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Check geosphere package distance function or fossil deg.dist function. You have data in degrees and need to translate it into meters or feet before doing clustering.


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The thing that you want to do is to solve Travelling Salesman Problem or one of its variations. 7000 points is a very difficult task. And you may spent a lot time calculating it (depending on software and algorithm) There is an open-source implementation of Genetic algorithm (not the most advanced one) in pgRouting extension for PostGIS. Another option is ...


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If you have the spatial analyst extension you can use the Euclidean Distance tool. 1. Convert the buffer to a raster with the "To Raster" tool (under conversion tools). 2. Then run the Euclidean Distance Tool which will create a raster with values representing distance from the cells that comprised the input raster. 3. Then if necessary, you can extract ...


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After a little editing i used this script and its working fine. Now i just have to do it for each ID and for each day :-) library(SDMTools) library(rgeos) library(maptools) library(lubridate) #points <- readShapePoints("C:/Users/mlra/Desktop/FunWithDistance/DummyBird.shp") points <- tellus_2012_2013 lon <- as.vector(points$long) lat <- ...


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It all depends on the level of accuracy that you need. A coarse approach would be the spherical law of cosines. This has issues with small distances - some say that it is around 1km, others say down to a few meters. A better approach would be the Haversine Formula. This works well, however, it doesn't take into account that the earth is not actually a ...


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You could rasterize (Polyline to Raster) the lines and use the Euclidean Distance tool as suggested. If the lines are in the same feature class, note that ED cell values will be distance from a line, so values will increase away from the two lines until the halfway point. If you want the distance from one line to another, you'd need to run ED on only one ...


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So I found this video, which has done just what I wanted to do. https://msdn.microsoft.com/en-gb/data/hh859721.aspx Hope it helps someone else. Thanks


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Create a Cost surface where your wetlands and the water area get the value 1 and all the other areas shall be NoData or extreme vals like 100,000. After that you can create an Cost Distance surface from your wetlands on the Cost Surface. Done. If the Cost Surface value is 1, the Cost Distance values will be map units. Cost Distance interpretation in ArcGIS ...


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I'm not clear from your question whether you're looking at cost paths or just a straight distance raster. In either case you probably need to create a mask raster and use that in whichever tool you're wanting or as an environment setting. You can use Con to create a mask that is/is not land, then use that mask as a parameter in the other tools to that it ...


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Please keep the site order to put only one question at a time. https://trac.osgeo.org/proj/wiki/GenParms will show you what the Proj.4 parameters mean. If you have input and output, you have to define projections on both sides. Changing the proj.4 definition does not change the coordinates automatically. Instead, you make simply 12 meters out of formerly ...


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Thanks guys, I found and easy and quick solution using the following query: SELECT * FROM mytable WHERE mytable.geom && ST_MakeEnvelope(minX, minY, maxX, maxY); My question was already answered somewhere else but I didn't use the correct search terms.


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If we look at postgis source code(geography,geometry) we see that the function ST_DWithin is defined as: CREATE OR REPLACE FUNCTION ST_DWithin(geom1 geometry, geom2 geometry, float8) RETURNS boolean AS 'SELECT $1 && ST_Expand($2,$3) AND $2 && ST_Expand($1,$3) AND _ST_DWithin($1, $2, $3)' LANGUAGE 'sql' IMMUTABLE; the ...


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Your Cost Surface could be the problem. The value of the Cost surface cell represents the cost for each map unit, not for the whole cell. The cost-values are multiplied with the resolution when using the cost-distance function.


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Your assumption concerning the output values is wrong. The cost-values are multiplied with the distance when using the cost-distance function. Hence 500m equal a cost-distance value of 5,000. I can only guess but I think the 70 may equal the cost-distance value from the center of the green pixel to the next cost-pixel. I see that there are high ...



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