How to remove Keys that would cause Collisions before executing Collectors.toMap()

Question

I have a stream of objects similar to this previous question, however, instead of ignoring duplicate values, I would like to remove any values from that stream beforehand and print them out.

For example, from this snippet:

Map<String, String> phoneBook = people.stream()
                                      .collect(toMap(Person::getName,
                                                     Person::getAddress));

If there were duplicate entries, it would cause a java.lang.IllegalStateException: Duplicate key error to be thrown.

The solution proposed in that question used a mergeFunction to keep the first entry if a collision was found.

Map<String, String> phoneBook = 
    people.stream()
          .collect(Collectors.toMap(
             Person::getName,
             Person::getAddress,
             (address1, address2) -> {
                 System.out.println("duplicate key found!");
                 return address1;
             }
          ));

Instead of keeping the first entry, if there is a collision from a duplicate key in the stream, I want to know which value caused the collision and make sure that there are no occurrences of that value within the resulting map.

I.e. if "Bob" appeared three times in the stream, it should not be in the map even once.

In the process of creating that map, I would like to filter out any duplicate names and record them some way.

I want to make sure that when creating the list there can be no duplicate entry and for there to be some way to know which entries had duplicate keys in incoming stream. I was thinking about using groupingBy and filter beforehand to find the duplicate keys, but I am not sure what the best way to do it is.

Answer 1

You can't know which keys are duplicates until you have processed the entire input stream. Therefore, any pre-processing step has to make a complete pass of the input before your main logic, which is wasteful.

An alternate approach could be:

1. Use the merge function to insert a dummy value for the offending key
2. At the same time, insert the offending key into a Set<K>
3. After the input stream is processed, iterate over the Set<K> to remove offending keys from the primary map.

Answer 2

I would like to remove any values from that stream beforehand.

As @JimGarrison has pointed out, preprocessing the data doesn't make sense.

You can't know it in advance whether a name is unique or not until the all data set has been processed.

Another thing that you have to consider that inside the stream pipeline (before the collector) you have knowledge on what data has been encountered previously. Because results of intermediate operations should not depend on any state.

In case if you are thinking that streams are acting like a sequence of loops and therefore assuming that it's possible to preprocess stream elements before collecting them, that's not correct. Elements of the stream pipeline are being processed lazily one at a time. I.e. all the operations in the pipeline will get applied on a single element and each operation will be applied only if it's needed (that's what laziness means).

For more information, have a look at this tutorial and API documentation

Implementations

You can segregate unique values and duplicates in a single stream statement by utilizing Collectors.teeing() and a custom object that will contain separate collections of duplicated and unique entries of the phone book.

Since the primarily function of this object only to carry the data I've implemented it as Java 16 record.

public record FilteredPhoneBook(Map<String, String> uniquePersonsAddressByName,
                                List<String> duplicatedNames) {}

Collector teeing() expects three arguments: two collectors and a function that merges the results produced by both collectors.

The map generated by the groupingBy() in conjunction with counting(), is meant to determine duplicated names.

Since there's no point to processing the data, toMap() which is used as the second collector will create a map containing all names.

When both collectors will hand out their results to the merger function, it will take care of removing the duplicates.

public static FilteredPhoneBook getFilteredPhoneBook(Collection<Person> people) {
    return people.stream()
        .collect(Collectors.teeing(
            Collectors.groupingBy(Person::getName, Collectors.counting()), // intermediate Map<String, Long>
            Collectors.toMap(                                              // intermediate Map<String, String>
                Person::getName,
                Person::getAddress,
                (left, right) -> left),
            (Map<String, Long> countByName, Map<String, String> addressByName) -> {
                countByName.values().removeIf(count -> count == 1);        // removing unique names
                addressByName.keySet().removeAll(countByName.keySet());    // removing all duplicates
                
                return new FilteredPhoneBook(addressByName, new ArrayList<>(countByName.keySet()));
            }
        ));
}

Another way to address this problem to utilize Map<String,Boolean> as the mean of discovering duplicates, as @Holger have suggested.

With the first collector will be written using toMap(). And it will associate true with a key that has been encountered only once, and its mergeFunction will assign the value of false if at least one duplicate was found.

The rest logic remains the same.

public static FilteredPhoneBook getFilteredPhoneBook(Collection<Person> people) {
    return people.stream()
        .collect(Collectors.teeing(
            Collectors.toMap(            // intermediate Map<String, Boolean>
                Person::getName,
                person -> true,          // not proved to be a duplicate and initially considered unique
                (left, right) -> false), // is a duplicate
            Collectors.toMap(            // intermediate Map<String, String>
                Person::getName,
                Person::getAddress,
                (left, right) -> left),
            (Map<String, Boolean> isUniqueByName, Map<String, String> addressByName) -> {
                isUniqueByName.values().removeIf(Boolean::booleanValue);   // removing unique names
                addressByName.keySet().removeAll(isUniqueByName.keySet()); // removing all duplicates

                return new FilteredPhoneBook(addressByName, new ArrayList<>(isUniqueByName.keySet()));
            }
        ));
}

main() - demo

public static void main(String[] args) {
    List<Person> people = List.of(
        new Person("Alise", "address1"),
        new Person("Bob", "address2"),
        new Person("Bob", "address3"),
        new Person("Carol", "address4"),
        new Person("Bob", "address5")
    );

   FilteredPhoneBook filteredPhoneBook = getFilteredPhoneBook(people);
        
    System.out.println("Unique entries:");
    filteredPhoneBook.uniquePersonsAddressByName.forEach((k, v) -> System.out.println(k + " : " + v));
    System.out.println("\nDuplicates:");
    filteredPhoneBook.duplicatedNames().forEach(System.out::println);
}

Output

Unique entries:
Alise : address1
Carol : address4

Duplicates:
Bob

Answer 3

In mathematical terms you want to partition your grouped aggregate and handle both parts separately.

Map<String, String> makePhoneBook(Collection<Person> people) {
    Map<Boolean, List<Person>> phoneBook = people.stream()
            .collect(Collectors.groupingBy(Person::getName))
            .values()
            .stream()
            .collect(Collectors.partitioningBy(list -> list.size() > 1,
            Collectors.mapping(r -> r.get(0),
                    Collectors.toList())));

    // handle duplicates
    phoneBook.get(true)
            .forEach(x -> System.out.println("duplicate found " + x));

    return phoneBook.get(false).stream()
            .collect(Collectors.toMap(
                    Person::getName,
                    Person::getAddress));
}