This is a very basic question, I'm just not that good with Java. I have a Map and I want to get a list or something of the keys in sorted order so I can iterate over them.
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4 Answers
89
Use a TreeMap, which is an implementation of the SortedMap interface. It presents its keys in sorted order.
Map<String, Object> map = new TreeMap<String, Object>();
/* Add entries to the map in any order. */
...
/* Now, iterate over the map's contents, sorted by key. */
for (Map.Entry<String, ?> entry : map.entrySet()) {
System.out.println(entry.getKey() + ": " + entry.getValue());
}
If you are working with another Map implementation that isn't sorted as you like, you can pass it to the constructor of TreeMap to create a new map with sorted keys.
void process(Map<String, Object> original) {
Map<String, Object> copy = new TreeMap<String, Object>(original);
/* Now use "copy", which will have keys in sorted order. */
...
}
A TreeMap works with any type of key that implements the Comparable interface, putting them in their "natural" order. For keys that aren't Comparable, or whose natural ordering isn't what you need, you can implement your own Comparator and specify that in the constructor.
erickson
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The code I'm consuming gives me a Map object, so how do I then convert it into a TreeMap or use a TreeMap to do the sorting? – Bialecki Feb 20 '09 at 22:03
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You can create the TreeMap using a constructor whose parameter is any Map. Also, congratulations erickson (I assume, since you're only 5 rep away from 10k). – Dan Lew Feb 20 '09 at 22:05
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1You would better off sorting keys (place in a TreeSet) then calling get on the original Map. Creating a new TreeMap results in everything getting rehashed. – Steve Kuo Feb 21 '09 at 01:03
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1Steve, that's incorrect. No hashing occurs with a TreeMap; it's a red-black tree. Anyway, a TreeSet delegates to an internal TreeMap, so creating a TreeSet, iterating over its keys, and using them to perform an O(log(n)) lookup in another tree would be confusing and inefficient. – erickson Feb 22 '09 at 04:06
39
You have several options. Listed in order of preference:
- Use a
SortedMap:
SortedMap<whatever> myNewMap = new TreeMap<whatever>(myOldMap);
This is vastly preferable if you want to iterate more than once. It keeps the keys sorted so you don't have to sort them before iterating. - There is no #2.
- There is no #3, either.
SortedSet<whatever> keys = new TreeSet<whatever>(myMap.keySet());List<whatever> keys = new ArrayList<whatever>(myMap.keySet()); Collections.sort(keys);
The last two will get you what you want, but should only be used if you only want to iterate once and then forget the whole thing.
Michael Myers
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In step #4, you could also create a TreeSet (a sorted set) instead of a list, which saves you from making an explicit call to sort(). – David Z Feb 20 '09 at 22:08
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@David: I did think of that, but for some reason I forgot that it was possible to iterate over a Set. It does require sorting every time, though. – Michael Myers Feb 20 '09 at 22:10
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hmm i like the #4 . why does it require more sorting than #1 ? i would have thought it was equally good – Johannes Schaub - litb Feb 20 '09 at 22:14
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@litb: If you're dumping the map into a TreeMap every time you want to iterate over it, then you're right, they are the same. – Michael Myers Feb 20 '09 at 22:17
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#4 isn't as good as #1 if you need the values from the map too. Iterating over entries saves the extra lookup for the value. – erickson Feb 20 '09 at 22:28
9
You can create a sorted collection when iterating but it make more sense to have a sorted map in the first place. (As has already been suggested)
All the same, here is how you do it.
Map<String, Object> map;
for(String key: new TreeSet<String>(map.keySet()) {
// accessed in sorted order.
}
Peter Lawrey
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7
Apart from the methods mentioned in other answers, with Java 8 streams, another shorthand to get a sorted key list from a map would be -
List<T> sortedKeys = myMap.keySet().stream().sorted().collect(Collectors.toList());
One could actually get stuff done after .sorted() as well (like using a .map(...) or a .forEach(...)), instead of collecting it in the list and then iterating over the list.
