Java Streams vs C# LINQ vs Java6

Jyri-Matti Lähteenmäki

2013-04-10

A while back I ran into an article comparing C# LINQ to the upcoming Java8 Streams API: http://blog.informatech.cr/2013/03/24/java-streams-preview-vs-net-linq/.

I'm not really experienced with C# but I have a feeling that the language as a whole is quite verbose and, well, bad? Except for LINQ which performs magic with monadic comprehension.

I've been coding in Java for the most of my career, so I know a thing or two about it:

it's really verbose.
it's really, really, verbose.
it's not nearly as verbose as you think. It's often just bad practice and inferior style that makes it so verbose and incomprehensible.

Since Java is still the language of my enterprise-day-job, I decided to ease my pain a bit, so I implemented my own functional utility library. I got a bit carried away, so I ended up with some annotation processors to bring something like poor-mans-first-class-functions into Java.

Since the library often provides rather clean ways to express ones intent, I wanted to see how it would compare to LINQ and Java Streams. So here it goes, examples from the Informatech blog supplemented with examples using plain old Java6 (released in 2007) using my functional library with annotation processors.

#Edit: I have written a follow-up with updated examples.

Challenge 1: Filtering

LINQ :

string[] names = { "Sam", "Pamela", "Dave", "Pascal", "Erik" };
List<string> filteredNames = names.Where(c => c.Contains("am"))
                                  .ToList();

Java Streams :

String[] names = {"Sam","Pamela", "Dave", "Pascal", "Erik"};
List<String> filteredNames = stream(names)
                 .filter(c -> c.contains("am"))
                 .collect(toList());

Java6 :

String[] names = { "Sam", "Pamela", "Dave", "Pascal", "Erik" };
List<string> filteredNames = newList(filter(names, contains("am")));

Challenge 2: Indexed Filtering

LINQ :

string[] names = { "Sam", "Pamela", "Dave", "Pascal", "Erik" };
var nameList = names.Where((c, index) => c.Length <= index + 1).ToList();

Java Streams :

String[] names = {"Sam","Pamela", "Dave", "Pascal", "Erik"};

List<String> nameList;
Stream<Integer> indices = intRange(1, names.length).boxed();
nameList = zip(indices, stream(names), SimpleEntry::new)
            .filter(e -> e.getValue().length() <= e.getKey())
            .map(Entry::getValue)
            .collect(toList());

Java6 :

String[] names = { "Sam", "Pamela", "Dave", "Pascal", "Erik" };
List<String> nameList = newList(map(filter(zipWithIndex(names), pred),
                                    Transformers.<String> right()));

static boolean pred(Map.Entry<Integer, String> candidate) {
    return candidate.getValue().length() <= candidate.getKey() + 1;
}

Challenge 3: Selecting/Mapping

LINQ :

List<string> nameList1 = new List(){ "Anders", "David", "James",
                                     "Jeff", "Joe", "Erik" };
nameList1.Select(c => "Hello! " + c).ToList()
         .ForEach(c => Console.WriteLine(c));

Java Streams :

List<String> nameList1 = asList("Anders", "David", "James",
                                "Jeff", "Joe", "Erik");
nameList1.stream()
     .map(c -> "Hello! " + c)
     .forEach(System.out::println);

Java6 :

List<String> nameList1 = newList("Anders", "David", "James", "Jeff", "Joe", "Erik");
foreach(map(nameList1, prepend("Hello! ")),
            PrintStream_.println8.apply(System.out));

Challenge 4: Selecting Many/Flattening

LINQ :

Dictionary<string, List<string>> map = new Dictionary<string,List<string>>();
map.Add("UK", new List<string>() {"Bermingham", "Bradford", "Liverpool"});
map.Add("USA", new List<string>() {"NYC", "New Jersey", "Boston", "Buffalo"});
var cities = map.SelectMany(c => c.Value).ToList();

Java Streams :

Map<String, List<String>> map = new LinkedHashMap<>();
map.put("UK", asList("Bermingham","Bradford","Liverpool"));
map.put("USA", asList("NYC","New Jersey","Boston","Buffalo"));

