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)
You can solve this using the recursive flatMap chain. First as we need to move back and forth by the map values, it’s better to copy them to the ArrayList (this is not the deep copy, in your case it’s ArrayList of 3 elements only, so the additional memory usage is low). Second, to maintain … Read more
If you don’t care about having duplicates, or maintaining indexes, then you can concatenate the two arrays in one line with: NEW=(“${OLD1[@]}” “${OLD2[@]}”) Full example: Unix=(‘Debian’ ‘Red hat’ ‘Ubuntu’ ‘Suse’ ‘Fedora’ ‘UTS’ ‘OpenLinux’); Shell=(‘bash’ ‘csh’ ‘jsh’ ‘rsh’ ‘ksh’ ‘rc’ ‘tcsh’); UnixShell=(“${Unix[@]}” “${Shell[@]}”) echo ${UnixShell[@]} echo ${#UnixShell[@]} Credit: http://www.thegeekstuff.com/2010/06/bash-array-tutorial/
>>> list(itertools.product(*arrays)) [(-1, -2, -3), (-1, -2, 3), (-1, 2, -3), (-1, 2, 3), (1, -2, -3), (1, -2, 3), (1, 2, -3), (1, 2, 3)] This will feed all the pairs as separate arguments to product, which will then give you the cartesian product of them. The reason your version isn’t working is that … Read more
The ndgrid function almost gives the answer, but has one caveat: n output variables must be explicitly defined to call it. Since n is arbitrary, the best way is to use a comma-separated list (generated from a cell array with ncells) to serve as output. The resulting n matrices are then concatenated into the desired … Read more
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
Edit: Previous solutions for two sets removed. See edit history for details. Here is a way to do it recursively for an arbitrary number of sets: public static Set<Set<Object>> cartesianProduct(Set<?>… sets) { if (sets.length < 2) throw new IllegalArgumentException( “Can’t have a product of fewer than two sets (got ” + sets.length + “)”); return … Read more
Here’s a solution I wouldn’t be ashamed to show. Rationale Assume that we have an input array $input with N sub-arrays, as in your example. Each sub-array has Cn items, where n is its index inside $input, and its key is Kn. I will refer to the ith item of the nth sub-array as Vn,i. … Read more
If I understand the question, you want the Cartesian Product of n sets of puppies. It is easy to get the Cartesian Product if you know at compile time how many sets there are: from p1 in dog1.Puppies from p2 in dog2.Puppies from p3 in dog3.Puppies select new {p1, p2, p3}; Suppose dog1 has puppies … Read more