java BlockingQueue does not have a blocking peek?

You could use a LinkedBlockingDeque and physically remove the item from the queue (using takeLast()) but replace it again at the end of the queue if processing fails using putLast(E e). Meanwhile your “producers” would add elements to the front of the queue using putFirst(E e). You could always encapsulate this behaviour within your own … Read more

Can I get an item from a PriorityQueue without removing it yet?

If a is a PriorityQueue object, You can use a.queue[0] to get the next item: from queue import PriorityQueue a = PriorityQueue() a.put((10, “a”)) a.put((4, “b”)) a.put((3,”c”)) print(a.queue[0]) print(a.queue) print(a.get()) print(a.queue) print(a.get()) print(a.queue) output is : (3, ‘c’) [(3, ‘c’), (10, ‘a’), (4, ‘b’)] (3, ‘c’) [(4, ‘b’), (10, ‘a’)] (4, ‘b’) [(10, ‘a’)] but … Read more

Peeking in a heap in python

Yes, you can make this assumption, because it is stated in the documentation: Heaps are arrays for which heap[k] <= heap[2*k+1] and heap[k] <= heap[2*k+2] for all k, counting elements from zero. For the sake of comparison, non-existing elements are considered to be infinite. The interesting property of a heap is that heap[0] is always … Read more

How to look ahead one element (peek) in a Python generator?

For sake of completeness, the more-itertools package (which should probably be part of any Python programmer’s toolbox) includes a peekable wrapper that implements this behavior. As the code example in the documentation shows: >>> p = peekable([‘a’, ‘b’]) >>> p.peek() ‘a’ >>> next(p) ‘a’ However, it’s often possible to rewrite code that would use this … Read more

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