coordinates

问题 I am developing a Spring app that has to search for persons using GPS coordinates. For each in the DB person I have his latitude and his longitude. The client pass a point and a max distance to the server which has to return all the clients around this point in this distance. The problem is that the database contains about 300k persons, so getting all the persons and iterating through this list to chck if they are near the point is really slow. Any idea to speed up this search ? 回答1: The best

问题 I want to do a search by coordinates, i.e. I want to have a function that works like this: function getLocationsInCircle($lat, $long, $minDist, $maxDist){ //return all the places that are at least $minDist //kilometers away and no more than $maxDist kilometers away } I have a "location" table that stores all location Ids and their latitude and longitude. The haversine formula is good enough for what I want to do 6371 * ACOS(SIN(RADIANS( $lat )) * SIN(RADIANS( latitude )) + COS(RADIANS( $lat )

问题 I have seen suggestions for this using JS but not Swift. How can I limit panning to the edges of an overlay image using Google Maps SDK for iOS with swift? I have a map overlay image covering a specific area, and only want the user to be able to pan to the edges of this overlay image. I thought this would be very simple, but apparently not! So if they pan right, and hit the east edge of the overlay image, they can't pan any further in that direction. So far, in the mapController, I have drawn

问题 I'm having trouble converting my OpenGL scene coordiates to my screen coordinates. I thought I needed to multiply my coordinates with the modelview matrix then with the projection matrix to get the ndc. But i'm getting weird coordinates. Here is my piece of code GLKVector3 coor = GLKVector3Make(point.x, point.y, 0); GLKMatrix4 modelview = GLKMatrix4MakeWithArray(glProjectionMatrix); GLKMatrix4 projetion = GLKMatrix4MakeWithArray(modelViewMatrix.data); GLKVector3 eyeCoor =

问题 Closed. This question is off-topic. It is not currently accepting answers. Want to improve this question? Update the question so it's on-topic for Stack Overflow. Closed 6 years ago . I have two points, lets say x1,y1 and x2,y2. and i have one more point x3,y3, here is the question, i need to check is x3,y3 coordinate occurs in the line between x1,y1 and x2,y2 or not occurs in javascript. Thanks in advance. 回答1: Try this: var point_a = [0, 0]; var point_b = [100, 200]; var coord = [50, 100];

问题 How can I calculate distance between two world map coordinates (latitude and longitude) using MATLAB R2015a (in meters)? 回答1: If you don't have access to the MATLAB Mapping toolbox then a simple approximation is to use the Haversine formula. Here is an excerpt from the link: The haversine formula is an equation important in navigation, giving great-circle distances between two points on a sphere from their longitudes and latitudes. It is a special case of a more general formula in spherical

问题 I've got following problem: I styled a TableViews Cells into Rectangles. Now I want to add Objects with an X and Y Coordinate in it ( obj.getAblaufX() and obj.getAblaufY() ). How can I add them correctly in JavaFX TableView? My Current Code: ObservableList<Event> eventList = loadEvents(); TableView<Event> tableView = new TableView<Event>(); tableView.getSelectionModel().setCellSelectionEnabled(true); tableView.getSelectionModel().setSelectionMode(SelectionMode.SINGLE); tableView.setItems

问题 I have 2 lists of coordinates of points in an image. Say, List1 = [[2,3],[4,5],[10,20],[45,60]] List2 = [[100,50],[65,48],[58,32],[98,45]...............[655,254],[232,545]] dist = 20 List1 would have 5 or 6 elements. List2 could have more than 1000 elements. I want to generate a list3 in which I have only the coordinates from List2 whose euclidean distance from all the points in List1 is more than dist=20 . Basically my aim is to remove all points from List2 which are near to points in List1

问题 I have 2 lists of coordinates of points in an image. Say, List1 = [[2,3],[4,5],[10,20],[45,60]] List2 = [[100,50],[65,48],[58,32],[98,45]...............[655,254],[232,545]] dist = 20 List1 would have 5 or 6 elements. List2 could have more than 1000 elements. I want to generate a list3 in which I have only the coordinates from List2 whose euclidean distance from all the points in List1 is more than dist=20 . Basically my aim is to remove all points from List2 which are near to points in List1

问题 I am doing some operations with vectors and physics in PyGame and the default coordinate system is inconvenient for me. Normally the (0, 0) point is at the top-left corner, but I would rather that the origin were at the bottom-left corner. I would rather change the coordinate system than converting every single thing I have to draw. Is it possible to change the coordinate system in PyGame to make it work like this? 回答1: Unfortunately, pygame does not provide any such functionality. The