38

I was wondering if in Java there is a function like the python range function.

range(4)

and it would return

[0,1,2,3]

This was an easy way to make for enhanced loops. It would be great to do this in Java because it would make for loops a lot easier. Is this possible?

Karl Knechtel
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Kacper Lubisz
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    Have you tried to write it yourself? It's not that hard – Barranka May 15 '13 at 16:15
  • Some nice solutions are also in this question: http://stackoverflow.com/questions/371026/shortest-way-to-get-an-iterator-over-a-range-of-integers-in-java – OndroMih Sep 01 '15 at 14:15
  • For a range of any `Comparable` s see [this](https://stackoverflow.com/a/50245738/3992939) answer – c0der May 09 '18 at 04:56

9 Answers9

71

Java 8 (2014) has added IntStream (similar to apache commons IntRange), so you don't need external lib now.

import java.util.stream.IntStream; 

IntStream.range(0, 3).forEachOrdered(n -> {
    System.out.println(n);
});

forEach can be used in place of forEachOrdered too if order is not important.

IntStream.range(0, 3).parallel() can be used for loops to run in parallel

Tinmarino
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zengr
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    Nice, but puts limitations on variables used in the lambda expression so might not be handy. – Arthur May 25 '22 at 14:40
24

Without an external library, you can do the following. It will consume significantly less memory for big ranges than the current accepted answer, as there is no array created.

Have a class like this:

class Range implements Iterable<Integer> {

    private int limit;

    public Range(int limit) {
        this.limit = limit;
    }

    @Override
    public Iterator<Integer> iterator() {
        final int max = limit;
        return new Iterator<Integer>() {

            private int current = 0;

            @Override
            public boolean hasNext() {
                return current < max;
            }

            @Override
            public Integer next() {
                if (hasNext()) {
                    return current++;   
                } else {
                    throw new NoSuchElementException("Range reached the end");
                }
            }

            @Override
            public void remove() {
                throw new UnsupportedOperationException("Can't remove values from a Range");
            }
        };
    }
}

and you can simply use it like this:

    for (int i : new Range(5)) {
        System.out.println(i);
    }

you can even reuse it:

    Range range5 = new Range(5);

    for (int i : range5) {
        System.out.println(i);
    }
    for (int i : range5) {
        System.out.println(i);
    }

As Henry Keiter pointed out in the comment below, we could add following method to the Range class (or anywhere else):

public static Range range(int max) {
    return new Range(max);
}

and then, in the other classes we can

import static package.name.Range.range;

and simply call

for (int i : range(5)) {
    System.out.println(i);
}
Community
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jlordo
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    +1 for actually writing out an Iterator for this! For fun: you could further wrap this in Python-like syntax with a simple getter method: `public static Range range(int max) { return new Range(max); }`, so your loop becomes `for (int i : range(5)) {}` – Henry Keiter May 15 '13 at 17:08
  • @HenryKeiter That's a nice suggestion, I like it so much I'll update my answer (and give you the credit). – jlordo May 15 '13 at 17:12
  • It is a nice answer, but not to the question. It simply implies that 'No, there is not a function like range, you must implement your own iterator'. Sorry, -1 for that. – OndroMih Sep 01 '15 at 14:06
15

Um... for (int i = 0; i < k; i++)? You don't have to write enhanced for loops all day, you know, although they are cool...

And just for the sake of argument:

for (int i : range(k)) char count: 22

for (int i = 0; i < k; i++) char count: 27

Discounting the implementation of range, it is pseudo even.

zw324
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  • Well this is the normal way to do it, but it would be a lot simpler if you could make an array for the loop to use. – Kacper Lubisz May 15 '13 at 16:17
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    This isn't what he's asking, though; he wants to know if there's an "easier" way. – Henry Keiter May 15 '13 at 16:19
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    @HenryKeiter:that's true, just trying to counter OP's argument: "It would be great to do this in java because it would makes for loops a lot easier". – zw324 May 15 '13 at 16:20
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    Well, I certainly can't disagree with you there. Seems like a lot of effort just to save a couple of characters (and ignore all conventions on the way), but I'm not here to judge what (anti-)idioms people might want to use :) – Henry Keiter May 15 '13 at 16:28
  • @ExotickBoyPl you can make somethink _like_ an array, just much more efficient, if you look at my answer ;) – jlordo May 15 '13 at 16:45
  • Just to be clear, range() is not a Java function, in case any idiots like me want to try it. :P – user124384 Sep 24 '15 at 02:43
11

Use Apache Commons Lang:

new IntRange(0, 3).toArray();

I wouldn't normally advocate introducing external libraries for something so simple, but Apache Commons are so widely used that you probably already have it in your project!

