Group object by field and nested object's fields with Java streams

What I would like to do is group elements of a List in order to create a map, based on specific fields. The desired output is the following: Map<String,Map<String,Map<String,MasterObject>>>. The order of the keys is date, type, color.

My code is as follows

public class MasterObject{
   private String date;
   private List<SubObject> subObject;
}

public class SubObject{
   private String type;
   private String color;
}

What I have tried and used is Collectors.groupingBy which works great if your fields are members of the same Object (ex. date), but haven't managed to make it work for containing objects (ex. subObject). My input is a List<MasterObject>. It could be done the hard way by using Map's put and get but maybe there is a much cleaner way to make it work with Java streams.

What I have tried thus far is the following:

    Map<String, Map<List<String>, List<MasterObject>>> collect = faList.stream().collect(
    Collectors.groupingBy(f -> f.getDate(),
    Collectors.groupingBy(f -> f.getSubObject().stream().map(z -> z.getType()).collect(Collectors.toList()))));

In my sample above, I haven't managed to achieve to group elements by type, instead my key is a List<String>.Also my list should have a group by color as well.

I would use a custom object to keep flattened values:

public class CustomObject {
private String date;
private String type;
private String color;
private MasterObject masterObj;
// constructor, getters, setters
}

Try this:

SubObject sub1 = new SubObject("typeA", "red");
SubObject sub2 = new SubObject("typeA", "blue");
SubObject sub3 = new SubObject("typeB", "green");
SubObject sub4 = new SubObject("typeB", "green");
SubObject sub5 = new SubObject("typeC", "red");
SubObject sub6 = new SubObject("typeC", "blue");
List<MasterObject> masterObjList = new ArrayList<>();
masterObjList.add(new MasterObject("01/01/2020", Arrays.asList(sub1, sub2, sub3)));
masterObjList.add(new MasterObject("02/01/2020", Arrays.asList(sub4, sub5, sub6)));
Map<String, Map<String, Map<String, MasterObject>>> result = masterObjList.stream().flatMap(
o -> o.getSubObject().stream().map(s -> new CustomObject(o.getDate(), s.getType(), s.getColor(), o)))
.collect(Collectors.groupingBy(CustomObject::getDate, Collectors.groupingBy(CustomObject::getType,
Collectors.toMap(CustomObject::getColor, CustomObject::getMasterObj))));
result.entrySet().forEach(e -> {
System.out.println(e.getKey());
e.getValue().entrySet().forEach(e1 -> {
System.out.println("\t" + e1.getKey());
e1.getValue().entrySet().forEach(e2 -> {
System.out.println("\t\t" + e2.getKey());
});
});
});

Output:

01/01/2020
typeB
green
typeA
red
blue
02/01/2020
typeC
red
blue
typeB
green

The solution by @Hülya is great, and scale better on complex data structure.

An alternative and more concise way to do it, potentially faster (if perf is important) due to avoiding the intermediate object, is to just look on the element and add in the map on the go:

public Map<String, Map<String, Map<String, MasterObject>>> group (List<MasterObject> objects) {
Map<String, Map<String, Map<String, MasterObject>>> byDate = new HashMap<>();
for (MasterObject m : objects) {
Map<String, Map<String, MasterObject>> byType = byDate.computeIfAbsent(m.date, k -> new HashMap<>();); 
for (SubObject sub : m.subObject) {          
Map<String, Map<String, MasterObject>> byColor = byType.computeIfAbsent(sub.type, k -> new HashMap<>(););
byColor.put(sub.color, m);
}
}
return byDate;
}

After all a groupBy is just that, iterating over all elements and put them in the right bucket. And sometime the iterative approach is shorter/smaller and as readable as the functional one.

Just for fun, in C++, lambda are annoying to use (it improved in recent versionsà and the imperative approach is even more concise than in java:

map<string, map<string, map<string, MasterObject>>> group (const vector<MasterObject> &objects) {
map<string, map<string, map<string, MasterObject>>> result;
for (auto&& m : objects) {
for (auto &&sub : m.subObject) {
result[m.date][sub.type][sub.color] = m;
}
}
return result;
}