问题
I have an instances of Student class.
class Student {
String name;
String addr;
String type;
public Student(String name, String addr, String type) {
super();
this.name = name;
this.addr = addr;
this.type = type;
}
@Override
public String toString() {
return "Student [name=" + name + ", addr=" + addr + "]";
}
public String getName() {
return name;
}
public String getAddr() {
return addr;
}
}
And I have a code to create a map , where it store the student name as the key and some processed addr values (a List since we have multiple addr values for the same student) as the value.
public class FilterId {
public static String getNum(String s) {
// should do some complex stuff, just for testing
return s.split(" ")[1];
}
public static void main(String[] args) {
List<Student> list = new ArrayList<Student>();
list.add(new Student("a", "test 1", "type 1"));
list.add(new Student("a", "test 1", "type 2"));
list.add(new Student("b", "test 1", "type 1"));
list.add(new Student("c", "test 1", "type 1"));
list.add(new Student("b", "test 1", "type 1"));
list.add(new Student("a", "test 1", "type 1"));
list.add(new Student("c", "test 3", "type 2"));
list.add(new Student("a", "test 2", "type 1"));
list.add(new Student("b", "test 2", "type 1"));
list.add(new Student("a", "test 3", "type 1"));
Map<String, List<String>> map = new HashMap<>();
// This will create a Map with Student names (distinct) and the test numbers (distinct List of tests numbers) associated with them.
for (Student student : list) {
if (map.containsKey(student.getName())) {
List<String> numList = map.get(student.getName());
String value = getNum(student.getAddr());
if (!numList.contains(value)) {
numList.add(value);
map.put(student.getName(), numList);
}
} else {
map.put(student.getName(), new ArrayList<>(Arrays.asList(getNum(student.getAddr()))));
}
}
System.out.println(map.toString());
}
}
Output would be :
{a=[1, 2, 3], b=[1, 2], c=[1, 3]}
How can I just do the same in java8 in a much more elegant way, may be using the streams ?
Found this Collectors.toMap in java 8 but could't find a way to actually do the same with this.
I was trying to map the elements as CSVs but that it didn't work since I couldn't figure out a way to remove the duplicates easily and the output is not what I need at the moment.
Map<String, String> map2 = new HashMap<>();
map2 = list.stream().collect(Collectors.toMap(Student::getName, Student::getAddr, (a, b) -> a + " , " + b));
System.out.println(map2.toString());
// {a=test 1 , test 1 , test 1 , test 2 , test 3, b=test 1 , test 1 , test 2, c=test 1 , test 3}
回答1:
With streams, you could use Collectors.groupingBy along with Collectors.mapping:
Map<String, Set<String>> map = list.stream()
.collect(Collectors.groupingBy(
Student::getName,
Collectors.mapping(student -> getNum(student.getAddr()),
Collectors.toSet())));
I've chosen to create a map of sets instead of a map of lists, as it seems that you don't want duplicates in the lists.
If you do need lists instead of sets, it's more efficient to first collect to sets and then convert the sets to lists:
Map<String, List<String>> map = list.stream()
.collect(Collectors.groupingBy(
Student::getName,
Collectors.mapping(s -> getNum(s.getAddr()),
Collectors.collectingAndThen(Collectors.toSet(), ArrayList::new))));
This uses Collectors.collectingAndThen, which first collects and then transforms the result.
Another more compact way, without streams:
Map<String, Set<String>> map = new HashMap<>(); // or LinkedHashMap
list.forEach(s ->
map.computeIfAbsent(s.getName(), k -> new HashSet<>()) // or LinkedHashSet
.add(getNum(s.getAddr())));
This variant uses Iterable.forEach to iterate the list and Map.computeIfAbsent to group transformed addresses by student name.
回答2:
First of all, the current solution is not really elegant, regardless of any streaming solution.
The pattern of
if (map.containsKey(k)) {
Value value = map.get(k);
...
