Java stream merge or reduce duplicate objects

问题

I need to generate a unique friend list from a list that can have duplicates by merging all duplicate entries into one object
Example - Friends are fetched from different social feeds and put into 1 big list
1. Friend - [name: "Johnny Depp", dob: "1970-11-10", source: "FB", fbAttribute: ".."]
2. Friend - [name: "Christian Bale", dob: "1970-01-01", source: "LI", liAttribute: ".."]
3. Friend - [name: "Johnny Depp", dob: "1970-11-10", source: "Twitter", twitterAttribute: ".."]
4. Friend - [name: "Johnny Depp", dob: "1970-11-10", source: "LinkedIn", liAttribute: ".."]
5. Friend - [name: "Christian Bale", dob: "1970-01-01", source: "LI", liAttribute: ".."]

Expected output
1. Friend - [name: "Christian Bale", dob: "1970-01-01", liAttribute: "..", fbAttribute: "..", twitterAttribute: ".."]
2. Friend - [name: "Johnny Depp", dob: "1970-11-10", liAttribute: "..", fbAttribute: "..", twitterAttribute: ".."]

Question - How can i merge without using any intermediate container? I can easily use an intermediate map and apply reduce on each value of the entry.

List<Friend> friends;
Map<String, List<Friend>> uniqueFriendMap
    = friends.stream().groupingBy(Friend::uniqueFunction);
List<Friend> mergedFriends = uniqueFriendMap.entrySet()
    .stream()
    .map(entry -> {
           return entry.getValue()
                .stream()
                .reduce((a,b) -> friendMergeFunction(a,b));
    })
    .filter(mergedPlace -> mergedPlace.isPresent())
    .collect(Collectors.toList());

I like to do this without using the intermediate Map uniqueFriendMap. Any suggestions?

回答1:

The groupingBy operation (or something similar) is unavoidable, the Map created by the operation is also used during the operation for looking up the grouping keys and finding the duplicates. But you can combine it with the reduction of the group elements:

Map<String, Friend> uniqueFriendMap = friends.stream()
    .collect(Collectors.groupingBy(Friend::uniqueFunction,
        Collectors.collectingAndThen(
            Collectors.reducing((a,b) -> friendMergeFunction(a,b)), Optional::get)));

The values of the map are already the resulting distinct friends. If you really need a List, you can create it with a plain Collection operation:

List<Friend> mergedFriends = new ArrayList<>(uniqueFriendMap.values());

If this second operation still annoys you, you can hide it within the collect operation:

List<Friend> mergedFriends = friends.stream()
    .collect(Collectors.collectingAndThen(
        Collectors.groupingBy(Friend::uniqueFunction, Collectors.collectingAndThen(
            Collectors.reducing((a,b) -> friendMergeFunction(a,b)), Optional::get)),
        m -> new ArrayList<>(m.values())));

Since the nested collector represents a Reduction (see also this answer), we can use toMap instead:

List<Friend> mergedFriends = friends.stream()
    .collect(Collectors.collectingAndThen(
        Collectors.toMap(Friend::uniqueFunction, Function.identity(),
            (a,b) -> friendMergeFunction(a,b)),
        m -> new ArrayList<>(m.values())));

Depending on whether friendMergeFunction is a static method or instance method, you may replace (a,b) -> friendMergeFunction(a,b) with DeclaringClass::friendMergeFunction or this::friendMergeFunction.

But note that even within your original approach, several simplifications are possible. When you only process the values of a Map, you don’t need to use the entrySet(), which requires you to call getValue() on each entry. You can process the values() in the first place. Then, you don’t need the verbose input -> { return expression; } syntax, as input -> expression is sufficient. Since the groups of the preceding grouping operation can not be empty, the filter step is obsolete. So your original approach would look like:

Map<String, List<Friend>> uniqueFriendMap
    = friends.stream().collect(Collectors.groupingBy(Friend::uniqueFunction));
List<Friend> mergedFriends = uniqueFriendMap.values().stream()
    .map(group -> group.stream().reduce((a,b) -> friendMergeFunction(a,b)).get())
    .collect(Collectors.toList());

which is not so bad. As said, the fused operation doesn’t skip the Map creation as that’s unavoidable. It only skips the creations of the Lists representing each group, as it will reduce them to a single Friend in-place.

回答2:

Just use the method Collectors.frequency.

来源：https://stackoverflow.com/questions/41539663/java-stream-merge-or-reduce-duplicate-objects

标签

java

list

merge

java-stream

reduce