To answer the question of “Why does Hoare partitioning work?”: Let’s simplify the values in the array to just three kinds: L values (those less than the pivot value), E values (those equal to the pivot value), and G value (those larger than the pivot value). We’ll also give a special name to one location … Read more
The segfault you are seeing is coming from next checking the range for you is an assertion in your Debug implementation to check against undefined behavior. The behavior of iterators and pointers is not defined beyond the their allocated range, and the “one past-the-end” element: Are iterators past the “one past-the-end” iterator undefined behavior? This … 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
[EDIT: This answer has been revised in accordance with the revision to the question.] The key to using jq to solve the problem is the -c command-line option, which produces output in JSON-Lines format (i.e., in the present case, one object per line). You can then use a tool such as awk or split to … 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
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
Making heavy use of some R idioms: > split(v, cumsum(c(1, diff(v) != 1))) $`1` [1] 1 $`2` [1] 3 4 5 $`3` [1] 9 10 $`4` [1] 17 $`5` [1] 29 30