coordinates

问题 Is there a Python module where I can create objects with a geographical location coordinate (latitude and longitude), and query all the objects for ones which are within a 5km distance (i.e. radius) of a given coordinate? I've been trying to store the latitude and longitude as keys in dictionaries (as they're indexed by key) and use some distance finding algorithms to query them. But this feels like a horrible hack. Essentially something like PostGIS for PostgreSQL, but all within my Python

问题 i would like to retrieve the element offset starting from his own x center coordinates. how can i do it? Actually i can find the window offset of an element but it retrieves the coordinates from the border of the element like this: var _position = $(this).offset(); 回答1: You have to use offset() to get the top and left position, then add half of the height() and width() values to them. That gives the center coordinates. var $this = $(this); var offset = $this.offset(); var width = $this.width(

问题 Using C++ / OpenCV I've drawn a line on an image using cv::line and now I'm trying to extract an array of its coordinates. I've tried assigning the line to cv::Mat but I get an error stating I cannot convert from void to cv::Mat . Is there a simple way to obtain these coordinates? Thanks for the help! 回答1: You have at least a couple of options. Assuming that you know the two endpoints A and B of the line: 1) Draw the line with line(...) on an zero initialized mask of the same size of your

问题 I know there are many many questions about sorting javascript arrays by multiple values, but none of the answers solved my problem. I have an array of coordinates like: x | y -------- 10 20 12 18 20 30 5 40 100 2 How can I get the coordinate that is closest to the point of origin? 回答1: Calculate the distance of each point using Math.sqrt( Math.pow(x, 2) + Math.pow(y, 2) ); Take the result that's the lowest var points = [ {x: 10, y: 20}, {x: 12, y: 18}, {x: 20, y: 30}, {x: 5, y: 40}, {x: 100,

问题 I'm developing an web application (jsp/strtus2) which requires the GPS coordinates of a place. What I want to know is, whether there are any free web services that I can get GPS coordinates of a place by passing address or zip code to the server. Thanks in advance, Nuwan 回答1: Yahoo's Geocoder is easy to use. Basically change a few query strings and it outputs an easy to parse xml file. Google's had a few issues in our area with the coordinates being returned being incredibly inaccurate at

问题 How do I get the coordinates for finger tapping in UIView? (I would prefer not to use a big array of buttons) Thank you 回答1: There’re two ways to accomplish this. If you’ve already got a subclass of UIView that you’re using, you can just override the -touchesEnded:withEvent: method on that subclass, like this: - (void)touchesEnded:(NSSet *)touches withEvent:(UIEvent *)event { UITouch *aTouch = [touches anyObject]; CGPoint point = [aTouch locationInView:self]; // point.x and point.y have the

问题 I have N GPS coordinates with N distances given to an unknown position which I wish to determine. My first approach was to use just three points and trilateration, exactly as described here. This approach was already quite accurate (best error~5km), but I would like to improve this and increase the robustness. Because the given distances are not very accurate to begin with, I thought about using multiple measurements and multilateration. However, it turned out that this approach is by far

问题 I have simple UIWebView with loaded html. I want to show the PopoverView pointed to the selected text. I can get the menuFrame of the UIMenuController and use its rect, but the result isn't as accurate at the corners of the screen as I need. I can calculate the screen size and get menuFrame rect and then I can suppose where the selection may be found, but I want exactly know where is the selected text. Is it generally possible? 回答1: You can use the javascript function getBoundingClientRect()

问题 I want to get a simple solution to calculate the angle of a line (like a pointer of a clock). I have 2 points: cX, cY - the center of the line. eX, eY - the end of the line. The result is angle (0 <= a < 360). Which function is able to provide this value? 回答1: You want the arctangent: dy = ey - cy dx = ex - cx theta = arctan(dy/dx) theta *= 180/pi // rads to degs Erm, note that the above is obviously not compiling Javascript code. You'll have to look through documentation for the arctangent

问题 I'm building an application based around finding a "convenient meeting point" given a set of locations. Currently I'm defining "convenient" as "minimising the total travel distance". This is a different problem from finding the centroid as illustrated by the following example (using cartesian coordinates rather than latitude and longitude for convenience): A is at (0,0) B is at (0,0) C is at (0,12) The location of minimum total travel for these points is at (0,0) with total travel distance of