How to obtain first 5 values from a LinkedHashSet?

Question

I have a LinkedHashSet which contains multiple number of values.

LinkedHashSet<String> lhs = new LinkedHashSet<String>();

I want to iterate through the set of values and display the first five values from the number of items stored in the set. I have used a for loop to iterate through values and display data, see below:

for (String sent : lhs) {        
    text.append(sent);                                          
}

This outputs all the values stored in the LinkedHashSet. What alterations should I make to my code in order to only get the first 5 values from the set.

How about breaking the loop after 5 iterations (controlled by `int` variable)? — Arrvi, Feb 22 '16 at 21:59
This depends on what version of Java you are using... With Java 7- the solution for this would be vastly different from what it would be with Java 8+. Also, you say you want the 5 first values: in what? A collection, an array? — fge, Feb 22 '16 at 21:59
@Mussa no it doesn't. Android, as far as I know, still doesn't have any viable JSR 203, or JSR 292, implementation — fge, Feb 22 '16 at 22:01
Can you please define "values"? You `.append()` to some `text` but do not even tell what this `text` is... — fge, Feb 22 '16 at 22:04

Pshemo · Answer 1 · 2016-02-22T22:12:05.780

6

You can get your sets Iterator

Iterator<String> it = yourSet.iterator();

and move to next() element N times (assuming that it still hasNext() element)

int counter = 0;
while(it.hasNext() && counter++ < N){
    String element = it.next();
    ...
}

edited Feb 22 '16 at 22:12

answered Feb 22 '16 at 21:59

Pshemo

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I'd make it a `for` loop. – Arrvi Feb 22 '16 at 22:04
@Arrvi Meh, that's really a matter of opinion. – Tunaki Feb 22 '16 at 22:05
@Arrvi why? This ultimately has no influence on the final result – fge Feb 22 '16 at 22:05
Meh, a `for` loop is really just a fancy way of saying a `while` loop. :P – ifly6 Feb 22 '16 at 22:06
It "hides" the _counter_ variable. – Arrvi Feb 22 '16 at 22:08

score 5 · Accepted Answer · answered Feb 22 '16 at 22:02

5

int i = 0;
for (String sentences : lhs) {       
    if (i > 4) break; 
    text.append(sentences);
    i++;
}

answered Feb 22 '16 at 22:02

Mussa

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Thank you very much, it works perfectly. This is exactly the type of solution that I was trying to figure out. – coder Feb 22 '16 at 22:10
@coder and what is supposed to happen if the input set has _less elements_ than what you are waiting for? – fge Feb 22 '16 at 22:12
@fge it would still work because it only stops if it is greater than 4. – coder Feb 22 '16 at 22:17
1

@fge it will stop if *i* is greater than 4 **or** if *lhs* runs out of items – Mussa Feb 22 '16 at 22:19

slartidan · Answer 3 · 2016-02-22T22:23:36.767

5

If you had java8, then I would suggest something like this:

yourLinkedHashSet
    .stream()
    .limit(5)
    .forEachOrdered(text::append);

Explanation:

stream() will take one String after another from the collection
limit(5) will stop the calculations after five elements are processed
forEachOrdered(...) takes an action for each item, one after another
text::append is the action to be done per item

edited Feb 22 '16 at 22:23

answered Feb 22 '16 at 22:12

slartidan

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score 0 · Answer 4 · answered Sep 01 '17 at 08:09

0

You can use subList without counter declaration

  Iterator<String> iter = new ArrayList<>(lhs).subList(5, lhs.size())
    .iterator();
  while (iter.hasNext()) {
    text.append(iter.next());
  }

answered Sep 01 '17 at 08:09

luk4

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How to obtain first 5 values from a LinkedHashSet?

4 Answers4