问题描述
我在一个类 (GlobalDataHolder.cardNameAndDescription) 中有一个 ArrayList()，我想按字母顺序排序，但我还希望在对前面提到的 arrayList 进行排序时对 ArrayList() (UserBoxGlbImageAdapter.mGLBIcons) 进行排序.
I have an ArrayList<String>() in a class (GlobalDataHolder.cardNameAndDescription) and i want to sort it alphabetically but i also want to sort an ArrayList<Integer>() (UserBoxGlbImageAdapter.mGLBIcons) while sorting the previously mentioned arrayList.
旁注:两个数组的大小均为 3(0-2).
Sidenote: Both arrays have a size of 3(0-2).
我正在编写自己的自定义 compare() 方法来执行此操作，但我没有实现我想要的.当我单击运行排序代码的按钮时，正确的顺序没有实现除非我点击按钮 3 次，尽管 String ArrayList 确实得到 按字母顺序 排序.所以我想我只需要按照数组的大小对数组进行排序(所以是 3 次).
I am writing my own custom compare() method to do this but i am not achieving what i'm looking for.When i click on the button that runs the sorting code, the correct order doesn't get achieved unless i click the button 3 times although the String ArrayList does get Alphabetically sorted. So i figured that i just need to sort the arrays as many times as the arrays's size is(so 3 times).
总而言之，String 和 Integer 数据应该按相同的顺序排列，因为它们依赖于其他数据，但我无法让它们同时适用于两个数组.
To sum up, the String and Integer data should be in the same order since they depend on it's other but i can't get that to work for both arrays.
这些都没有用.有人能告诉我我在这里用第二个数组的排序做错了什么吗?这是我的代码:
None of that worked. Can someone tell me what i'm doing wrong here with the 2nd array's sorting? Here's my code:
public class SortingDialog extends DialogFragment {
@NonNull
@Override
public Dialog onCreateDialog(Bundle savedInstanceState) {
// Create a builder to make the dialog building process easier
AlertDialog.Builder builder = new AlertDialog.Builder(getActivity());
builder.setTitle("Sorting Dialog");
builder.setSingleChoiceItems(R.array.sorting_options, 0,
new DialogInterface.OnClickListener() {
@Override
public void onClick(DialogInterface dialogInterface, int i) {
if (i == 1) {
Toast.makeText(getActivity(), "2nd option Clicked", Toast.LENGTH_SHORT).show();
if (getActivity().getSupportFragmentManager().findFragmentByTag("GLOBAL_FRAGMENT") != null) {
sortGlobalListsBasedOnNameAndDesc();
}
}
for (int j = 0; j < GlobalDataHolder.cardNameAndDescription.size(); j++) {
Log.v("card_names", GlobalDataHolder.cardNameAndDescription.get(j));
}
}
});
builder.setPositiveButton("OK", new DialogInterface.OnClickListener() {
@Override
public void onClick(DialogInterface dialogInterface, int i) {
createToast();
dismiss();
}
});
builder.setNegativeButton("Cancel", new DialogInterface.OnClickListener() {
@Override
public void onClick(DialogInterface dialogInterface, int i) {
dismiss();
}
});
return builder.create();
}
private void sortGlobalListsBasedOnNameAndDesc() {
if (Build.VERSION.SDK_INT >= Build.VERSION_CODES.N) {
GlobalDataHolder.cardNameAndDescription.sort(new Comparator<String>() {
@Override
public int compare(String s1, String s2) {
int id1 = GlobalDataHolder.cardNameAndDescription.indexOf(s1);
int id2 = GlobalDataHolder.cardNameAndDescription.indexOf(s2);
if (s1.equals(s2)) {
return 0;
} else if (s1.compareToIgnoreCase(s2) > 0) { //s1 is greater
//Collections.swap(UserBoxGlbImageAdapter.mGLBIcons,id2,id1);
swap(UserBoxGlbImageAdapter.mGLBIcons,id2,id1);
swap(GlobalDataHolder.cardNameAndDescription,id2,id1);
Log.d("case1","Called 1 time");
return 1;
} else if (s1.compareToIgnoreCase(s2) < 0) { //s1 is smaller
//Collections.swap(UserBoxGlbImageAdapter.mGLBIcons,id1,id2);
swap(UserBoxGlbImageAdapter.mGLBIcons,id1,id2);
swap(GlobalDataHolder.cardNameAndDescription,id1,id2);
Log.d("case2","Called 1 time");
return -1;
} else {
return 0;
}
}
});
}
}
private void swap(List list,int objIndex1, int objIndex2) {
for (int i=0;i < list.size(); i++) {
Collections.swap(list,objIndex1,objIndex2);
UserBoxGlbImageAdapter.refreshFragmentView(UserBoxGLBFragment.getUserBoxAdapter());
}
}
private void createToast() {
Toast.makeText(getActivity(), "Cards sorted based on AVG Stats", Toast.LENGTH_SHORT).show();
}
}
推荐答案
排序索引列表而不是列表本身并不困难.从中您可以轻松地对列表重新排序.
It is not difficult to sort a list of indexes instead of the list itself. From that you can easily reorder the list.
public class Test {
List<String> testStrings = Arrays.asList(new String[]{"One", "Two", "Three", "Four", "Five", "Six", "Seven", "Eight", "Nine", "Ten"});
List<Integer> testNumbers = Arrays.asList(new Integer[]{1, 2, 3, 4, 5, 6, 7, 8, 9, 10});
static <T extends Comparable<T>> List<Integer> getSortOrder(List<T> list) {
// Ints in increasing order from 0. One for each entry in the list.
List<Integer> order = IntStream.rangeClosed(0, list.size() - 1).boxed().collect(Collectors.toList());
Collections.sort(order, new Comparator<Integer>() {
@Override
public int compare(Integer o1, Integer o2) {
// Comparing the contents of the list at the position of the integer.
return list.get(o1).compareTo(list.get(o2));
}
});
return order;
}
static <T> List<T> reorder(List<T> list, List<Integer> order) {
return order.stream().map(i -> list.get(i)).collect(Collectors.toList());
}
public void test() {
System.out.println("The strings: " + testStrings);
List<Integer> sortOrder = getSortOrder(testStrings);
System.out.println("The order they would be if they were sorted: " + sortOrder + " i.e. " + reorder(testStrings, sortOrder));
List<Integer> reordered = reorder(testNumbers, sortOrder);
System.out.println("Numbers in alphabetical order of their names: " + reordered);
}
public static void main(String[] args) {
new Test().test();
System.out.println("Hello");
}
}
印刷品:
字符串:[一、二、三、四、五、六、七、八、九、十]
排序后的顺序:[7, 4, 3, 8, 0, 6, 5, 9, 2, 1] 即[八，五，四，九，一，七，六，十、三、二]
The order they would be if they were sorted: [7, 4, 3, 8, 0, 6, 5, 9, 2, 1] i.e. [Eight, Five, Four, Nine, One, Seven, Six, Ten, Three, Two]
按姓名字母顺序排列的数字:[8, 5, 4, 9, 1, 7, 6, 10, 3, 2]
Numbers in alphabetical order of their names: [8, 5, 4, 9, 1, 7, 6, 10, 3, 2]
如果您觉得需要，我让您自行添加自定义比较器.
I leave it up to you to add a custom comparator if you feel you need one.
这篇关于根据另一个 ArrayList 的排序对 ArrayList 进行排序的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！