originalList.size();
int chunkSize = Length / batch_size;
int residual = Length-chunkSize*batch_size;
List<Integer> list_nums = new ArrayList<Integer>();
for (int i = 0; i < batch_size; i++) {
list_nums.add(chunkSize);
}
for (int i = 0; i < residual; i++) {
list_nums.set(i, list_nums.get(i) + 1);
}
List<Integer> list_index = new ArrayList<Integer>();
int cumulative = 0;
for (int i = 0; i < batch_size; i++) {
list_index.add(cumulative);
cumulative += list_nums.get(i);
}
list_index.add(cumulative);
List<List<Object>> listOfChunks = new ArrayList<List<Object>>();
for (int i = 0; i < batch_size; i++) {
listOfChunks.add(originalList.subList(list_index.get(i),
list_index.get(i + 1)));
}
return listOfChunks;
}
产生以下输出：
//[0,..,99] equally partition into 6 batch
// result:batch_size=[17,17,17,17,16,16]
//Continually partition into 6 batch, and residual also equally
//partition into top n batch
// Output:
[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16]
[17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33]
[34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50]
[51,52,53,54,55,56,57,58,59,60,61,62,63,64,65,66,67]
[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]
通用方法为您提供帮助：
private static List<List<Object>> createBatch(List<Object> originalList,
int chunkSize) {
List<List<Object>> listOfChunks = new ArrayList<List<Object>>();
for (int i = 0; i < originalList.size() / chunkSize; i++) {
listOfChunks.add(originalList.subList(i * chunkSize, i * chunkSize
+ chunkSize));
}
if (originalList.size() % chunkSize != 0) {
listOfChunks.add(originalList.subList(originalList.size()
- originalList.size() % chunkSize, originalList.size()));
}
return listOfChunks;