Map<Boolean, List<Integer>> partitioned = set.stream().collect(Collectors.partitioningBy(x -> x%2 == 0)); The elements in partitioned.get(true) are even; the elements in partitioned.get(false) are odd. Unlike doing this using groupingBy, it is guaranteed that both true and false lists will be present in the map even if they are empty. (Not documented in Java 8, but it was true; … Read more
Here is an extension method that will do what you want: public static IEnumerable<List<T>> Partition<T>(this IList<T> source, Int32 size) { for (int i = 0; i < (source.Count / size) + (source.Count % size > 0 ? 1 : 0); i++) yield return new List<T>(source.Skip(size * i).Take(size)); } Edit: Here is a much cleaner version … Read more
Watch out: I believe the accepted answer is not quite right! I’m glad you ask this question, because the behavior of these similarly-named functions differs in important and unexpected ways that are not well documented in the official spark documentation. The first part of the accepted answer is correct: calling df.repartition(COL, numPartitions=k) will create a … Read more
Try this: def partition(pred, iterable): trues = [] falses = [] for item in iterable: if pred(item): trues.append(item) else: falses.append(item) return trues, falses Usage: >>> trues, falses = partition(lambda x: x > 10, [1,4,12,7,42]) >>> trues [12, 42] >>> falses [1, 4, 7] There is also an implementation suggestion in itertools recipes: from itertools import … Read more