javascript,arrays,enumerate,cartesian,powerset

var obj = {"varA":[1,2,3], "varB":['good','bad'], "varC":[0,100], "varD":['low','med','high'] } // flatten the object into an array so it's easier to work with var obj2list = function(obj) { var list = []; for (var key in obj) { if (obj.hasOwnProperty(key)) { list.push({ name: key, val: obj[key] }); } } return list;...

When you are appending to the result, this will invalidate iterators when the reallocation occurs and reallocation occurs when you exceed vector capacity. for(vector<int> subSet:result){ subSet.push_back(i); result.push_back(subSet); } If you reserve enough space in the vector result at the beginning via reserve () member function, it should work. I didn't...

Power set of a set A is the set of all of the subsets of A. Not the most friendly definition in the world, but an example will help : Eg. for {1, 2}, the subsets are : {}, {1}, {2}, {1, 2} Thus, the power set is {{}, {1},...

algorithm,permutation,powerset

Algorithm to generate all possible permutations of a list? It's a complicated algorithm any way you want to do it, I think recursively is the fun way. The recursive way basically removes one element, calls the recursion on the shortened list, then returns the recursive result with the removed element...

For the specific case of a space of natural numbers, you want np.indices: >>> np.indices((4, 4)).reshape(2,-1).T array([[0, 0], [0, 1], [0, 2], [0, 3], [1, 0], [1, 1], [1, 2], [1, 3], [2, 0], [2, 1], [2, 2], [2, 3], [3, 0], [3, 1], [3, 2], [3, 3]]) (numpy actually...

c++,string,combinations,powerset,lexicographic

This is a simple recursive approach: #include <string> #include <iostream> using namespace std; void get( string str, string res ) { cout << res << endl; for( int i = 0; i < str.length(); i++ ) get( string(str).erase(i,1), res + str[i] ); } int main( int argc, char **argv) {...