FlatMapper<Entry<String, List<String>>,String> flattener;
flattener = (entry,consumer) -> { entry.getValue().forEach(consumer); };

List<String> cities = map.entrySet()
             .stream()
             .flatMap( flattener )
             .collect(toList());

Java6 :

Map<String, List<String>> map = newMap(
    Pair.of("UK", newList("Bermingham", "Bradford", "Liverpool")),
    Pair.of("USA", newList("NYC", "New Jersey", "Boston", "Buffalo")));
List<String> cities = newList(flatten(map.values()));

Challenge 5: Taking an Arbitrary Number of Items

LINQ :

int[] numbers = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13 };
var first4 = numbers.Take(4).ToList();

Java Streams :

int[] numbers = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12,13 };

List<Integer> firstFour;
firstFour = stream(numbers).limit(4)
                           .boxed()
                           .collect(toList());

Java6 :

int[] numbers = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13 };
List<Integer> firstFour = newList(take(newArray(numbers), 4));

Challenge 6: Taking Items Based on Predicate

LINQ :

string[] moreNames = { "Sam", "Samuel", "Dave", "Pascal", "Erik",  "Sid" };
var sNames = moreNames.TakeWhile(c => c.StartsWith("S"));

Java Streams :

String[] names  = { "Sam","Samuel","Dave","Pascal","Erik","Sid" };

List<String> found;
found = stream(names).collect(partitioningBy( c -> c.startsWith("S")))
                     .get(true);

Java6 :

String[] names = { "Sam", "Samuel", "Dave", "Pascal", "Erik", "Sid" };
List<String> found = newList(takeWhile(names, startsWith("S")));

Challenge 7: Skipping an Arbitrary Number of Items

LINQ :

string[] vipNames = { "Sam", "Samuel", "Samu", "Remo", "Arnold","Terry" };
var skippedList = vipNames.Skip(3).ToList();//Leaving the first 3.

Java Streams :

String[] vipNames = { "Sam", "Samuel", "Samu", "Remo", "Arnold","Terry" };

List<String> skippedList;
skippedList = stream(vipNames).substream(3).collect(toList());

Java6 :

String[] vipNames = { "Sam", "Samuel", "Samu", "Remo", "Arnold", "Terry" };
List<String> skippedList = newList(drop(vipNames, 3));

Challenge 8: Skipping Items Based on Predicate

LINQ :

int[] numbers = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 20 };
var skippedList = numbers.SkipWhile(c => c < 10);

Java Streams :

//With current streams API I found no way to implement this idiom.

Java6 :

int[] numbers = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 20 };
List<Integer> skippedList = newList(dropWhile(newArray(numbers), lessThan(10)));

Challenge 9: Ordering/Sorting Elements

LINQ :

string[] friends = { "Sam", "Pamela", "Dave", "Anders", "Erik" };
friends = friends.OrderBy(c => c).ToArray();

Java Streams :

String[] friends = { "Sam", "Pamela", "Dave", "Anders", "Erik" };
friends = stream(friends).sorted().toArray(String[]::new);

Java6 :

String[] friends = { "Sam", "Pamela", "Dave", "Anders", "Erik" };
friends = newArray(sort(friends), String.class);

Challenge 10: Ordering/Sorting Elements by Specific Criterium

LINQ :

string[] friends = { "Sam", "Pamela", "Dave", "Anders", "Erik" };
friends = friends.OrderBy(c => c.Length).ToArray();

Java Streams :

String[] friends = { "Sam", "Pamela", "Dave", "Anders", "Erik" };
friends = stream(friends)
           .sorted(comparing((ToIntFunction<String>)String::length))
           .toArray(String[]::new);

Java6 :

String[] friends = { "Sam", "Pamela", "Dave", "Anders", "Erik" };
friends = newArray(sort(friends, by(String_.length)), String.class);

