Grouping by object - Java streams

Question

I have a list of data like the one given below:

List<Data> data = new ArrayList<Data>();
data.add(new Data("d1", "option1"));
data.add(new Data("d2", "option1"));
data.add(new Data("d1", "option2"));
data.add(new Data("d3", "option1"));
data.add(new Data("d3", "option2"));
data.add(new Data("d3", "option3"));

The structure looks like this:

class Data {
    private String name;
    private String option;
    private List<String> options = new ArrayList<>();
    public Data(String name, String option) {
        this.name = name;
        this.option = option;
    }

    public void addOption(String option) {
        options.add(option);
    }
}

How to group the items to a new array based on the name with its options,

[
"d1": {
    "name": "d1",
    "options": ["option1", "option2"]
},
"d2": {
    "name": "d2",
    "options": ["option1"]
},
"d3": {
    "name": "d3",
    "options": ["option1", "option2", "option3"]
}
]

To clarify: Do you want the output as a JSON-String? Or a way to get all option fields in the options field of a new instance of Data? For the latter I would already recommend to change something because it's weird to have both a list of options and a single String option in your class if they both basically mean the same. — Ben
– Ben, Commented May 30, 2018 at 6:22
You can't instantiate List; it's an interface. Why does Data have both a single option and a list of options? You never call addOption. Also, you can't group them to an array. Do you mean a Map? — David Conrad
– David Conrad, Commented May 30, 2018 at 6:26
No JSON, just for brevity I have given the object structure like that. — vvra
– vvra, Commented May 30, 2018 at 6:52

Eran · Accepted Answer · 2018-05-30 06:26:36Z

4

You can use a Collectors.toMap collector:

Map<String,Data>
    grouped = data.stream()
                  .collect(Collectors.toMap(Data::getName,
                                            d -> new Data(d.getName(),d.getOption()),
                                            (d1,d2) -> {d1.addOptions(d2.getOptions()); return d1;});

This will require changing the Data constructor to add the passed option to the options List as well as adding an addOptions method that receives a list of options and adds all of them to the options List of the current instance.

answered May 30, 2018 at 6:26

Eran

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1 Comment

vvra Over a year ago

This approach looks convenient for me, I get collection of type "Data" as values from the result map, than of type List<String>.

Ravindra Ranwala · Accepted Answer · 2018-05-30 06:56:56Z

3

Try this out,

final Map<String, List<String>> optionsByName = dataList.stream().collect(
        Collectors.groupingBy(Data::getName, Collectors.mapping(Data::getOption, Collectors.toList())));

I would suggest you to use this map as a source to create the DTO you need, without cluttering your code to get the exact result you want.

edited May 30, 2018 at 6:56

answered May 30, 2018 at 6:27

Ravindra Ranwala

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Comments

ernest_k · Accepted Answer · 2018-05-30 06:37:20Z

You can use a simple grouping collector on the stream:

Map<String, List<String>> optionMap = data.stream()
            .collect(Collectors.groupingBy(Data::getName, 
                     Collectors.mapping(Data::getOption, Collectors.toList())));

When tested with your test list, the above produces this output:

{d1=[option1, option2], d2=[option1], d3=[option1, option2, option3]}

The above seems like a better, type-safe alternative to your desired map. Additionally, it avoids unnecessary duplication.

Your final map can nonetheless be computed based on the above:

Map<String, Map<String, Object>> m = new HashMap<>();
optionMap.entrySet().forEach(entry -> {
    Map<String, Object> map = new HashMap<>();
    map.put("name", entry.getKey());
    map.put("options", entry.getValue());

    m.put(entry.getKey(), map);
});

The m map above looks like:

{d1={name=d1, options=[option1, option2]}, 
 d2={name=d2, options=[option1]}, 
 d3={name=d3, options=[option1, option2, option3]}}

This is what you need, but it seems that it contains duplicated data, and it's less type-safe than the simple map in the previous result.

sudhansu · Accepted Answer · 2025-03-22 07:18:23Z

0


public class Test2 {

    static List<Employee> getAllEmployees() {
        List<Employee> employeeList = new ArrayList<Employee>();
        Employee e1= new Employee("sudhansu",46,135000.00,"JAVA");
        Employee e2= new Employee("Himanshu",26,135000.00,"JAVA");
        Employee e3= new Employee("Pabitra",53,135000.00,"ORACLE");
        Employee e4= new Employee("Rajan",36,135000.00,"Hibernate");
        Employee e5= new Employee("Raman",36,135000.00,"JAVA");
        Employee e6= new Employee("Sambhu",26,135000.00,"MYSQL");
        Employee e7= new Employee("Sagar",37,135000.00,"ORACLE");
        Employee e8= new Employee("Ranjeet",26,135000.00,".NET");
        Employee e9= new Employee("Chitra",46,135000.00,".NET");
        Employee e10= new Employee("Minu",28,135000.00,".NET");
        employeeList.add(e1);
        employeeList.add(e2);
        employeeList.add(e3);
        employeeList.add(e4);
        employeeList.add(e5);
        employeeList.add(e6);
        employeeList.add(e7);
        employeeList.add(e8);
        employeeList.add(e9);
        employeeList.add(e10);
        return employeeList;
    }
    public static void main(String[] args) {
        // TODO Auto-generated method stub
        List<Employee> empList= getAllEmployees();
        Map<String, List<String>> map1=empList.stream().collect(Collectors.groupingBy(e->e.getSkills(),LinkedHashMap::new,
                Collectors.mapping(Employee::getName,Collectors.toList())));
        System.out.println("employee map"+map1);
    }

}

answered Mar 22 at 7:18

sudhansu

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2 Comments

sudhansu Mar 22 at 7:19

same.....almost

Basil Bourque Mar 22 at 7:47

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