} else {
map.put(k, new Value());
}
can often be simplified with Map#computeIfAbsent. In your example, this would be
// This will create a Map with Student names (distinct) and the test
// numbers (distinct List of tests numbers) associated with them.
for (Student student : list)
{
List<String> numList = map.computeIfAbsent(
student.getName(), s -> new ArrayList<String>());
String value = getNum(student.getAddr());
if (!numList.contains(value))
{
numList.add(value);
}
}
(This is a Java 8 function, but it is still unrelated to streams).
Next, the data structure that you want to build there does not seem to be the most appropriate one. In general, the pattern of
if (!list.contains(someValue)) {
list.add(someValue);
}
is a strong sign that you should not use a List, but a Set. The set will contain each element only once, and you will avoid the contains calls on the list, which are O(n) and thus may be expensive for larger lists.
Even if you really need a List in the end, it is often more elegant and efficient to first collect the elements in a Set, and afterwards convert this Set into a List in one dedicated step.
So the first part could be solved like this:
// This will create a Map with Student names (distinct) and the test
// numbers (distinct List of tests numbers) associated with them.
Map<String, Collection<String>> map = new HashMap<>();
for (Student student : list)
{
String value = getNum(student.getAddr());
map.computeIfAbsent(student.getName(), s -> new LinkedHashSet<String>())
.add(value);
}
It will create a Map<String, Collection<String>>. This can then be converted into a Map<String, List<String>> :
// Convert the 'Collection' values of the map into 'List' values
Map<String, List<String>> result =
map.entrySet().stream().collect(Collectors.toMap(
Entry::getKey, e -> new ArrayList<String>(e.getValue())));
Or, more generically, using a utility method for this:
private static <K, V> Map<K, List<V>> convertValuesToLists(
Map<K, ? extends Collection<? extends V>> map)
{
return map.entrySet().stream().collect(Collectors.toMap(
Entry::getKey, e -> new ArrayList<V>(e.getValue())));
}
I do not recommend this, but you also could convert the for loop into a stream operation:
Map<String, Set<String>> map =
list.stream().collect(Collectors.groupingBy(
Student::getName, LinkedHashMap::new,
Collectors.mapping(
s -> getNum(s.getAddr()), Collectors.toSet())));
Alternatively, you could do the "grouping by" and the conversion from Set to List in one step:
Map<String, List<String>> result =
list.stream().collect(Collectors.groupingBy(
Student::getName, LinkedHashMap::new,
Collectors.mapping(
s -> getNum(s.getAddr()),
Collectors.collectingAndThen(
Collectors.toSet(), ArrayList<String>::new))));
Or you could introduce an own collector, that does the List#contains call, but all this tends to be far less readable than the other solutions...
回答3:
I think you are looking for something like below
Map<String,Set<String>> map = list.stream().
collect(Collectors.groupingBy(
Student::getName,
Collectors.mapping(e->getNum(e.getAddr()), Collectors.toSet())
));
System.out.println("Map : "+map);
回答4:
Here is a version that collects everything in sets, and converts the final result to array lists:
/*
import java.util.*;
import java.util.stream.*;
import static java.util.stream.Collectors.*;
import java.util.function.*;
*/
Map<String, List<String>> map2 = list.stream().collect(groupingBy(
Student::getName, // we will group the students by name
Collector.of(
HashSet::new, // for each student name, we will collect result in a hash set
(arr, student) -> arr.add(getNum(student.getAddr())), // which we fill with processed addresses
(left, right) -> { left.addAll(right); return left; }, // we merge subresults like this
(Function<HashSet<String>, List<String>>) ArrayList::new // finish by converting to List
)
));
System.out.println(map2);
// Output:
// {a=[1, 2, 3], b=[1, 2], c=[1, 3]}
EDIT: made the finisher shorter using Marco13's hint.
来源：https://stackoverflow.com/questions/48587730/converting-listmyobject-to-mapstring-liststring-in-java-8-when-we-have-du