Challenge 11: Ordering/Sorting Elements by Multiple Criteria

LINQ :

string[] fruits = {"grape", "passionfruit", "banana",
                   "apple", "orange", "raspberry",
                   "mango", "blueberry" };

//Sort the strings first by their length and then alphabetically.
//preserving the first order.
var sortedFruits = fruits.OrderBy(fruit =>fruit.Length)
                         .ThenBy(fruit => fruit);

Java Streams :

String[] fruits = {"grape", "passionfruit", "banana","apple",
                   "orange", "raspberry","mango", "blueberry" };

Comparator<String> comparator;
comparator = comparing((Function<String,Integer>)String::length,
                       Integer::compare)
            .thenComparing((Comparator<String>)String::compareTo);

fruits = stream(fruits) .sorted(comparator)
                        .toArray(String[]::new);

Java6 :

String[] fruits = { "grape", "passionfruit", "banana", "apple",
                    "orange", "raspberry", "mango", "blueberry" };
fruits = newArray(sort(fruits, by(String_.length).then(
                               byNatural())), String.class);

Challenge 12: Grouping by a Criterium

LINQ :

string[] names = {"Sam", "Samuel", "Samu", "Ravi", "Ratna",  "Barsha"};
var groups = names.GroupBy(c => c.Length);

Java Streams :

String[] names = {"Sam", "Samuel", "Samu", "Ravi", "Ratna",  "Barsha"};

Map<Integer,List<String>> groups;
groups = stream(names).collect(groupingBy(String::length));

Java6 :

String[] names = { "Sam", "Samuel", "Samu", "Ravi", "Ratna", "Barsha" };
Map<Integer, List<String>> groups = groupBy(names, String_.length);

Challenge 13: Filter Distinct Elements

LINQ :

string[] songIds = {"Song#1", "Song#2", "Song#2", "Song#2", "Song#3", "Song#1"};
//This will work as strings implement IComparable
var uniqueSongIds = songIds.Distinct();

Java Streams :

String[] songIds = {"Song#1", "Song#2", "Song#2", "Song#2", "Song#3", "Song#1"};
//according to Object.equals
stream(songIds).distinct();

Java6 :

String[] songIds = { "Song#1", "Song#2", "Song#2", "Song#2", "Song#3", "Song#1" };
newSet(songIds);

Challenge 14: Union of Two Sets

LINQ :

List<string> friends1 = new List<string>() {"Anders", "David","James",
                                            "Jeff", "Joe", "Erik"};
List<string> friends2 = new List<string>() { "Erik", "David", "Derik" };
var allMyFriends = friends1.Union(friends2);

Java Streams :

List<String> friends1 = asList("Anders","David","James","Jeff","Joe","Erik");
List<String> friends2 = asList("Erik","David","Derik");
Stream<String> allMyFriends = concat(friends1.stream(),
                                     friends2.stream()).distinct();

Java6 :

List<String> friends1 = newList("Anders", "David", "James", "Jeff", "Joe", "Erik");
List<String> friends2 = newList("Erik", "David", "Derik");
Set<String> allMyFriends = union(newSet(friends1), newSet(friends2));

Challenge 15: First Element

LINQ :

string[] otherFriends = {"Sam", "Danny", "Jeff", "Erik", "Anders","Derik"};
string firstName = otherFriends.First();
string firstNameConditional = otherFriends.First(c => c.Length == 5);

Java Streams :

String[] otherFriends = {"Sam", "Danny", "Jeff", "Erik", "Anders","Derik"};
Optional<String> found = stream(otherFriends).findFirst();

Optional<String> maybe = stream(otherFriends).filter(c -> c.length() == 5)
                                             .findFirst();
if(maybe.isPresent()) {
   //do something with found data
}

Java6 :

String[] otherFriends = { "Sam", "Danny", "Jeff", "Erik", "Anders", "Derik" };
Option<String> found = headOption(otherFriends);
Option<String> maybe = find(otherFriends, String_.length.andThen(equalTo(5)));
for (String m: maybe) {
    // ...
}

