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javafunctionfor-loop
Kacper Lubisz
Kacper Lubisz

Reputation: 800

For loop - like Python range function

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?

Upvotes: 40

Views: 102418

Answers (9)

Anvith
Anvith

Reputation: 1

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.

Upvotes: 0

zengr
zengr

Reputation: 38899

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

Upvotes: 74

Zutty
Zutty

Reputation: 5377

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 .

Upvotes: 11

jlordo
jlordo

Reputation: 37813

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);
}

Upvotes: 24

&#211;scar L&#243;pez
&#211;scar L&#243;pez

Reputation: 236004

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.

Upvotes: 2

Henry Keiter
Henry Keiter

Reputation: 17168

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
}

Upvotes: 0

Barranka
Barranka

Reputation: 21047

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;
}

Upvotes: 0

Subhrajyoti Majumder
Subhrajyoti Majumder

Reputation: 41200

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;
}

Upvotes: 2

zw324
zw324

Reputation: 27190

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.

Upvotes: 15

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