Flatten a Map> to Map with stream

Question

I would like to flatten a Map which associates an Integer key to a list of String, without losing the key mapping. I am curious as though it is possible and useful to do so with stream and lambda.

We start with something like this:

Map<Integer, List<String>> mapFrom = new HashMap<>();

Let's assume that mapFrom is populated somewhere, and looks like:

1: a,b,c
2: d,e,f
etc.

Let's also assume that the values in the lists are unique.

Now, I want to "unfold" it to get a second map like:

a: 1
b: 1
c: 1
d: 2
e: 2
f: 2
etc.

I could do it like this (or very similarly, using foreach):

Map<String, Integer> mapTo = new HashMap<>();
for (Map.Entry<Integer, List<String>> entry: mapFrom.entrySet()) {
    for (String s: entry.getValue()) {
        mapTo.put(s, entry.getKey());
    }
}

Now let's assume that I want to use lambda instead of nested for loops. I would probably do something like this:

Map<String, Integer> mapTo = mapFrom.entrySet().stream().map(e -> {
    e.getValue().stream().?
    // Here I can iterate on each List, 
    // but my best try would only give me a flat map for each key, 
    // that I wouldn't know how to flatten.
}).collect(Collectors.toMap(/*A String value*/,/*An Integer key*/))

I also gave a try to flatMap, but I don't think that it is the right way to go, because although it helps me get rid of the dimensionality issue, I lose the key in the process.

In a nutshell, my two questions are :

• Is it possible to use streams and lambda to achieve this?
• Is is useful (performance, readability) to do so?

Answer 1

You should use flatMap as follows:

entrySet.stream()
        .flatMap(e -> e.getValue().stream()
                       .map(s -> new SimpleImmutableEntry(e.getKey(), s)));

SimpleImmutableEntry is a nested class in AbstractMap.

Answer 2

This should work. Please notice that you lost some keys from List.

Map<Integer, List<String>> mapFrom = new HashMap<>();
Map<String, Integer> mapTo = mapFrom.entrySet().stream()
        .flatMap(integerListEntry -> integerListEntry.getValue()
                .stream()
                .map(listItem -> new AbstractMap.SimpleEntry<>(listItem, integerListEntry.getKey())))
        .collect(Collectors.toMap(AbstractMap.SimpleEntry::getKey, AbstractMap.SimpleEntry::getValue));

Answer 3

You need to use flatMap to flatten the values into a new stream, but since you still need the original keys for collecting into a Map, you have to map to a temporary object holding key and value, e.g.

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

The Map.Entry is a stand-in for the nonexistent tuple type, any other type capable of holding two objects of different type is sufficient.

An alternative not requiring these temporary objects, is a custom collector:

Map<String, Integer> mapTo = mapFrom.entrySet().stream().collect(
    HashMap::new, (m,e)->e.getValue().forEach(v->m.put(v, e.getKey())), Map::putAll);

This differs from toMap in overwriting duplicate keys silently, whereas toMap without a merger function will throw an exception, if there is a duplicate key. Basically, this custom collector is a parallel capable variant of

Map<String, Integer> mapTo = new HashMap<>();
mapFrom.forEach((k, l) -> l.forEach(v -> mapTo.put(v, k)));

But note that this task wouldn’t benefit from parallel processing, even with a very large input map. Only if there were additional computational intense task within the stream pipeline that could benefit from SMP, there was a chance of getting a benefit from parallel streams. So perhaps, the concise, sequential Collection API solution is preferable.

Answer 4

Hope this would do it in simplest way. :))

mapFrom.forEach((key, values) -> values.forEach(value -> mapTo.put(value, key)));