Is there a convenient way for such conversion other than a for loop such as
List<Map.Entry<String, Integer>> entryList = new List<>(//initialization);
List<String>> keyList = new List<>(entryList.size());
for (Map.Entry<String, Integer> e : entryList) {
keyList.add(e.getKey());
}
I would like the order to be preserved.
Use java 8 streams to convert this:
List<Map.Entry<String, ?>> entryList = new List<>(//initialization);
List<String> stringList = entryList.stream().map(Entry::getKey).collect(Collectors.toList());
This makes a stream of the entries, then uses the map method to convert them to strings, then collects it to a list using Collectors.toList().
Alternatively, this method can be changed in a helper function if you need it more times like this:
public static <K> List<K> getKeys(List<Map.Entry<K,?>> entryList) {
return entryList.stream().map(Entry::getKey).collect(Collectors.toList());
}
public static <V> List<V> getValues(List<Map.Entry<?,V>> entryList) {
return entryList.stream().map(Entry::getValue).collect(Collectors.toList());
}
While the above code works, you can also get a List<K> from a map by doing new ArrayList<>(map.keySet()), with this having the advantage than you don't need to convert the entryset to a list, before converted to a stream, and then back a list again.
return entryList.stream().map(e->e.getKey() + "=" + e.getValue()).collect(Collectors.toList());If you do not really need to make a copy of the list you can implement a wrapper arround the list like this, with the adicional bonus that changes made to the entryList are automatically reflected in the stringList. Bear in mind that this simple wrapper is read only.
List<Map.Entry<String, ?>> entryList = new List<>(//initialization);
List<String> stringList = new AbstractList<String>() {
List<Map.Entry<String, Integer>> internal = entryList;
public String get(int index) {
return internal.get(index).getKey();
}
public int size() {
return internal.size();
}
};
entryList.stream().map(Map.Entry::getKey).collect(Collectors.toList())Nicely made question, by the way.