I am trying to extract decimal degrees data from a shapefile using the QGIS software but no luck. I am using the field calculator method where i create a new double field then select Geometry then $x and $y but am still getting data in this format 247152.338941123. Have set the CRS layer to WGS84 but no luck. Please if you know what am doing wrong, help me. Thanks
From the values returned it seems that your layer is stored as a projected coordinate system. Try saving the layer as a geographic coordinate system (EPSG:4326) and it should work.
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I've been learning how to use Cartopy and Matplotlib to plot map. But I have a question regarding the argument transform. According to the Cartopy document, transform specifies "what coordinate system your data are defined in". Suppose I am going to plot temperatures of an area, and the area has been split into several grid cells. Each grid cells has a corresponding coordinate defined in lat and lon (Geodetic Systems). Based on the Cartopy document, I need to use crs.PlateCarree() instead of crs.Geodetic(). I'm a bit confused about it. Because,I think the PlateCarree is a way of projection. In other words, coordinates defined in PlateCarree projections are projected data. But latitude and longitude should be unprojected data. Can anyone help me with it? Thanks!
I am using this query to extract the geometries of all countries using BigQuery public dataset, see question here
how to extract all countries geometry from Openstreet map dataset in BigQuery
I use R to draw the results
I tried Kepler.GL and gave me the same results
Something is wrong with Russia and the USA
I know little about R visualization, but what is probably happening is you getting WKT text from BigQuery, and feeding it to R, which has different assumptions.
Tthe issue is your R package probably treats WKT differently than BigQuery. WKT semantics depends on the spatial reference system (SRS) used, which could be geographic (non-projected, using sphere or ellipsoid) or projected (flat map). BigQuery uses geographic system, so edge between points A and B is the shortest geodesic path. Most visualization systems use projected coordinates, and assume flat map. Edge between A and B is shortest straight line on the flat map.
While this does not matter too much in many cases, it still does affect precision when you have long edges. But when an edge crosses anti-meridian (180 degree meridian) you get big problem. An edge between (-169, 66) (eastern edge of Russia) and say (176, 70) (a nearby point on Chukchi sea) is relatively short on the sphere, it crosses anti-meridian, and spans 15 degrees longitude. But the same edge on flat map span 145 degrees longitude and crosses most of the map! These are the long near-horizontal lines you see.
What should you do?
If R has a packet that supports geographic SRS (it is sometime an option to use geodesic edges), you could try it.
Or you can also let BigQuery convert geography from geographic SRS to flat map, that R would understand, using ST_AsGeoJson function. GeoJson is defined on flat map, so BigQuery ST_AsGeoJson converts the semantics from geographic SRS to flat map SRS. You then visualize GeoJson string instead of WKT string in R.
ST_AsGeoJson does a lot of work, to make result conformant to GeoJson spec and flat map. It splits parts of geography that lay east and west of anti-meridian, so you don't get edges that cross it. It also approximates geodesic edges with flat map edges. But it makes visualization system much easier.
I am trying to figure out a way to convert the result of presto geo spatial function ST_DISTANCE to meters.
If I run the this example query:
SELECT ST_Distance(ST_Point(0.3476, 32.5825),ST_Point(0.0512, 32.4637))
The result I get from Presto is: 0.3193217812802629. The actual distance between these two places is 40,000m.
The presto documentation states that ST_DISTANCE: Returns the 2-dimensional cartesian minimum distance (based on spatial ref) between two geometries in projected units.
What I can understand about spatial ref is at links such as these:
http://webhelp.esri.com/arcgiSDEsktop/9.3/index.cfm?TopicName=Defining_a_spatial_reference
Which leads me to believe I need to figure you what spatial-ref Presto is using.
If I check the prest docs here:
https://github.com/prestodb/presto/blob/master/presto-geospatial/src/main/java/com/facebook/presto/geospatial/GeoFunctions.java
I can guess that is using the ESRI libraries so I assume the ESRI spatial ref? But that is where I get a bit lost as to where to proceed?
Thank you for your help..
I would recommend using Presto’s great_circle_distance() function instead of ST_Distance(). It will interpret your coordinates as WGS84 (aka EPSG:4326), and find the distance between them in kilometres by treating the shape of the earth as a sphere.
ST_Distance() would be appropriate if the coordinate system being used was already projected into a system that used metres or miles or some other unit, but there's no trivial way to do that in Presto.
From looking at the docs, it appears that presto supports a geometry type but not a geography type. That means it's not working with Latitude and Longitude, which is what I assume you're supplying as those point parameters. It's just an arbitrary 2D grid and so the resulting units are in whatever units you supplied as input.
The distance, in meters, between two points which are both approximately 32.5 meters "up" from the origin and about 0.5 meters "left" from the origin (how presto will have interpreted your points) is, indeed, 0.3193217812802629, the value that was returned to you.
I'm working with a postgresql table that contains many rows with a GEOMETRY(Point, 4326). Using the ST_SnapToGrid function and a DISTINCT select, I only extract a subset of rows depending on the displayed map zoom level. I'm having trouble finetuning the ST_SnapToGrid function, as I don't not what unit the size parameter is in?
The size is specified in the same units as the SRID of the geometry.
In the case of SRID 4326 this is decimal degrees. The actual distance
that is of course varies depending on the actual latitude and longitude
of the point in question. If this matters, if might help to work in
some projected coordinates and do the rounding there.
I am new to VB.NET and I am trying to write code in vb.net to find a equation when data points are given. For example (1,5),(2,6) etc.
I need to find a equation(not necessarily always linear) from the given points.
I tried to use the help given in
How do I calculate a trendline for a graph?
but couldn't figure out how to get equation.
Any help will be highly appreciated.
Thanks in advance.
What you're looking for is called Interpolation
Basically it's a field in numerical analysis, and it helps you create a polynomial representation of the data points.