How to Convert a Map<String, List<String>> to Map<String, String> in java 8 functional APIs

Question

I have a map like bellow,

 [key = "car", value = ["bmw", "toyota"]]
 [key = "bike", value = ["honda", "kawasaki"]]

I want to convert it to another map using java 8 functional apis like bellow,

 [key = "bmw", value = "car"]
 [key = "toyota", value = "car"]
 [key = "honda", value = "bike"]
 [key = "kawasaki", value = "bike"]

Doing it using iteration is not a problem, want to do it with functional apis — faisalbegins
– faisalbegins, Commented Nov 10, 2017 at 2:01
@shmosel lets assume there will be no duplicate values. I just wanted to check how complicated it becomes with functional apis — faisalbegins
– faisalbegins, Commented Nov 10, 2017 at 2:32
@nullpointer: you don’t need to store the key list into a local variable: for(String k: map.keySet()) for(String key: map.get(k)) result.put(key, k);, but in either case, it’s doing an unnecessary lookup on the map it’s iterating over, so for(Map.Entry<String,List<String>> e: map.entrySet()) for(String key: e.getValue()) result.put(key, e.getKey()); is preferable… — Holger
– Holger, Commented Nov 10, 2017 at 9:13

teppic · Accepted Answer · 2017-11-10 03:05:42Z

8

Flatten the map values to entries then collect them:

Map<String, String> m2 = map
    .entrySet()
    .stream()
    .flatMap(e -> e.getValue().stream().map(v -> new AbstractMap.SimpleEntry<>(v, e.getKey())))
    .collect(Collectors.toMap(Map.Entry::getKey, Map.Entry::getValue));

This can be shortened by importing AbstractMap.SimpleEntry and Map.Entry.

edited Nov 10, 2017 at 3:05

answered Nov 10, 2017 at 2:03

teppic

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3 Comments

shmosel Over a year ago

You can shorten the final line by using Map.Entry (or just Entry) instead of AbstractMap.SimpleEntry.

Eugene Over a year ago

@teppic in 9 there is Map.entry method to use instead of a Pair or AbstractMap.SimpleEntry

Holger Over a year ago

@Eugene: but everyone should be aware that they are not identical, i.e. Map.entry does not allow null, neither for key nor for value. If that is understood and not a problem for the actual use case, it might be the preferable option.

Holger · Accepted Answer · 2017-11-10 09:23:56Z

2

A solution that doesn’t need constructing temporary Map.Entry instances, is:

Map<String, String> result = source.entrySet().stream()
  .collect(HashMap::new, (m,e)->e.getValue().forEach(k->m.put(k,e.getKey())), Map::putAll);

You might notice the similarity to the non-stream solution

Map<String, String> result = new HashMap<>();
source.forEach((key, value) -> value.forEach(k -> result.put(k, key)));

or the pre-Java 8 solution

Map<String, String> result = new HashMap<>();
for(Map.Entry<String,List<String>> e: source.entrySet())
    for(String key: e.getValue()) result.put(key, e.getKey());

answered Nov 10, 2017 at 9:23

Holger

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