coordinates

So here is my problem, I have to make a picture for my CS class and it is really frustrating estimating in turtle. I planed to use .onclick() to show me to position. import turtle as t def getPos(x,y): print("(", x, "," ,y,")") return def main(): t.onclick(getPos) t.mainloop() main() The turtle documentation seems to say that the onclick will pass the coordinates in a function that takes in two variables. http://docs.python.org/3.1/library/turtle.html#turtle.onclick NOTE: It works when I click on the arrow head, but thats not what I want. I want to click some other position on the screen to

Is there a way to use FQL to select all my friends Checkins that are in range of a specified location? I can use FQL to return my friends checkin coords: https://api.facebook.com/method/fql.query?access_token={0}&query=SELECT coords FROM checkin WHERE author_uid IN (SELECT uid2 FROM friend WHERE uid1 = me()) but how can I add to the above query to do something like... Where coords Within 10 miles of [(New York City, NY) | (long, lat) | (user.current_location)] Any help would be greatly appreciated. Thanks in advance. EDIT (3/9 3:05pm est): I'm mainly just looking for an end result - doesn't

I have a set of coordinates that I turn into an svg path (using cubic beziers to make it smooth). When I apply a certain stroke width, I get the following result (the blue dots are my coordinates) What I am interested in is to get a path that runs around the gray shape (like: pick any point on the gray/white border, and round around the shape until you're back at the starting point). How would I go about computing such a path? for reference, this is my svg info: <g> <title>number 3</title> <path d="m238,50c5.67569,-1.01351 11.8327,-3.8229 20.92029,-2.14724c8.68106,0.69732 14.21173,4.90255 18

I have to get the real world coordinates (x, y, z) using Kinect. Actually, I want the x, y, z distance (in meters) from Kinect. I have to get these coordinates from a unique object (e.g. a little yellow box) in the scenario, colored in a distinct color. Here you can see an example of the scenario I want the distance (x, y, z in meters) of the yellow object in the shelf. Note that is not required a person (skeleton) in the scenario. First of all, I would like to know if it is possible and simple to do? So, I would appreciate if you send some links/code that could help me doing this task. Hayko

I would like to perform blinear interpolation using python. Example gps point for which I want to interpolate height is: B = 54.4786674627 L = 17.0470721369 using four adjacent points with known coordinates and height values: n = [(54.5, 17.041667, 31.993), (54.5, 17.083333, 31.911), (54.458333, 17.041667, 31.945), (54.458333, 17.083333, 31.866)] z01 z11 z z00 z10 and here's my primitive attempt: import math z00 = n[0][2] z01 = n[1][2] z10 = n[2][2] z11 = n[3][2] c = 0.016667 #grid spacing x0 = 56 #latitude of origin of grid y0 = 13 #longitude of origin of grid i = math.floor((L-y0)/c) j =

I want to implement parallel coordinates for my muldimensional result. Does anyone have a good link to its implementation in matlab or R? Furthermore, are there any suggestions regarding the best tool to use for producing the parallel coordinates? VitoshKa R solution lattice package comes with R and includes parallel function: parallel(~iris[1:4] | Species, iris) ggplot2 is also your friend here: D <- data.frame(Gain = rnorm(20), Trader = factor(LETTERS[1:4]), Day = factor(rep(1:5, each = 4))) ggplot(D) + geom_line(aes(x = Trader, y = Gain, group = Day, color = Day)) lattice and ggplot require

I am trying to roll up 19 lines of code into a single for loop, but I am feeling a bit stumped. The reason I ask, is because I want the grid to be other sizes instead of 5. In Main::drawHexGridAdvanced() , I am trying to deduce the similarities between each line as opposed to Main::drawHexGridBasic() where I am hard-coding values. I am not sure how to determine the start of the x for each column in each row, because the pattern for n == 5 is 0, -1 -2 -2 -2 after that each consecutive column is just incremented except when the loop reaches the halfway point... Information and Understanding `n`