英文:
how to sort map based on key length
问题
我有这个地图：Map<String, String> unsortMap = new HashMap<String, String>();unsortMap.put("./cq:dialog/content/items/tabs/items/tab1/items/columns/items", "40");unsortMap.put("./cq:dialog/content", "80");unsortMap.put("./cq:dialog", "75");unsortMap.put("./cq:dialog/content/items/tabs/items/tab2/items/columns/items", "40");unsortMap.put("./cq:dialog/content/sling:resourcetype", "granite/ui/components/coral/foundation/container");我想根据键的长度对其进行排序，例如，如果键的“/”数量较少，则它将位于顶部。期望的输出：./cq:dialog./cq:dialog/content/./cq:dialog/content/sling:resourcetype./cq:dialog/content/items/tabs/items/tab1/items/columns/items./cq:dialog/content/items/tabs/items/tab2/items/columns/items为此，我写了以下代码：Map<String, String> sortedMap = unsortMap.entrySet().stream().sorted((entry1, entry2) -> {int slashes1 = entry1.getKey().length() - entry1.getKey().replace("/", "").length();int slashes2 = entry2.getKey().length() - entry2.getKey().replace("/", "").length();return Integer.compare(slashes1, slashes2);}).collect(Collectors.toMap(Map.Entry::getKey, Map.Entry::getValue,(oldValue, newValue) -> oldValue, LinkedHashMap::new));这样做是因为它会基于“/”的数量进行比较。我尝试了自定义比较器，但没有获得预期的结果。
英文:
I have this map:
Map < String, String > unsortMap = new HashMap < String, String > ();unsortMap.put("./cq:dialog/content/items/tabs/items/tab1/items/columns/items", "40");unsortMap.put("./cq:dialog/content", "80");unsortMap.put("./cq:dialog", "75");unsortMap.put("./cq:dialog/content/items/tabs/items/tab2/items/columns/items", "40");unsortMap.put("./cq:dialog/content/sling:resourcetype", "granite/ui/components/coral/foundation/container");
I would like to sort it as based on the length of the key, for example if the key has less number of "/" then it will be the on top.
Expected output:
./cq:dialog./cq:dialog/content/./cq:dialog/content/sling:resourcetype./cq:dialog/content/items/tabs/items/tab1/items/columns/items./cq:dialog/content/items/tabs/items/tab2/items/columns/items
So for this, I write like this:
Map<String, String> sortedMap = unsortMap.entrySet().stream().sorted(Map.Entry.comparingByKey()).collect(Collectors.toMap(Map.Entry::getKey, Map.Entry::getValue,(oldValue, newValue) -> oldValue, LinkedHashMap::new));
It does not give output as expected because it does not compare based on "/". I have tried custom comparator but could not get the expected result.
答案1
得分: 3
仅比较长度时，我们可以在自定义的比较器中使用 Integer.compare。<sup>示例</sup>
Map<String, String> sortedMap = unsortMap.entrySet().stream().sorted((a, b) -> Integer.compare(a.getKey().length(), b.getKey().length())).collect(Collectors.toMap(Map.Entry::getKey,Map.Entry::getValue, (oldValue, newValue) -> oldValue, LinkedHashMap::new));
当然，你也可以使用这个自定义比较器创建一个 TreeMap，这样新的键仍然会被正确排序。
final NavigableMap<String, String> sortedMap = new TreeMap<>((a, b) -> a.length() != b.length() ? Integer.compare(a.length(), b.length()) : a.compareTo(b));sortedMap.putAll(unsortMap);
<hr>
为了比较斜杠的数量，我们可以在两个字符串的字符上进行过滤。
final NavigableMap<String, String> sortedMap = new TreeMap<>((a, b) -> {final long slashes1 = a.chars().filter(c -> c == '/').count();final long slashes2 = b.chars().filter(c -> c == '/').count();return slashes1 != slashes2 ? Long.compare(slashes1, slashes2) : a.compareTo(b);});sortedMap.putAll(unsortMap);
正如Andreas建议的那样，可以使用.thenComparing简化比较器：
final NavigableMap<String, String> sortedMap = new TreeMap<>(Comparator.comparingLong((String s) -> s.chars().filter(c -> c == '/').count()).thenComparing(Comparator.naturalOrder()));
或者：
Map<String, String> sortedMap = unsortMap.entrySet().stream().sorted((a, b) -> Long.compare(a.getKey().chars().filter(c -> c == '/').count(), b.getKey().chars().filter(c -> c == '/').count())).collect(Collectors.toMap(Map.Entry::getKey, Map.Entry::getValue, (oldValue, newValue) -> oldValue,LinkedHashMap::new));
英文:
To only compare the lengths, we can use Integer.compare in a custom Comparator. <sup>Demo</sup>
Map<String, String> sortedMap = unsortMap.entrySet().stream().sorted((a, b) -> Integer.compare(a.getKey().length(), b.getKey().length())).collect(Collectors.toMap(Map.Entry::getKey,Map.Entry::getValue, (oldValue, newValue) -> oldValue, LinkedHashMap::new));
Of course, you could also create a TreeMap with this custom Comparator so that new keys will still be ordered correctly.
final NavigableMap<String, String> sortedMap = new TreeMap<>((a, b) -> a.length() != b.length() ? Integer.compare(a.length(), b.length()) : a.compareTo(b));sortedMap.putAll(unsortMap);
<hr>
In order to compare the number of slashes, we can filter over the characters of both Strings.
final NavigableMap<String, String> sortedMap = new TreeMap<>((a, b) -> {final long slashes1 = a.chars().filter(c -> c == '/').count();final long slashes2 = b.chars().filter(c -> c == '/').count();return slashes1 != slashes2 ? Long.compare(slashes1, slashes2) : a.compareTo(b);});sortedMap.putAll(unsortMap);
As Andreas suggested, the comparator can be simplified using .thenComparing:
final NavigableMap<String, String> sortedMap = new TreeMap<>(Comparator.comparingLong((String s) -> s.chars().filter(c -> c == '/').count()).thenComparing(Comparator.naturalOrder()));
Alternatively:
Map<String, String> sortedMap = unsortMap.entrySet().stream().sorted((a, b) -> Long.compare(a.getKey().chars().filter(c->c=='/').count(), b.getKey().chars().filter(c->c=='/').count())).collect(Collectors.toMap(Map.Entry::getKey, Map.Entry::getValue, (oldValue, newValue) -> oldValue,LinkedHashMap::new));
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