distance

问题 I am trying to find the nearest distance of locations in dataset 1 to dataset 2. Both data sets are different sizes. Ive looked into using the Haversine function but I'm unsure what I need to do after. 回答1: Since you have not provided a sample of your data, I am going to use the oregon.tract data set from the UScensus2000tract library as a reproducible example. Here is a solution based on fast data.table that I get from this other answer here. # load libraries library(data.table) library

问题 I have two point of a line like p1(a,b) and p2(c,d) my point is X(x,y) I've searched and find like here but it isn't php can anyone help me 回答1: DISCLAIMER: I assumed the JS code from the linked answer works. Below is my attempt to convert the javascript code from here to PHP. function sqr($x) { return $x * $x; } function dist2($v, $w) { return sqr($v->x - $w->x) + sqr($v->y - $w->y); } function distToSegmentSquared($p, $v, $w) { $l2 = dist2($v, $w); if ($l2 == 0) return dist2($p, $v); $t = (

问题 For my project I am using parts of the next code: link. To track objects of a specific color I implemented this method: My question is: How can I calculate the distance to the tracked colored objects? Thank you in advance! *The application calls the method for the left and right frame. This is not efficient... **I need to calculate detectedObject.Zcor DetectedObject Detect(IplImage *frame) { //Track object (left frame and right frame) //Calculate average position //Show X,Y,Z coordinate and

问题 I found a lot of solutions to calculate the distance between two locations, just like how to find the distance between two geopoints? or Calculating distance between two geographic locations or to be more specific, here's a code for the .distanceTo method: Location locationA = new Location("point A"); locationA.setLatitude(pointA.getLatitudeE6() / 1E6); locationA.setLongitude(pointA.getLongitudeE6() / 1E6); Location locationB = new Location("point B"); locationB.setLatitude(pointB

问题 I've centroids for customer A and customer B. Now I need to calculate distance between customer A and B into miles using centroids. How can I do it in Oracle? Right now, I'm using alteryx "distance" spatial tool for centroid distance calculation but need to convert it into oracle query. Thanks! 回答1: If you have Cartesian coordinates then the approximate distance if given by the Pythagorean theorem as already provided by Matthew. For Lat/Lon values you should use the Oracle build in SDO_GEOM

问题 I have a data frame with a mixture of numeric (15 fields) and categorical (5 fields) data. I can create a complete distance matrix of the numeric fields following create distance matrix using own calculation pandas I want to include the categorical fields as well. Using as template: import scipy from scipy.spatial import distance_matrix from scipy.spatial.distance import squareform from scipy.spatial.distance import pdist df2=pd.DataFrame({'col1':[1,2,3,4],'col2':[5,6,7,8],'col3':['cat','cat'

问题 Visual aid here Please see visual aid. Following the gif, there are two GUIs. The GUI on the left is scaling properly based on the cameras distance to the object. The GUI on the right is scaling, seen clearly in the 'Scene' view, but not even noticeable in game view. The application is for the Hololens as the users head controls the camera within the Unity scene. The problem lies in setting the objects scaling. Been racking my brain all day trying to replicate the type of scaling shown on the

问题 I have two lists. Each list has three lines. The coordinate system is (x,y,z) from up to down for each list. I tried to use array but it didn't work. Here are my codes. import numpy as np p1 = np.array([list(marker_11_x['11:-.X']), list(marker_11_y['11:-.Y']), list(marker_11_z['11:-.Z']) ]) p2 = np.array([list(original_x['13:-.X']), list(original_y['13:-.Y']), list(original_z['13:-.Z'])]) squared_dist = np.sum(((p1[0]-p2[0])**2+(p1[1]-p2[1] )**2+(p1[3]-p2[3] )**2), axis=0) dist = np.sqrt

问题 I'm going to build an app where the users can see points of interest in a predefined radius around their location. My first idea was to store the latitude and longitude of all POI's in a database and comparing the users location with the POI's location via SQL. The problem is the performance I think. If there are thousands of POI's and thousands of user requests with their location, it wouldn't be very economically or is this no problem for todays servers? My next approach was to divide the

问题 I need to find the minimum distance between patches in front of my agent to a certain patch (goal), in order to select the patch that would create the most optimal (shortest) path. The primitive distance only requires one argument so I can't use it as is for this function. 回答1: The distance primitive only requires one argument, yes, but it is a "patch or turtle primitive": it must be run in the context of a particular agent by "asking" it for its distance to another, so you can think of the