The best reusable option is to implement the interface Iterable and override the method iterator(). Here’s an example of a an ArrayList like class implementing the interface, in which you override the method Iterator(). import java.util.Iterator; public class SOList<Type> implements Iterable<Type> { private Type[] arrayList; private int currentSize; public SOList(Type[] newArray) { this.arrayList = newArray; … Read more
From this site: for i in $(seq 1 10); do echo $i done
To read only the first row of the csv file use next() on the reader object. with open(‘some.csv’, newline=””) as f: reader = csv.reader(f) row1 = next(reader) # gets the first line # now do something here # if first row is the header, then you can do one more next() to get the next … Read more
range is a class of immutable iterable objects. Their iteration behavior can be compared to lists: you can’t call next directly on them; you have to get an iterator by using iter. So no, range is not a generator. You may be thinking, “why didn’t they make it an iterator”? Well, ranges have some useful … Read more
Using Guava library: int size = Iterators.size(iterator); Internally it just iterates over all elements so its just for convenience.
Just use zip >>> l = [1, 7, 3, 5] >>> for first, second in zip(l, l[1:]): … print first, second … 1 7 7 3 3 5 If you use Python 2 (not suggested) you might consider using the izip function in itertools for very long lists where you don’t want to create a … Read more
I had the same question and found using a ListIterator worked. Similar to the test above: List<String> list = Arrays.asList(“zero”, “one”, “two”); ListIterator<String> iter = list.listIterator(); while (iter.hasNext()) { System.out.println(“index: ” + iter.nextIndex() + ” value: ” + iter.next()); } Make sure you call the nextIndex() before you actually get the next().
iter() is an iterator over a sequence. [x] * n produces a list containing n quantity of x, i.e. a list of length n, where each element is x. *arg unpacks a sequence into arguments for a function call. Therefore you’re passing the same iterator 3 times to zip(), and it pulls an item from … Read more
In a linked list, each element has a pointer to the next element: head -> item1 -> item2 -> item3 -> etc. To access item3, you can see clearly that you need to walk from the head through every node until you reach item3, since you cannot jump directly. Thus, if I wanted to print … Read more
scala> 10 to 1 by -1 res1: scala.collection.immutable.Range = Range(10, 9, 8, 7, 6, 5, 4, 3, 2, 1)