128

Does Java have an equivalent to Python's range(int, int) method?

4
  • 1
    Do you mean a Python 2.x range which returns a list, or a Python 3.x range which returns an iterator (equivalent to the 2.x xrange)? The former is relatively simple to implement as others have done below, but the iterator version is a bit more tricky. Commented Sep 24, 2010 at 21:13
  • stackoverflow.com/questions/16570091/… Commented Aug 19, 2015 at 23:32
  • For a range of any Comparable s see this answer Commented May 9, 2018 at 4:55
  • For a range with stepping see this Commented Sep 23, 2019 at 10:08

14 Answers 14

261

Old question, new answer (for Java 8)

IntStream.range(0, 10).forEach(n -> System.out.println(n));

or with method references:

IntStream.range(0, 10).forEach(System.out::println);
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4 Comments

What package? how do I import it?
@Justin That is java 8 language feature.
the first example could just be: Intstream.range(0,10).forEach( n -> System.out.println(n));
30

Guava also provides something similar to Python's range:

Range.closed(1, 5).asSet(DiscreteDomains.integers());

You can also implement a fairly simple iterator to do the same sort of thing using Guava's AbstractIterator:

return new AbstractIterator<Integer>() {
  int next = getStart();

  @Override protected Integer computeNext() {
    if (isBeyondEnd(next)) {
      return endOfData();
    }
    Integer result = next;
    next = next + getStep();
    return result;
  }
};

4 Comments

I wish I had seen this before I implemented my own generic range. It's nice but still another reminder of how clunky Java can be compared to more functional languages.
Range#asSet is seems to have become deprecated. You now need to do this: ContiguousSet.create(Range.closed(low, high), DiscreteDomain.integers())
what library to i need to import to use this?
From Java 8, IntStream and LongStream have methods range and rangeClosed.
17

I'm working on a little Java utils library called Jools, and it contains a class Range which provides the functionality you need (there's a downloadable JAR).
Constructors are either Range(int stop), Range(int start, int stop), or Range(int start, int stop, int step) (similiar to a for loop) and you can either iterate through it, which used lazy evaluation, or you can use its toList() method to explicitly get the range list.

for (int i : new Range(10)) {...} // i = 0,1,2,3,4,5,6,7,8,9

for (int i : new Range(4,10)) {...} // i = 4,5,6,7,8,9

for (int i : new Range(0,10,2)) {...} // i = 0,2,4,6,8

Range range = new Range(0,10,2);
range.toList(); // [0,2,4,6,8]

Comments

16

Since Guava 15.0, Range.asSet() has been deprecated and is scheduled to be removed in version 16. Use the following instead:

ContiguousSet.create(Range.closed(1, 5), DiscreteDomain.integers());

Comments

16

You can use the following code snippet in order to get a range set of integers:

    Set<Integer> iset = IntStream.rangeClosed(1, 5).boxed().collect
            (Collectors.toSet());

Comments

15
public int[] range(int start, int stop)
{
   int[] result = new int[stop-start];

   for(int i=0;i<stop-start;i++)
      result[i] = start+i;

   return result;
}

Forgive any syntax or style errors; I normally program in C#.

2 Comments

given that Vivien Barousse beat you to an answer, why don't you delete yours to avoid any dup. Unless you really plan to nicely flesh it out of course.
They're similar; I think mine's a little more readable. His use of "length" is misleading, and I don't think his meets the Python spec (he includes the upper bound, which network-theory.co.uk/docs/pytut/rangeFunction.html says doesn't happen in Python). If you think one's a dupe, I believe you have sufficient reputation to deal with it yourself.
9
public int[] range(int start, int length) {
    int[] range = new int[length - start + 1];
    for (int i = start; i <= length; i++) {
        range[i - start] = i;
    }
    return range;
}

(Long answer just to say "No")

1 Comment

Also, see that "range" in python 3 and the preferred "xrange" in Python 2 return a "live" object that does not use up memory for each item it contains. That would be even bigger to implement in Java.
4

Java 9 - IntStream::iterate

Since Java 9 you can use IntStream::iterate and you can even customize the step. For example if you want int array :

public static int[] getInRange(final int min, final int max, final int step) {
    return IntStream.iterate(min, i -> i < max, i -> i + step)
            .toArray();
}

or List :

public static List<Integer> getInRange(final int min, final int max, final int step) {
    return IntStream.iterate(min, i -> i < max, i -> i + step)
            .boxed()
            .collect(Collectors.toList());
}

And then use it :

int[] range = getInRange(0, 10, 1);

Comments

3
IntStream.range(0, 10).boxed().collect(Collectors.toUnmodifiableList());

1 Comment

It would help if you explained how this code solves the problem.
2

Groovy's nifty Range class can be used from Java, though it's certainly not as groovy.

Comments

2

The "Functional Java" library allows to program in such a way to a limited degree, it has a range() method creating an fj.data.Array instance.

See:

Similarly the "Totally Lazy" library offers a lazy range method: http://code.google.com/p/totallylazy/

Comments

2

I know this is an old post but if you are looking for a solution that returns an object stream and don't want to or can't use any additional dependencies:

Stream.iterate(start, n -> n + 1).limit(stop);

start - inclusive stop - exclusive

Comments

1

If you mean to use it like you would in a Python loop, Java loops nicely with the for statement, which renders this structure unnecessary for that purpose.

2 Comments

You don't usually use it for a loop in python either. There's almost always a cleaner way to iterate.
Well, range is usually used in a for loop. But for loops are often used without range.
1

Java 8

private static int[] range(int start, int stop, int step) {
    int[] result = new int[(stop-start)%step == 0 ? (stop-start)/step : (stop-start)/step+1];
    int count = 0;
    Function<Integer, Boolean> condition = step > 0 ? (x) -> x < stop : (x) -> x > stop;
    for (int i = start; condition.apply(i); i += step) {
        result[count] = i;
        count++;
    }
    return result;
}

1 Comment

Welcome to StackOverflow. While this code may solve the question, including an explanation of how and why this solves the problem would really help to improve the quality of your post, and probably result in more up-votes. Remember that you are answering the question for readers in the future, not just the person asking now. Please edit your answer to add explanations and give an indication of what limitations and assumptions apply.

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