It results in an error because according to the SQL standard, the JOIN keyword has higher precedence than the comma. The sticky point is that table aliases are not usable until after the corresponding table has been evaluated in the FROM clause. So when you reference e1 in your JOIN…ON expression, e1 doesn’t exist yet. … Read more
>>> 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
Ok, I wrote an example for myself reflecting your structure and this should work: int projectId = 1; // replace that with the id you want // required for the joins in QueryOver Project pAlias = null; Partner paAlias = null; PartnerCosts pcAlias = null; Address aAlias = null; Money mAlias = null; // Query … Read more
It’s quite similar to this answer I gave to another question: var combinations = from a in A from b in B from c in C orderby a, b, c select new List<int> { a, b, c }; var x = combinations.ToList(); For a variable number of inputs, now with added generics: var x = … Read more
First, I’ll show you a recursive version. // Cartesion product of vector of vectors #include <vector> #include <iostream> #include <iterator> // Types to hold vector-of-ints (Vi) and vector-of-vector-of-ints (Vvi) typedef std::vector<int> Vi; typedef std::vector<Vi> Vvi; // Just for the sample — populate the intput data set Vvi build_input() { Vvi vvi; for(int i = 0; … Read more
You need recursion: Let’s say all your lists are in lists, which is a list of lists. Let result be the list of your required permutations. You could implement it like this: void generatePermutations(List<List<Character>> lists, List<String> result, int depth, String current) { if (depth == lists.size()) { result.add(current); return; } for (int i = 0; … Read more
Sure thing. It is a bit tricky to do this with LINQ but certainly possible using only the standard query operators. UPDATE: This is the subject of my blog on Monday June 28th 2010; thanks for the great question. Also, a commenter on my blog noted that there is an even more elegant query than … Read more
This is very easy with list comprehensions. To get the cartesian product of the lists xs and ys, we just need to take the tuple (x,y) for each element x in xs and each element y in ys. This gives us the following list comprehension: cartProd xs ys = [(x,y) | x <- xs, y … 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
Consider this solution using the NDGRID function: sets = {[1 2], [1 2], [4 5]}; [x y z] = ndgrid(sets{:}); cartProd = [x(:) y(:) z(:)]; cartProd = 1 1 4 2 1 4 1 2 4 2 2 4 1 1 5 2 1 5 1 2 5 2 2 5 Or if you want … Read more