Multiple operations on java object using Stream and lambda

Question

i have class A {int sNo; String name;} with valid constructor. i need to trim name and sort on the basis of name using stream API.

public class Test1 {
    public static void main(String[] args) {



    ArrayList<A> l = new ArrayList();
    l.add(new A(1, " 1name "));
    l.add(new A(7, " 3name "));
    l.add(new A(6, ""));
    l.add(new A(5, " 2name "));
    l.add(new A(4, " 5name "));
    l.add(new A(2, ""));

    List<String> i = l.stream()
            .filter(s -> !s.name.isEmpty())
            .map(s -> s.name.trim())
            .sorted()
            .collect(Collectors.toList());
    System.out.println(i); 
}

}

it is returning only name in sorted order

[, 1name, 2name, 3name, 5name]

but i need entire object.

i know s -> s.name.trim() is the reason its storing name only,

how can apply operation on particular filed but store entire object.

thanks @janos for your reply , yes its working fine but i have a small doubt. s -> new A(s.sNo, s.name.trim())) is not scalable , i mean in this example class A containing 2 variables only so its ok to call its constructor, but if class A has lets say 20-30 variables then it is not wise to call constructor with all same values and just i updated 1, is there any work around?? — shrikant.sharma
– shrikant.sharma, Commented Nov 7, 2017 at 6:18

apophis · Accepted Answer · 2017-11-06 19:49:57Z

2

Just for fun, not because it's particularly nice or even recommended. One could accomplish that also with anonymous Objects:

    List<A> i = l.stream()
            .filter(s -> !s.name.isEmpty())
            .map(s -> new Object() {A a = s; String name = s.name.trim();})
            .sorted(Comparator.comparing(o -> o.name))
            .map(o -> o.a)
            .collect(Collectors.toList());

Not every compiler will be happy about this (javac will), but it's valid Java code.

edited Nov 6, 2017 at 19:49

answered Nov 5, 2017 at 16:34

apophis

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Comments

developer_hatch · Accepted Answer · 2017-11-05 06:40:53Z

1

When you use the map function you are converting all your objects into strings, so instead of map, use for each to transform the strings inside the object but not returning a lists of strings :

 class A {
 int sNo; 
 String name;

 public int getName(){ return name}



 public void setName(String name){ this.name = name;}

  }

i.forEach(s -> s.setName(s.getName().trim()));

List<A> i = l.stream()
        .filter(s -> !s.name.isEmpty())       
        .sorted(Comparator.comparing(a -> a.name))
        .collect(Collectors.toList());

edited Nov 5, 2017 at 6:40

answered Nov 5, 2017 at 6:14

developer_hatch

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4 Comments

Andreas Over a year ago

What do you believe forEach(s -> s.name.trim()) does? I mean, other than use up the CPU?

developer_hatch Over a year ago

@Andreas you're right, forgot to set the value to update the object state

Andreas Over a year ago

s doesn't have a setName() method. Or, if it does, it likely also has a getName() method.

developer_hatch Over a year ago

@Andreas of course of course, forgot to add that important detail too, I updated again. Thanks a lot

janos · Accepted Answer · 2017-11-05 06:43:21Z

1

it is returning only name in sorted order [...] but i need entire object.

The .map(s -> s.name.trim()) step transforms your stream of A into a stream of String. That's why you're left with strings, instead of entire objects.

If you want a sorted list of objects, then the result should be in a List<A> and not in a List<String>.

To sort a list of objects by a field, you can pass a Comparator to sorted.

List<A> result = l.stream()
    .filter(s -> !s.name.isEmpty())
    .map(s -> new A(s.sNo, s.name.trim()))
    .sorted(Comparator.comparing(a -> a.name))
    .collect(Collectors.toList());

If you don't want to create a new A with the trimmed name, but want to modify the original object (though I seriously doubt that), you can replace the map step with peek and modify the object:

.peek(s -> s.name = s.name.trim())

However, note that as @Mustafa pointed out, peek is intended only for debugging. And in fact if you want to modify the underlying objects, it would be better to that in a step separate from the filtering and sorting, as it is a logically unrelated step anyway (it looks like data cleaning). That step can be implemented using forEach.

If you don't actually need to trim the name (thanks @Andreas for the hint), only for the purpose of sorting, then you can drop the map step entirely, and write the sorted step like this:

.sorted(Comparator.comparing(a -> a.name.trim()))

edited Nov 5, 2017 at 6:43

answered Nov 5, 2017 at 6:16

janos

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4 Comments

Andreas Over a year ago

You could just do sorted(Comparator.comparing(a -> a.name.trim())). Sure, it's not good for performance, but it works and eliminates the need for map().

janos Over a year ago

@Andreas I was under the impression that OP actually wants to trim the string, not only sort on the trimmed value. But your idea is more likely, thanks, I updated my answer.

Mustafa Over a year ago

Creating new objects is better, otherwise the use of forEach is recommended over peek. According to the documentation, peek is only meant for debugging.

shrikant.sharma Over a year ago

thanks @janos for your reply , yes its working fine but i have a small doubt. s -> new A(s.sNo, s.name.trim())) is not scalable , i mean in this example class A containing 2 variables only so its ok to call its constructor, but if class A has lets say 20-30 variables then it is not wise to call constructor with all same values and just i updated 1, is there any work around??

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