How to filter a map with Java stream api?

Question

Map<Integer, String> map = new HashMap<>();
  map.put(1, "f");
  map.put(2, "I");
  map.put(3, "a");
  map.put(4, "c");....etc

Now I have a list:

List<Integer> picks = {1,3}

I would like to get back a list of Strings, ie, values from map that matches the key values, found in the 'pick' list.So, I am expecting to get back {"f", "a"} as result. Is there a way to use java stream api to do it in elegant way?

When there is one value , I am doing it this way:

map.entrySet().stream()
   .filter(entry -> "a".equals(entry.getValue()))
   .map(entry -> entry.getValue())
   .collect(Collectors.toList())

But getting hard time when there is a list of keys/picks to filter with.

Akif Hadziabdic · Accepted Answer · 2021-04-18 20:36:19Z

8

You can use List.contains() in the Stream#filter to only accept those values which are present in the list:

List<String> result = map.entrySet()
   .stream()
   .filter(ent -> picks.contains(ent.getKey()))
   .map(Map.Entry::getValue)     
   .collect(Collectors.toList());

edited Apr 18, 2021 at 20:36

Akif Hadziabdic

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answered Aug 21, 2019 at 18:47

Amardeep Bhowmick

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Comments

Michał Krzywański · Accepted Answer · 2019-08-22 03:59:55Z

8

You can achieve this by using something like this :

List<String> values = map.entrySet()
                .stream()
                .filter(entry -> picks.contains(entry.getKey()))
                .map(Map.Entry::getValue)
                .collect(Collectors.toList());

values.forEach(System.out::println);

Output:

f
a

However it might not be efficient as List::contains is O(N). Consider using a Set (for example HashSet) instead of List for picks as HashSet::contains is O(1).

edited Aug 22, 2019 at 3:59

answered Aug 21, 2019 at 18:45

Michał Krzywański

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Comments

Ravindra Ranwala · Accepted Answer · 2019-08-24 13:57:29Z

5

You don't need to traverse the entire map just to collect picked values. Rather iterate over the required keys and grab the relevant values from the map. If the map is far more larger compared to the values you need to pick which is usually the case, then this approach should outperform the other. Moreover this solution is much more compact and has less visual clutter to me. Here's how it looks.

List<String> pickedValues = picks.stream().map(map::get)
    .filter(Objects::nonNull)
    .collect(Collectors.toList());

edited Aug 24, 2019 at 13:57

answered Aug 22, 2019 at 2:27

Ravindra Ranwala

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1 Comment

Michał Krzywański Over a year ago

This is a very good solution. However you should also filter out null values after map operation as Map::get() will return null for non existing keys in the map for values from picks.

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