geo

I insert many polygons into MongoDB(2.4.8), and hope find the polygon a specified Point sits in. It seems a common question. But after reading all docs from google, I didn't get the result. So create this question. e.g. db.addr_poly.insert( { loc : { type : "Polygon" , coordinates : [ [ [ 0 , 0 ] , [ 3 , 6 ] , [ 6 , 1 ] , [ 0 , 0 ] ] ] } }) db.addr_poly.insert( { loc : { type : "Polygon" , coordinates : [ [ [ 0 , 0 ] , [ -3 , -6 ] , [ -6 , -1 ] , [ 0 , 0 ] ] ] } }) Now how to find the polygon which contains Point(1,1)? Anybody can help me? Thanks a lot! Use the $geoIntersects operator . It

目录 GEO指令 GEOADD GEODIST GEOPOP GEOHASH GEORADIUS GEORADIUSBYMEMBER 指令补充 删除操作 避免单集合数量过多 存储原理 GEOADD存储原理 GEOHASH精度问题 Redis 3.2版本新增GEO(地理位置)。 GEO指令 GEOADD 命令： GEOADD key longitude latitude member [longitude latitude member ...] 命令描述：将指定的地理空间位置（纬度、经度、名称）添加到指定的 key 中。 返回值：添加到sorted set元素的数目，但不包括已更新score的元素。 时间复杂度：O(log(N)) GEODIST 命令： GEODIST key member1 member2 [unit] 命令描述： 返回两个给定位置之间的距离。如果两个位置之间的其中一个不存在， 那么命令返回空值。指定单位的参数 unit 必须是以下单位的其中一个： m 米 km 千米 mi 英里 ft 英尺 时间复杂度：O(log(N)) GEOPOP 命令： GEOPOS key member [member ...] 命令描述：从key里返回所有给定位置元素的位置（经度和纬度）。 返回值：GEOPOS 命令返回一个数组， 数组中的每个项都由两个元素组成：

What is the maximum length (in kilometers or miles - but please specify) that one degree of latitude and longitude can have in the Earth surface? I'm not sure if I'm being clear enough, let me rephrase that. The Earth is not a perfect circle, as we all know, and a change of 1.0 in the latitude / longitude on the equator (or in Ecuador) can mean one distance while the same change at the poles can mean another completely different distance. I'm trying to shrink down the number of results returned by the database (in this case MySQL) so that I can calculate the distances between several points

Background / Concept: I want to create an application that can help user searching for specific things. Let say nearby restaurants. Here I want to include an offline map covering 1 km area around that restaurant, (its easy i can take snap of google map). Problem: is it possible to indicate current user position on that static image file? I have some idea that I note the Lat / Long of all four corners of that image, take user current position, do some calculation and show the current spot.. But I m not sure where to start and if there is already a good optimized solution. If I can show current

In the image below, you can see that there 3 lines drawn from one point (the black circle) to its 3 related points (). IMAGE QUESTION How to calculate latitude and longitude point between points along each line, using a percentage of the distance between the two points? For example, if I wanted to get the position to be able to draw additional circles along each line with a 20% difference? WHAT CODE I HAVE NOW var data = [ { "coords" : [ 53.409045, -2.985406 ]}, { "coords" : [ 53.408747, -2.982862 ]}, { "coords" : [ 53.407630, -2.984136 ]}, { "coords" : [ 53.407142, -2.986931 ]} ]; var pointA

I have a Spark dataframe in R as follows head(df) Lat1 Lng1 Lat2 Lng2 23.123 24.234 25.345 26.456 ... ... ... ... The DataFrame contains two points Latitude and Longitude I would like to calculate the Geo distance between the nodes in each row and add it to a new column. In R I am using distCosine function from geosphere library. df$dist = distCosine(cbind(df$lng1,df$lat1),cbind(df$lng2,df$lat2)) I am wondering how I should calculate it in SparkR. SparkR produces the following error, Error in as.integer(length(x) > 0L) : cannot coerce type 'S4' to vector of type 'integer' You cannot use

7.7、陆地与波浪演示程序 在本小节中，我们将展示如何构建“陆地与波浪”演示程序： 构建一个三角形栅格（grid），并通过将其中的顶点设置到不同的位置来创建 地形 再构建一个三角形栅格（grid），通过动态的改变其顶点高度来创建 波浪 此示例还会 针对不同的常量缓冲区来切换所使用的根描述符 ，这可以让我们不需要设置CBV描述符堆 7.7.1、生成栅格顶点 为了生成地形和波浪，我们首先要构建三角形栅格（grid），这里我们不进行详细的数学分析，直接给出结论： \[ 在xz平面上，第i行、第j列栅格顶点的坐标可以表示为：v = [-0.5w + j*dx, 0.0f, 0.5d - i*dz] \] 其中w为栅格的总宽度（宽），dx为每一个小四边形的宽度，d为栅格的总深度（长），dz为每一个小四边形的深度， （其实可以把栅格理解成一个由许多小四边形组成的大四边形） 下面是生成栅格顶点的代码： /* ** Summary：生成栅格顶点 ** Parameters： ** width：栅格的宽度 ** depth：栅格的深度 ** m：行数 ** n：列数 ** Return：栅格的网格数据 */ GeometryGenerator::MeshData GeometryGenerator::CreateGrid(float width, float depth, uint32 m,