Challenge 16: Generate a Range of Numbers

LINQ :

var multiplesOfEleven = Enumerable.Range(1, 100).Where(c => c % 11 == 0);

Java Streams :

IntStream multiplesOfEleven = intRange(1,100).filter(n -> n % 11 == 0);

Java6 :

Iterable<Integer> multiplesOfEleven = filter(range(1, 100), mod(11).andThen(equalTo(0)));

Challenge 17: All

LINQ :

string[] persons = {"Sam", "Danny", "Jeff", "Erik", "Anders","Derik"};
bool x = persons.All(c => c.Length == 5);

Java Streams :

String[] persons = {"Sam", "Danny", "Jeff", "Erik", "Anders","Derik"};
boolean x = stream(persons).allMatch(c -> c.length() == 5);

Java6 :

String[] persons = { "Sam", "Danny", "Jeff", "Erik", "Anders", "Derik" };
boolean x = forAll(persons, String_.length.andThen(equalTo(5)));

Challenge 18: Any

LINQ :

string[] persons = {"Sam", "Danny", "Jeff", "Erik", "Anders","Derik"};
bool x = persons.Any(c => c.Length == 5);

Java Streams :

String[] persons = {"Sam", "Danny", "Jeff", "Erik", "Anders","Derik"};
boolean x = stream(persons).anyMatch(c -> c.length() == 5);

Java6 :

String[] persons = { "Sam", "Danny", "Jeff", "Erik", "Anders", "Derik" };
boolean x = exists(persons, String_.length.andThen(equalTo(5)));

Challenge 19: Zip

LINQ :

string[] salutations = {"Mr.", "Mrs.", "Ms", "Master"};
string[] firstNames = {"Samuel", "Jenny", "Joyace", "Sam"};
string lastName = "McEnzie";

salutations.Zip(firstNames, (sal, first) => sal + " " + first)
           .ToList()
           .ForEach(c => Console.WriteLine(c + " " + lastName));

Java Streams :

String[] salutations = {"Mr.", "Mrs.", "Ms", "Master"};
String[] firstNames = {"Samuel", "Jenny", "Joyace", "Sam"};
String lastName = "McEnzie";

zip(
    stream(salutations),
    stream(firstNames),
    (sal,first) -> sal + " " +first)
.forEach(c -> { System.out.println(c + " " + lastName); });

Java6 :

String[] salutations = { "Mr.", "Mrs.", "Ms", "Master" };
String[] firstNames = { "Samuel", "Jenny", "Joyace", "Sam" };
String lastName = "McEnzie";

foreach(map(zip(salutations, firstNames, repeat(lastName)), mkString(" ")),
        PrintStream_.println8.apply(System.out));

Conclusion

Based on these examples I have a funny feeling that Java8 Streams API is going to be a failure. And since developers will not be able to extend it with useful constructs, it may well end up being just another nail in the coffin.

Of these examples, personally, I find the Java6 code to be the most readable. Even with its oddities, of which most are caused by the original authors decision to use ints (instead of Integers) and Lists (instead of Iterables). The ability to do this has been around since 2007, and Java8 will be released in... 2014?

I'm a bit biased, though, so what do you think?

Architecturally Elegant - lahteenmaki.net

Challenge 1: Filtering

Challenge 2: Indexed Filtering

Challenge 3: Selecting/Mapping

Challenge 4: Selecting Many/Flattening

Challenge 5: Taking an Arbitrary Number of Items

Challenge 6: Taking Items Based on Predicate

Challenge 7: Skipping an Arbitrary Number of Items

Challenge 8: Skipping Items Based on Predicate

Challenge 9: Ordering/Sorting Elements

Challenge 10: Ordering/Sorting Elements by Specific Criterium

Challenge 11: Ordering/Sorting Elements by Multiple Criteria

Challenge 12: Grouping by a Criterium

Challenge 13: Filter Distinct Elements

Challenge 14: Union of Two Sets

Challenge 15: First Element

Challenge 16: Generate a Range of Numbers

Challenge 17: All

Challenge 18: Any

Challenge 19: Zip

Conclusion