Fetching Collections is a difficult operation. It has many side effects (as you realized, when there are fetched more collections). But even with fetching one collection, we are loading many duplicated rows. In general, for collections loading, I would suggest to use the batch processing. This will execute more SQL queries… but not so much, … 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
In recent versions of Pandas (>= 1.2) this is built into merge so you can do: from pandas import DataFrame df1 = DataFrame({‘col1′:[1,2],’col2’:[3,4]}) df2 = DataFrame({‘col3′:[5,6]}) df1.merge(df2, how=’cross’) This is equivalent to the previous pandas < 1.2 answer but is easier to read. For pandas < 1.2: If you have a key that is repeated … Read more
In newer version of numpy (>1.8.x), numpy.meshgrid() provides a much faster implementation: @pv’s solution In [113]: %timeit cartesian(([1, 2, 3], [4, 5], [6, 7])) 10000 loops, best of 3: 135 µs per loop In [114]: cartesian(([1, 2, 3], [4, 5], [6, 7])) Out[114]: array([[1, 4, 6], [1, 4, 7], [1, 5, 6], [1, 5, 7], … 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
A canonical cartesian_product (almost) There are many approaches to this problem with different properties. Some are faster than others, and some are more general-purpose. After a lot of testing and tweaking, I’ve found that the following function, which calculates an n-dimensional cartesian_product, is faster than most others for many inputs. For a pair of approaches … Read more
Cartesian product of multiple arrays in JavaScript
itertools.product Available from Python 2.6. import itertools somelists = [ [1, 2, 3], [‘a’, ‘b’], [4, 5] ] for element in itertools.product(*somelists): print(element) Which is the same as, for element in itertools.product([1, 2, 3], [‘a’, ‘b’], [4, 5]): print(element)