They are very different. One set changes the set in place, while the other leaves the original set alone, and returns a copy instead. >>> s = {1, 2, 3} >>> news = s | {4} >>> s set([1, 2, 3]) >>> news set([1, 2, 3, 4]) Note how s has remained unchanged. >>> s.update({4}) … Read more
You could define three functions inBoth, inFirstOnly, and inSecondOnly which all take two lists as arguments, and return a list as can be understood from the function name. The main logic could be put in a common function operation that all three rely on. Here are a few implementations for that operation to choose from, … Read more
** in argument lists has a special meaning, as covered in section 4.7 of the tutorial. The dictionary (or dictionary-like) object passed with **kwargs is expanded into keyword arguments to the callable, much like *args is expanded into separate positional arguments.
Yes, Swift has the Set class. let array1 = [“a”, “b”, “c”] let array2 = [“a”, “b”, “d”] let set1:Set<String> = Set(array1) let set2:Set<String> = Set(array2) Swift 3.0+ can do operations on sets as: firstSet.union(secondSet)// Union of two sets firstSet.intersection(secondSet)// Intersection of two sets firstSet.symmetricDifference(secondSet)// exclusiveOr Swift 2.0 can calculate on array arguments: set1.union(array2) // … Read more
This question provides an idiom. You use one of the dicts as keyword arguments to the dict() constructor: dict(y, **x) Duplicates are resolved in favor of the value in x; for example dict({‘a’ : ‘y[a]’}, **{‘a’, ‘x[a]’}) == {‘a’ : ‘x[a]’}