Edit: I know its not necessarily as simple or fast as a for loop, but its a nice bit of syntactic sugar that makes the intent clear.

Edit: See @zengr's answer using IntStream in Java 8 .

Community
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Zutty
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  • Well its not ,but I was expecting it to be, its still go to know – Kacper Lubisz May 15 '13 at 16:24
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    This will also be much slower than the traditional for loop. – Louis Wasserman May 15 '13 at 16:26
  • This solution is pretty inefficient - you need a big array for big ranges. [Extending AbstractList](http://stackoverflow.com/questions/371026/shortest-way-to-get-an-iterator-over-a-range-of-integers-in-java#answer-6828887) is the simplest efficient solution I can think of. – OndroMih Sep 01 '15 at 14:12
2

If you really, really want to obtain an equivalent result in Java, you'll have to do some more work:

public int[] range(int start, int end, int step) {
    int n = (int) Math.ceil((end-start)/(double)step);
    int[] arange = new int[n];
    for (int i = 0; i < n; i++)
        arange[i] = i*step+start;
    return arange;
}

Now range(0, 4, 1) will return the expected value, just like Python: [0, 1, 2, 3]. Sadly there isn't a simpler way in Java, it's not a very expressive language, like Python.

Óscar López
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  • The implementation has some flaws. Run `int[] range = range(5, 33, 5); System.out.println(Arrays.toString(range));` – jlordo May 15 '13 at 16:47
2

Its not available that true. But you make a static method and use it -

public static int[] range(int index){
    int[] arr = new int[index];
    for(int i=0;i<index;i++){
        arr[i]=i;
    }
    return arr;
}
Subhrajyoti Majumder
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0

There's no Java equivalent to the range function, but there is an enhanced for-loop:

for (String s : strings) {
    // Do stuff
}

You could also roll your own range function, if you're really attached to the syntax, but it seems a little silly.

public static int[] range(int length) {
    int[] r = new int[length];
    for (int i = 0; i < length; i++) {
        r[i] = i;
    }
    return r;
}

// ...

String s;
for (int i : range(arrayOfStrings.length)) {
    s = arrayOfStrings[i];
    // Do stuff
}
Henry Keiter
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  • That will create a **huge** array, if you want a range to `Integer.MAX_VALUE` ;) – jlordo May 15 '13 at 16:48
  • @jlordo Well that's certainly true ;-) But the same is true of the `range` function in Python (2), and I felt like writing up an Iterator was overkill for just trying to avoid writing a basic loop (+1 to your answer for actually doing it though, lol). – Henry Keiter May 15 '13 at 17:04
  • certainly overkill for a simple loop. Although you could modify the iterator to use all of the other options the Python `range()` method has and use much less memory ;) Thanks for +1 – jlordo May 15 '13 at 17:06
0

As far as I know, there's not an equivalent function in java. But you can write it yourself:

public static int[] range(int n) {
    int[] ans = new int[n];
    int i;
    for(i = 0; i < n; i++) {
        ans[i] = i;
    }
    return ans;
}
Barranka
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0

What you can do to substitute the range in python in java can be done with the following code. NOTE: I am not going off of your code, I am just writing a small piece of code and showing it to you.

in python you would do.. . .

if -2 <= x <= 10:
     print(x) 

in java you would substitute this range and do. . ..

if(x >= -2 && x <= 10){
System.out.println("x")
}

By doing the above code in java, you don't need a range, but you have the -2 <= x <=10 range and split it into x >= -2 and x <= 10. It means the same thing, but the one I explained in java may take the compiler a longer time to read. So if you are using python go with the former's code format, and if you are using java, use the latter's code format.

  • Welcome to Stackoverflow and congratulations on your first answer. Here we try to keep answers focused on the specific question asked -- it is not immediately clear how what you wrote relates to the specific question of whether there was an equivalent of `range(,,)` in Java. You may want to enhance your answer by linking it more directly to the specific question – piterbarg Nov 29 '20 at 00:32