How to get data from Set in one go

[Bhagat Singh Rawat]

Hi Ranchers,

I want to get all values of a Set interface in one go in Comma Separated String.

For Example:

Set<String> fruits= new HashSet<String>();

fruits.add("Apple");

fruits.add("Banana");

fruits.add("Orange");

if I print Set as value.toString then the output would be:

[Apple, Banana, Orange]

But my requirement is

Apple, Banana, Orange

without [] squire brackets.

Please suggest.

[armando fonseca]

is this a homework? or an assignment ? there is a policy about question regarding those scenario. You can use String methods, or regex if you are comfortable with it.

[Bhagat Singh Rawat]

armando fonseca wrote:is this a homework? or an assignment ? there is a policy about question regarding those scenario. You can use String methods, or regex if you are comfortable with it.

Its homework in terms of performance, I have 3 solutions:
1-

	public String collectionToCommaSeparatedString(Collection pCollection) {

		Iterator<String> iterate = pCollection.iterator();
		StringBuilder sb = new StringBuilder();
		while (iterate.hasNext() == true) {
			sb.append(iterate.next());
			if (iterate.hasNext())
				sb.append(",");
		}
		return sb.toString();
	}

2-

fruits.toString().substring(1, fruits.toString().length - 1);

3-
String allFruits = StringUtils.join(fruits, ", ");
Note: StringUtils is class from Apache Commons Lang Library

Now you tell me which one is good in terms of performance?

[Deepak Bala]

Hard to tell unless you actually clock it. Any algorithm would take atleast O(N) time.

[Jesper de Jong]

Bhagat Singh Rawat wrote:Its homework in terms of performance, I have 3 solutions: ...

There is really only one way to do this: go through all the values in the set and concatenate them together. All three of your solutions come down to the same, there will not be a major performance difference between them. I bet that if you look up the source code of the StringUtils method from Apache Commons, it will be very similar to your first solution.

Performance-wise, you're already doing the right thing in your own first solution in that you are using a StringBuilder to concatenate all the parts together, instead of using String objects and concatenating them with the + operator (that would make a lot of unnecessary temporary String objects and unnecessary copying of data).

By the way, this is not really a question about the SCJP. I will move this topic to a more appropriate forum for you.

[Ernest Friedman-Hill]

Note that there's a very easy way to double the performance of method #2 -- but I'll leave that to your imagination...

[Rob Spoor]

Bhagat Singh Rawat wrote:1-

	public String collectionToCommaSeparatedString(Collection pCollection) {

		Iterator<String> iterate = pCollection.iterator();
		StringBuilder sb = new StringBuilder();
		while (iterate.hasNext() == true) {
			sb.append(iterate.next());
			if (iterate.hasNext())
				sb.append(",");
		}
		return sb.toString();
	}

That's more or less what Set's toString does (or actually AbstractCollection):

public String toString() {
    Iterator<E> i = iterator();
    if (! i.hasNext())
        return "[]";

    StringBuilder sb = new StringBuilder();
    sb.append('[');
    for (;;) {
        E e = i.next();
        sb.append(e == this ? "(this Collection)" : e);
        if (! i.hasNext())
            return sb.append(']').toString();
        sb.append(", ");
    }
}

I agree that using your while code looks much nicer than this though.

And to quote Campbell: never use "== true" or "== false". You risk using an assignment if you omit one of the = symbols.

[Bhagat Singh Rawat]

Ernest Friedman-Hill wrote:Note that there's a very easy way to double the performance of method #2 -- but I'll leave that to your imagination...

Humm, I think that is out of reach . Please tell me how can I do that.

[Brian Legg]

I would imagine is has something to do with you calling "fruits.toString()" 2 times in the same method call. A doubling of performance would occur if you could somehow omit one

[Brian Legg]

wow... took me 4 tries to post that. The boards are really acting up today

[Bhagat Singh Rawat]

Brian Legg wrote:I would imagine is has something to do with you calling "fruits.toString()" 2 times in the same method call. A doubling of performance would occur if you could somehow omit one

You mean if I change my code as

String str = fruits.toString();
str = str.substring(1, str.length - 1);

would be faster than the earlier code?

[David Marco]

fruits.toString().substring(1, fruits.toString().length - 1);

above code appears to create three string objects "from side to side" whereas

1. String str = fruits.toString();
2. str = str.substring(1, str.length - 1);

creates two Strings. Also the first solution leaves all the three objects unreferenced on the heap (a memory wast?) whereas the last solution only leaves one unreferenced String.

Correct me if I'm wrong.

[Rob Spoor]

David Marco wrote:

fruits.toString().substring(1, fruits.toString().length - 1);

above code appears to create three string objects "from side to side" whereas

1. String str = fruits.toString();
2. str = str.substring(1, str.length - 1);

creates two Strings. Also the first solution leaves all the three objects unreferenced on the heap (a memory wast?) whereas the last solution only leaves one unreferenced String.

Correct me if I'm wrong.

Technically you are right. However, the String returned by String.substring shares the string data (a char[]) with the original String. So yes, there is a little "waist", but that amounts to one object reference with a few primitive fields. Nothing to worry about unless resources are scarce.