You can do this pretty straightforwardly with an implicit class and a for-comprehension in Scala 2.10: implicit class Crossable[X](xs: Traversable[X]) { def cross[Y](ys: Traversable[Y]) = for { x <- xs; y <- ys } yield (x, y) } val xs = Seq(1, 2) val ys = List(“hello”, “world”, “bye”) And now: scala> xs cross ys … Read more
Simplifying Amro’s answer, you could use this: %// Sample data x = ‘ABCD’; %// Set of possible letters K = 3; %// Length of each permutation %// Create all possible permutations (with repetition) of letters stored in x C = cell(K, 1); %// Preallocate a cell array [C{:}] = ndgrid(x); %// Create K grids of … Read more
You could create an extension method like the following: public static class EnumerableExtensions { public static IEnumerable<TValue []> Permutations<TKey, TValue>(this IEnumerable<TKey> keys, Func<TKey, IEnumerable<TValue>> selector) { var keyArray = keys.ToArray(); if (keyArray.Length < 1) yield break; TValue [] values = new TValue[keyArray.Length]; foreach (var array in Permutations(keyArray, 0, selector, values)) yield return array; } static … Read more
The complexity of cartesian product is O(n2), there is no shortcut. In specific cases, the order you iterate your two axis is important. For example, if your code is visiting every slot in an array — or every pixel in an in image — then you should try to visit the slots in natural order. … Read more
I think you want the Cartesian product, not a permutation, in which case itertools.product can help: >>> from itertools import product >>> d = {‘Color’: [‘Red’, ‘Yellow’], ‘Size’: [‘Small’, ‘Medium’, ‘Large’]} >>> [dict(zip(d, v)) for v in product(*d.values())] [{‘Color’: ‘Red’, ‘Size’: ‘Small’}, {‘Color’: ‘Red’, ‘Size’: ‘Medium’}, {‘Color’: ‘Red’, ‘Size’: ‘Large’}, {‘Color’: ‘Yellow’, ‘Size’: ‘Small’}, {‘Color’: … Read more
You are looking for the Cartesian Product. In mathematics, a Cartesian product (or product set) is the direct product of two sets. In your case, this would be {1, 2, 3, 4, 5, 6} x {1, 2, 3, 4, 5, 6}. itertools can help you there: import itertools x = [1, 2, 3, 4, 5, … Read more
Use itertools.product, which has been available since Python 2.6. import itertools somelists = [ [1, 2, 3], [‘a’, ‘b’], [4, 5] ] for element in itertools.product(*somelists): print(element) This is the same as: for element in itertools.product([1, 2, 3], [‘a’, ‘b’], [4, 5]): print(element)
I’ve written a solution that doesn’t require you to fill up a large collection in memory. Unfortunately, the code required is hundreds of lines long. You may have to wait until it appears in the Guava project (https://github.com/google/guava), which I hope will be by the end of the year. Sorry. 🙁 Note that you may … Read more
Use itertools.product, which has been available since Python 2.6. import itertools somelists = [ [1, 2, 3], [‘a’, ‘b’], [4, 5] ] for element in itertools.product(*somelists): print(element) This is the same as: for element in itertools.product([1, 2, 3], [‘a’, ‘b’], [4, 5]): print(element)