Using Streams filter Map based on a list of keys

Question

I have a particular problem and was wondering whether the Java 8 Streams API could solve it. I know that this can be done outside of using Streams API but I don't want to add all the boilerplate code associated with trying to achieve that, if it can be done using Streams. I have a map

Map<String, String> greetings = new HashMap<>();
greetings.put("abc", "Hello");
greetings.put("def", "Goodbye");
greetings.put("ghi", "Ciao");
greetings.put("xyz", "Bonsoir");

and a list of keys:

List<String> keys = Arrays.asList("def", "zxy");

and using the above with Streams API, is it possible to filter that down to:

Map<String, String> filteredGreetings = new HashMap<>();
filteredGreetings.put("def", "Goodbye");
filteredGreetings.put("xyz", "Bonsoir");

Hopefully this makes sense what I am trying to achieve.

So far I have got this to work only when specifying the exact key which to filter the map's keySet on, but then this would only return a single entry set. I am interested in a completely filtered down map and I am struggling to achieve that.

Is it just a coincidental, typo or a requirement, I get to notice the key "zxy" and the result still put("xyz", "Bonsoir")? — Naman
– Naman, Commented Jun 14, 2019 at 18:48

Naman · Accepted Answer · 2019-06-14 18:51:56Z

4

If the input and the expected output in the question is not a typo, you can also retain the keys of the input map as:

Map<String, String> futureGreetings = new HashMap<>(greetings);
futureGreetings.keySet().retainAll(keys);

answered Jun 14, 2019 at 18:51

Naman

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2 Comments

Naman Over a year ago

Of course, this doesn't use streams, but one can compare the complexity and readability for both the solutions before opting.

Samuel Philipp Over a year ago

That's the best and most simple solution. Much better than mine ;)

Samuel Philipp · Accepted Answer · 2019-06-14 17:54:53Z

3

Try this:

Map<String, String> result = keys.stream()
        .filter(greetings::containsKey)
        .collect(Collectors.toMap(Function.identity(), greetings::get));

Or the other way round:

Map<String, String> result = greetings.entrySet().stream()
        .filter(e -> keys.contains(e.getKey()))
        .collect(Collectors.toMap(Map.Entry::getKey, Map.Entry::getValue));

For the second approach I would recommend using a Set<String> for a larger list of keys because it has a O(1) time complexity for .contains() because it does not contain any duplicates:

Set<String> keys = new HashSet<>(Arrays.asList("def", "zxy"));

edited Jun 14, 2019 at 17:54

answered Jun 14, 2019 at 16:48

Samuel Philipp

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4 Comments

Holger Over a year ago

For a list of two elements, the time complexity really doesn’t matter. In fact, the overhead of hashing will be higher than the linear lookup.

Samuel Philipp Over a year ago

@Holger Yes that's true. Just want to add that information for general.

Richard C Over a year ago

@SamuelPhilipp spot-on - thanks a lot! I wasn't aware of "identity" either. I need to read up a bit more on this.

Samuel Philipp Over a year ago

@RichardC Function.identity() is a synonym for i -> i.

Hadi · Accepted Answer · 2019-06-14 16:53:20Z

0

Try this

filteredGreetings= keys.stream()
                  .collect(Collectors.toMap(Function.identity(),key->greetings.get(key)));

or even for absent key you can use getOrDefault() method.

... .collect(Collectors.toMap(Function.identity(),key->greetings.getOrDefault(key,"")));

edited Jun 14, 2019 at 16:53

answered Jun 14, 2019 at 16:48

Hadi

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1 Comment

Hadi Over a year ago

Because that I added second approach. Though I think he did not mean the keys were absent

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