Two easy ways to accomplish just the filtering are: Using filter: names = filter(lambda name: name[-5:] != “Smith”, names) Using list comprehensions: names = [name for name in names if name[-5:] != “Smith”] Note that both cases keep the values for which the predicate function evaluates to True, so you have to reverse the logic … Read more
You can use an enhanced for loop: Set<String> set = new HashSet<String>(); //populate set for (String s : set) { System.out.println(s); } Or with Java 8: set.forEach(System.out::println);
Two options that don’t require copying the whole set: for e in s: break # e is now an element from s Or… e = next(iter(s)) But in general, sets don’t support indexing or slicing.
You need to create an alternation group out of the set with String.join(“|”, toDelete) and use as line = line.replaceAll(“\\b(?:”+String.join(“|”, toDelete)+”)\\b”, “”); The pattern will look like \b(?:end|something)\b See the regex demo. Here, (?:…) is a non-capturing group that is used to group several alternatives without creating a memory buffer for the capture (you do … Read more
You could create a dictionary from your elements, with whatever you wanted to be unique as the key, then extracting the values. This works for anything where the ‘unique’ sub-element is hashable. Integers are hashable: def unique_by_key(elements, key=None): if key is None: # no key: the whole element must be unique key = lambda e: … Read more
The fastest way is by using itertools, especially chain and combinations: >>> from itertools import chain, combinations >>> i = set([1, 2, 3]) >>> for z in chain.from_iterable(combinations(i, r) for r in range(len(i)+1)): print z () (1,) (2,) (3,) (1, 2) (1, 3) (2, 3) (1, 2, 3) >>> If you need a generator just … Read more
Here is a solution inspired by MapReduce: Simplified Data Processing on Large Clusters, which can therefore be written in a distributed way if you want. Map all of your elements in sets to pairs [element, set]. Group this list by element. You can just sort and pull out elements. Or you can create a hash … Read more
If you put your objects in a TreeSet, you need to either provide an implementation of the Comparator interface in the constructor, or you need your objects to be of a class that implements Comparable. You said you implement compareTo from the Comparable interface, but in your comment you say that you didn’t, so am … Read more
Passing the streams in your example is never better than passing Lists: private static <T> Stream<T> cartesian(BinaryOperator<T> aggregator, List<T>… lists) { … } And use it like this: Stream<String> result = cartesian( (a, b) -> a + b, Arrays.asList(“A”, “B”), Arrays.asList(“K”, “L”), Arrays.asList(“X”, “Y”) ); In both cases you create an implicit array from varargs … Read more
One of the properties of NAN is that NAN != NAN, unlike all other numbers. However, the implementation of set first checks to see if id(x) matches the existing member at a hash index before it tries to insert a new one. If you have two objects with different ids that both have the value … Read more