Is there a concurrent List in Java’s JDK?

ConcurrentLinkedQueue If you don’t care about having index-based access and just want the insertion-order-preserving characteristics of a List, you could consider a java.util.concurrent.ConcurrentLinkedQueue. Since it implements Iterable, once you’ve finished adding all the items, you can loop over the contents using the enhanced for syntax: Queue<String> globalQueue = new ConcurrentLinkedQueue<String>(); //Multiple threads can safely call … Read more

Add list to set

Adding the contents of a list Use set.update() or the |= operator: >>> a = set(‘abc’) >>> a {‘a’, ‘b’, ‘c’} >>> xs = [‘d’, ‘e’] >>> a.update(xs) >>> a {‘e’, ‘b’, ‘c’, ‘d’, ‘a’} >>> xs = [‘f’, ‘g’] >>> a |= set(xs) >>> a {‘e’, ‘b’, ‘f’, ‘c’, ‘d’, ‘g’, ‘a’} Adding the … Read more

Convert a list with strings all to lowercase or uppercase

It can be done with list comprehensions >>> [x.lower() for x in [“A”, “B”, “C”]] [‘a’, ‘b’, ‘c’] >>> [x.upper() for x in [“a”, “b”, “c”]] [‘A’, ‘B’, ‘C’] or with the map function >>> list(map(lambda x: x.lower(), [“A”, “B”, “C”])) [‘a’, ‘b’, ‘c’] >>> list(map(lambda x: x.upper(), [“a”, “b”, “c”])) [‘A’, ‘B’, ‘C’]

Find object in list that has attribute equal to some value (that meets any condition)

next((x for x in test_list if x.value == value), None) This gets the first item from the list that matches the condition, and returns None if no item matches. It’s my preferred single-expression form. However, for x in test_list: if x.value == value: print(“i found it!”) break The naive loop-break version, is perfectly Pythonic — … Read more

Why doesn’t list have safe “get” method like dictionary?

Ultimately it probably doesn’t have a safe .get method because a dict is an associative collection (values are associated with names) where it is inefficient to check if a key is present (and return its value) without throwing an exception, while it is super trivial to avoid exceptions accessing list elements (as the len method … Read more

How to empty a list?

This actually removes the contents from the list, but doesn’t replace the old label with a new empty list: del lst[:] Here’s an example: lst1 = [1, 2, 3] lst2 = lst1 del lst1[:] print(lst2) For the sake of completeness, the slice assignment has the same effect: lst[:] = [] It can also be used … Read more

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