Java stream: Difficulty in object property update

Question

I have one list List with a quite a number of fields, need to create a List in a single pass using stream.

We need to mark the field current true in the List, for the least positive integer in the input List

Input:

[PT(pn=-1, endDate=2019-01-11, dp=MAR 2019,..),           
PT(pn=4, endDate=null, dp=APR 2019,..),        
PT(pn=6, endDate=2019-05-11, dp=MAY 2019,..)]

Output:

[OP(pn=11, current=false, dp=MAR 2019),
 OP(pn=4, current=true, dp=APR 2019),
 OP(pn=6, current=false, dp=MAY 2019)]

class OP{       
   Integer pn;      
   Boolean current;      
   String dp;       
}

List<OP> filteredList11 = availablePT.get()
    .stream()
    .map(e->new OP(e.getPn(),**,e.getDp()))
    .collect(Collectors.toList());

I am bit confused how to derive the logic to update the current flag(**) as true for the least positive integer in a single loop.

You want to achieve both the conversion and the update in one single loop? And get back the one object that has smallest positive number? Or do you need the full list converted? — MadaManu
– MadaManu, Commented Jan 7, 2020 at 3:20
@Madamanu, Thanks, the program should return all the elements in the list. — busywithJava
– busywithJava, Commented Jan 7, 2020 at 3:25
If any of the answers below worked for you, please do accept it as the right answer. Thanks. — MadaManu
– MadaManu, Commented May 20, 2021 at 18:39

MadaManu · Accepted Answer · 2020-01-07 03:38:55Z

1

Here is another approach. Have a OP constructor that takes in a PT and maps all the fields, plus the extra boolean set by default to false.

Then do something like this:

List<OP> ops = availablePT.stream().map(dp -> {return new OP(dp);})
 .filter(line -> line.getPn()!=null && line.getPn().doubleValue()>=0)
 .sorted(Comparator.comparingInt(OP::getPn))
 .collect(Collectors.toList());
if(ops!=null && !ops.isEmpty()) {
   ops.get(0).setCurrent(true);
}

This basically creates the new objects, then filters the negatives out, then sorts the possitives, and then you're guaranteed that the op with smallest dp value is first, so set it's boolean to true

edited Jan 7, 2020 at 3:38

answered Jan 7, 2020 at 3:32

MadaManu

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2 Comments

MadaManu Over a year ago

Updated - had the wrong way around the signs for comparison of the filter. Updated now

busywithJava Over a year ago

MadaManu, Thanks a lot

Hadi · Accepted Answer · 2020-01-07 08:57:53Z

You can do something like this :

first find the least positive PT object.

Optional<PT> latestPositive = availablePT.stream()
                             .filter(pt -> pt.getPn() > 0)
                             .min(Comparator.comparingInt(PT::getPn));

then by stream over availablePT list create OP object.

List<OP> filteredList11 = availablePT.stream()
            .map(pt -> new OP(pt.getPn(),
                              pt.equals(latestPositive.orElse(null)), pt.getDp()))
            .collect(Collectors.toList());

If you are interested in to use single stream do this:

first write a custom comparator.

 Comparator<PT> ptComparator = (o1, o2) -> {
        if (o2.getPn() < 0) {
            if (o1.getPn() > 0)
                return -1;
        } else if (o2.getPn() > 0) {
            if (o1.getPn() < 0)
                return 1;
        }
        return Integer.compare(o1.getPn(), o2.getPn());
    };

then sort list with your comparator

AtomicInteger integer = new AtomicInteger(0);
 List<OP> filteredList11 = availablePT.stream()
          .sorted(ptComparator)
          .map(pt -> new OP(pt.getPn(), integer.incrementAndGet()==1, pt.getDp()))
          .collect(Collectors.toList());

Cường Lư Quốc · Accepted Answer · 2020-01-07 03:18:19Z

0

Here is your required:

AtomicBoolean isFirst = new AtomicBoolean(true);
    List<OP> filteredList11 = availablePT.get()
        .stream()
        .map(e->new OP(e.getPn(),**,e.getDp()))
        .sorted(Comparator.comparingInt(OP::getPn))
        .peek(e -> {
                        if (e.getPn() % 2 == 0) {
                            e.setCurrent(isFirst.getAndSet(false));
                        }
                    })
        .collect(Collectors.toList());

answered Jan 7, 2020 at 3:18

Cường Lư Quốc

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