distance

问题 For some reason a friend and myself were talking about calculating the distance between yourself (laptop, phone, etc.) and an AP based of the info you get from the devices (RSSI, freq, SNR, etc...). So, after spending sometime researching about trilateration, triangulation, and free-space path loss. (with the help of some blog posts and wiki) I was able to get a distance in meters from the AP to my laptop and the results were a lot better then what I thought they would be. Whenever I'm in the

问题 If you are given say a list of 10 vectors, called A that represent different groups. Then you have a time series of vectors v1,v2,...,vn, each being a vector as well. I was wondering if there was a way to find the "closest" vector in A for each v1,v2,...,vn if you define some distance metric? Is there a quick way to do this besides for looping through and just comparing all entries? Edit: No I am not asking how to do k-means or something like that. 回答1: You can use the spatial KDtree in scipy

问题 I need to calculate the distance on a grid between 2 points. The movement allowed is horizontal and vertical as well diagonal to the next neighbor (so 45 degree rotations). So Manhattan distance is not an option. Also Euclidean distance is not an option cause then it does not move correct along the grid which can result in a to low value (as in the red line). I'm looking to get the distance as in the green line where it moves from cell to cell. It's preferred that the formula is fast. 回答1:

问题 I want to calculate a euclidean distance between points that the user enter,so as you can see here : static void Main(string[] args) { int numtest = int.Parse(Console.ReadLine()); int[,] points=new int[10,2]; for (int i = 0; i < numtest; i++) { Console.WriteLine("point " +(i+1).ToString()+" x: "); points[i, 0] = int.Parse(Console.ReadLine()); Console.WriteLine("point " + (i + 1).ToString() + " y: "); points[i, 1] = int.Parse(Console.ReadLine()); } } public float[] calculate(int[,] points) {

问题 I am trying to find the minimum distance from a point (x0,y0,z0) to a line joined by (x1,y1,z1) and (x2,y2,z2) using numpy or anything in python. Unfortunately, all i can find on the net is related to 2d spaces and i am fairly new to python. Any help will be appreciated. Thanks in advance! 回答1: StackOverflow doesn't support Latex, so I'm going to gloss over some of the math. One solution comes from the idea that if your line spans the points p and q , then every point on that line can be

问题 I am using geolocation to get the users current location and monitor it using the watchPosition method. However, is there a way of calculating the distance between the users starting position and current position? Below is my code: var x = document.getElementById("info"); function getLocation() { if(navigator.geolocation) { navigator.geolocation.watchPosition(showPosition, showError, { enableHighAccuracy: true, maximumAge: 60000, timeout: 27000 }) } else { x.innerHTML = "Geolocation is not

问题 I want to calculate the distance between several GPS points. I tried distm(c(lon1,lat1), c(lon2,lat2), fun = distHaversine) which worked for one point, but not for the columns in my data frame. So I tried as recommended here: Calculate distance between 2 lat longs But I do get different results for these two calculations: df <- read.table(sep=",", col.names=c("lat1", "lon1", "lat2", "lon2"),text=" 7.348687,53.36575,7.348940,53.36507 7.348940, 53.36507,7.350939,53.36484") # as recommended in

问题 I would like to build a distance matrix using Pearson correlation distance. I first tried the scipy.spatial.distance.pdist(df,'correlation') which is very fast for my 5000 rows * 20 features dataset. Since I want to build a recommender, I wanted to slightly change the distance, only considering features which are distinct for NaN for both users. Indeed, scipy.spatial.distance.pdist(df,'correlation') output NaN when it meets any feature whose value is float('nan'). Here is my code, df being my

问题 Im wondering if someone has built a raster of the continents of the world where each cell equals the distance of that cell cell to the nearest shore. This map would highlight the land areas that are most isolated inland. I would imagine this would simply rasterize a shapefile of the global boundaries and then calculate the distances. 回答1: You can do this with raster::distance , which calculates the distance from each NA cell to the closest non- NA cell. You just need to create a raster that

问题 My assignment is to create main class in which I initialize the value of any point to be at (0,0,0) and to be able to access and mutate all three values (x,y,z) individually. To do this I have used getters and setters. My next task is to create a method within my main class (which I shall call "distanceTo") that calculates the distance between two points. How do I go about creating the method " distanceTo " that calculates the distance between two points by taking in the x,y,z coordinates ? I