Group List Of Objects based on attributes populated

Question

I have a class with 10 attributes. I want to group list of objects of that class based on attributes populated or not. Precisely I will get 2^n groups. I can iterate the list and write 2^n conditions and group them into new lists. Are there any other better ways to implement this functionality instead of writing those switch conditions ?

Wouldn't ten binary values lead to 1024 permutations? You could collect this as bits (each field having its own bit), which could be directly used as a grouping index. — vlumi
– vlumi, Commented Jun 17, 2019 at 22:38

Varun Upadhyay · Accepted Answer · 2019-06-17 23:31:08Z

0

You can avoid writing multiple switch statements and have an identifier/key in the class itself which helps in understanding if certain properties are set or not.

Example:

DemoClass (3 attributes for simplification)

public class DemoClass {
    public String property1 = null;
    public String property2 = null;
    public String property3 = null;

    public String getKey() {
        StringBuilder sb = new StringBuilder();
        sb.append(this.property1 != null ? "PROPERTY1" : "").append("|");
        sb.append(this.property2 != null ? "PROPERTY2" : "").append("|");
        sb.append(this.property3 != null ? "PROPERTY3" : "").append("|");

        return sb.toString();
    }
}

Above the fields can be populated in various ways such as overloaded constructors or setter methods.

Now when you want to separate into groups, you can use a HashMap and add the objects of the class to the list based on the value of their keys as below

Map<String, List<DemoClass>> map = new HashMap<>();
for (DemoClass demoClass : demoClasses) {
    map.computeIfAbsent(demoClass.getKey(), k -> new ArrayList<>()).add(demoClass);
}

answered Jun 17, 2019 at 23:31

Varun Upadhyay

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2 Comments

Varun Upadhyay Over a year ago

We cannot use demoClass instead as it is an instance of DemoClass and not of the HashMap. Also computeIfAbsent is a nice API provided by Java and it makes the code much easier to understand as compared to doing a getOrDefault which is useful if the value is not a Collection

Varun Upadhyay Over a year ago

A BitSet can also be used but I felt that the above solution of having a String key which gives info about the set attributes is much more readable and can be extended further to use cases such as finding out which attributes are populated.

Prasanth · Accepted Answer · 2019-06-18 01:11:59Z

0

I ended up getting a bit value of the attributes of the Record (1's and 0's for every attribute (1 means value exists and 0 means it does not exist)) in a string and grouped by them by String.

records.stream().map(record -> new ExportDto.Builder().buildUpon(record).bitValue(getBitValueOfExportRecord(record)).build())
.collect(Collectors.groupingBy(ExportDto::getBitValue))

answered Jun 18, 2019 at 1:11

Prasanth

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