distance

I have an input of 36,742 points which means if I wanted to calculate the lower triangle of a distance matrix (using the vincenty approximation) I would need to generate 36,742*36,741*0.5 = 1,349,974,563 distances. I want to keep the pair combinations which are within 50km of each other. My current set-up is as follows shops= [[id,lat,lon]...] def lower_triangle_mat(points): for i in range(len(shops)-1): for j in range(i+1,len(shops)): yield [shops[i],shops[j]] def return_stores_cutoff(points,cutoff_km=0): below_cut = [] counter = 0 for x in lower_triangle_mat(points): dist_km = vincenty(x[0]

I've got an sqlite db with long and lat of shops and I want to find out the closest 5 shops. So the following code works fine. if(sqlite3_prepare_v2(db, sqlStatement, -1, &compiledStatement, NULL) == SQLITE_OK) { while (sqlite3_step(compiledStatement) == SQLITE_ROW) { NSString *branchStr = [NSString stringWithUTF8String:(char *)sqlite3_column_text(compiledStatement, 0)]; NSNumber *fLat = [NSNumber numberWithFloat:(float)sqlite3_column_double(compiledStatement, 1)]; NSNumber *fLong = [NSNumber numberWithFloat:(float)sqlite3_column_double(compiledStatement, 2)]; NSLog(@"Address %@, Lat = %@,

OK, so I've written most of a program that will allow me to determine if two circles overlap. I have no problems whatsoever with my program aside from one issue: the program won't accept the code I've written for the distance between the two center points. I can figure out the if/else logic to tell the user what happens depending on the value of distance later, but I want to know what's wrong now. Eclipse, the program I'm coding on, is telling me that distance should be resolved to an array, but I've already told you that it's an int. Here is my code: package circles; import java.util.Scanner;

I have a hexagon grid: with template type coordinates T. How I can calculate distance between two hexagons? For example: dist((3,3), (5,5)) = 3 dist((1,2), (1,4)) = 2 First apply the transform (y, x) |-> (u, v) = (x, y + floor(x / 2)). Now the facial adjacency looks like 0 1 2 3 0*-*-*-* |\|\|\| 1*-*-*-* |\|\|\| 2*-*-*-* Let the points be (u1, v1) and (u2, v2). Let du = u2 - u1 and dv = v2 - v1. The distance is if du and dv have the same sign: max(|du|, |dv|), by using the diagonals if du and dv have different signs: |du| + |dv|, because the diagonals are unproductive In Python: def dist(p1,

Given a quaternion value, I would like to find its nearest neighbour in a set of quaternions. To do this, I clearly need a way to compare the "distance" between two quaternions. What distance representation is needed for such a comparison and how is it computed? Thanks, Josh Olhovsky Is your quaternion just a point in 3D space with an orientation? Then the distance between two quaternions x1,y1,z1,w1 and x2,y2,x2,w2 is given by: distance = sqrt((x1-x2)^2 + (y1-y2)^2 + (z1-z2)^2) , assuming that the w component is used for orientation. I.e. this is the same as the distance between two 3D points