intersection

I have two object literals like so: var firstObject = { x: 0, y: 1, z: 2, a: 10, b: 20, e: 30 } var secondObject = { x: 0, y: 1, z: 2, a: 10, c: 20, d: 30 } I want to get the intersection of the keys these two object literals have like so: var intersectionKeys = ['x', 'y', 'z', 'a'] I can obviously do a loop and see if a key with the same name exists in the other object, but I am wondering if this would be a good case for some functional programming and map / filter / reduce usage? I myself have not done that much functional programming, but I have a feeling, that there could exist a clean and

On my machine (Quad core, 8gb ram), running Vista x64 Business, with Visual Studio 2008 SP1, I am trying to intersect two sets of numbers very quickly. I've implemented two approaches in C++, and one in C#. The C# approach is faster so far, I'd like to improve the C++ approach so its faster than C#, which I expect C++ can do. Here is the C# output: (Release build) Found the intersection 1000 times, in 4741.407 ms Here is the initial C++ output, for two different approaches (Release x64 build): Found the intersection (using unordered_map) 1000 times, in 21580.7ms Found the intersection (using

I'm calculating intersection of 2 sets of sorted numbers in a time-critical part of my application. This calculation is the biggest bottleneck of the whole application so I need to speed it up. I've tried a bunch of simple options and am currently using this: foreach (var index in firstSet) { if (secondSet.BinarySearch(index) < 0) continue; //do stuff } Both firstSet and secondSet are of type List. I've also tried using LINQ: var intersection = firstSet.Where(t => secondSet.BinarySearch(t) >= 0).ToList(); and then looping through intersection . But as both of these sets are sorted I feel there

I want to create different methods for a class called Multiset . I have all the required methods, but I'm unsure of how to write intersection, union, and subset methods. For intersection and union, my code starts like this: def intersect(var) x = Multiset.new end Here is an example: X = [1, 1, 2, 4] Y = [1, 2, 2, 2] then the intersection of X and Y is [1, 2] . Mike Lewis Utilizing the fact that you can do set operations on arrays by doing & (intersection), - (difference), and | (union). Obviously I didn't implement the MultiSet to spec, but this should get you started: class MultiSet attr

I have got a 2D array of cells with size 10x10, and many points that are pairs of floating point values, like: (1.6, 1.54), (4.53, 3.23). The pairs (x,y) are such that x<10 and y<10 Each cell takes points whose coordinates have the same integer part as the cell coordinates. So arr[3][7] will take points with x={3...3.99(9)} and y={7... 7.99(9)}, for example (3.5, 7.1) or (3.2, 7.6). Similarly (1.6, 1.54) is in arr[1][1] , (4.53, 3.23) is in arr[4][3], etc. Each point has got a designated place in the array that is easy to find, because I just have to cast x and y to int to get rid of the

I have two lists, one with old IDs and one with new IDs. I want to get items in common and items not common. The new_Items list has all new ones. The old_Items has the old ones. I suppose that when I calculate the ones in common plus the in new items list but not in old items list, I get the actual number of new items. Here is the code and the output. print(old_Items) print(new_Items) common = set(new_Items) & set(old_Items) not_common = set(new_Items) - set(old_Items) print(len(old_Items)) print(len(new_Items)) print(len(common)) print(len(not_common)) output ['312064913440', '312062038159',

Anyone have a simple algorithm for this? No need for rotation or anything. Just finding if a line segment made from two points intersects a square Eric This code should do the trick. It checks where the line intersects the sides, then checks if that is within the width of the square. The number of intesections is returned. float CalcY(float xval, float x0, float y0, float x1, float y1) { if(x1 == x0) return NaN; return y0 + (xval - x0)*(y1 - y0)/(x1 - x0); } float CalcX(float yval, float x0, float y0, float x1, float y1) { if(x1 == x0) return NaN; return x0 + (yval - y0)*(y1 - y0)/(x1 - x0); }