I need Lat/LON from GIS data
I have data files from
http://www.mngeo.state.mn.us/chouse/land_own_property.html
given in the format of
.dbf, .prj, .sbn, .sbx, .shp, and .shx
in the .dbf I see
PIN, Shape_area, Shape_len
PARC_CODE Parcel Polygon to Parcel Point numeric 2
and PIN Relationship Code
and in the .prj
PROJCS["NAD_1983_UTM_Zone_15N",GEOGCS["GCS_North_American_1983",DATUM["D_North_American_1983",SPHEROID["GRS_1980",6378137.0,298.257222101]],PRIMEM["Greenwich",0.0],UNIT["Degree",0.0174532925199433]],PROJECTION["Transverse_Mercator"],PARAMETER["False_Easting",500000.0],PARAMETER["False_Northing",0.0],PARAMETER["Central_Meridian",-93.0],PARAMETER["Scale_Factor",0.9996],PARAMETER["Latitude_Of_Origin",0.0],UNIT["Meter",1.0]]
I also know the polygon points for each county
polygons points
Anoka 129139 129138
Carver 38134 38133
Dakota 135925 150294
Hennepin 422976 446623
Ramsey 149169 168233
Scott 55191 55191
Washington 98915 103915
and I know the bounding coordinates
-94.012
-92.732
45.415
44.471
there seems to be tons of software applications for GIS
http://en.wikipedia.org/wiki/List_of_geographic_information_systems_software
but what do I need to do?
I want the lat, lon of every house
Is there a library that will do this for me?
What is the data I need?
I think you need to install one GIS software. You can try open-source Qgis.
Because, firstly your data is not in long/lat (geographic) coordinates. Your .prj part of the shapefile (yes, all .dbf, .prj, .sbn, .sbx, .shp, and .shx files with the same name are one shapefile for GIS) says that the data are in the projected coordinate system NAD 1983 UTM Zone 15N. So, you need to transform your data to geographic system. This you easy can do in GIS, or programmatically by proj.4 library. (In Qgis add the shapefile to the project, then select it in the table of contents, right mouse button and choose "save as...". It will ask you for the target coordinate system.) Note, that you need to decide which geographic coordinates you wish, because your data are in the North American Datum (NAD 1983), but the most common worldwide now is WGS 1984.
Secondly, in GIS you will see your data, are they really points, or maybe polygons. (In case your houses are polygons you will need to get centroids of them, in Qgis menu Vector - Geometry Tools - Polygon Centroids).
Finally, when you really have your houses as points in geographic coordinates, you can get their coordinates, for example using advices from these questions Get list of coordinates for points in a layer and How do I calculate the latitude and longitude of points using QGIS.
Besides, there is a good library to work with GIS vector data, OGR, which can be used by many programming languages.
The file extensions above show, that the files are in ESRI Shape File format. In Java you could use GeoTools libraries, to read that.
The example below shows the first lines, search Internet for a more complete example.
// init shapefile
File shpFile = new File(fileName);
if (!shpFile.exists()) {
LOGGER.fatal(fileName + " does not exist");
}
Map<String, URL> connect = new HashMap<String, URL>();
FeatureCollection collection = null;
FeatureIterator iterator = null;
try {
connect.put("url", shpFile.toURI().toURL());
DataStore dataStore = DataStoreFinder.getDataStore(connect);
String typeName = dataStore.getTypeNames()[0];
"I want the lat, lon of every house" suggests that what you want to do is the process called geocoding. There are services you can use for that, some free (for limited uses) some not. You could start by looking at the List of geocoding systems to get an idea of where to start. I don't think you want to start by learning GIS or shapefiles, other than to extract the addresses you are trying to geocode.
You could estimate the lat/lon of each house by computing the centroid of each parcel. You could more roughly estimate the lat/lon of each house by calculating the centroid of the bounding rectangle of each parcel. Either of those would require extracting the parcel coordinates. If you are doing that for every house in Minnesota you will processing lots of data. A geocoding service would be cheaper. The Census Geocoder might help.
Related
I have a pdb file that is a subset of a much larger system. This pdb is special because I have some vectors based on this file's coordinate system. I want to draw these vectors and the basis vector of that system onto the pdb. Eventually I would like to visualize the vector and basis vectors as it moves through some MD simulation where I an update the vector position based on the trajectory over time.
To start, I would like to read a pdb that has coordinates that define the basis vectors that further define the other vectors I want to visualize. Right now I'm using this class in MDAnalysis:
https://docs.mdanalysis.org/1.0.0/_modules/MDAnalysis/coordinates/PDB.html#PDBReader
molecule=mda.coordinates.PDB.PDBReader('molecule.pdb')
This works and it reads the pdb just fine, but returns with this variable type
coordinates.PDB.Reader
I suppose this isn't a surprise, but I want to be able to print this variable and get some array of coordinate positions and atom types. I'd love to see the bonds as well but that's not necessary. Now when I print I get
<PDBReader molecule.pdb with 1 frames of 60 atoms>
I want something that would look like
[atomtype1,x1,y1,z1]...[atomtypen,xn,yn,zn]
Thank you,
Loading data: Universe
To get the coordinates in MDAnalysis you first load a Universe (you don't normally use the coordinate readers directly):
import MDAnalysis as mda
u = mda.Universe('molecule.pdb')
The Universe contains "topology" (atom types, bonds if available, etc) and the "trajectory" (i.e., coordinates).
Accessing per-atom data: Universe.atoms
All the atoms are stored in u.atoms (they form an AtomGroup). All information about the atoms is available from an AtomGroup. For example, all positions are available as a numpy array with
u.atoms.positions
The names are
u.atoms.names
and there are many more attributes. (The same works for residues: u.residues or u.atoms.residues gives the residues. You can also slice/index an AtomGroup to get a new AtomGroup. For example, the residues that belong to the first 20 atoms are u.atoms[:20].residues... AtomGroups are the key to working with MDAnalysis.)
Extracting atom names and positions
To build the list that you asked for:
names_positions = [[at] + list(pos) for at, pos in zip(u.atoms.names, u.atoms.positions)]
For example, with the test file PDB that is included with the MDAnalysisTests package:
import MDAnalysis as mda
from MDAnalysisTests.datafiles import PDB
u = mda.Universe(PDB)
names_positions = [[at] + list(pos) for at, pos in zip(u.atoms.names, u.atoms.positions)]
# show the first three entries
print(names_positions[:3])
gives
[['N', 52.017, 43.56, 31.555], ['H1', 51.188, 44.112, 31.722], ['H2', 51.551, 42.828, 31.039]]
Learning more...
For a rapid introduction to MDAnalysis have a look at the Quickstart Guide, which explains most of these things in more detail. It also tells you how to select specific atoms and form new AtomGroups.
Then you can have a look at the rest of the User Guide.
Bonds are a bit more tricky (you can get them with u.atoms.bonds but if you want to use them you have to learn more about how MDAnalysis represents topologies — I'd suggest you start by asking on user mailing list, see participating in MDAnalysis because this is where MDAnalysis developers primarily answer questions.)
How would I go about programming a function to interact with Apple's MapKit so that when a specific location(latitude/longitude) is given, the program will search for certain aspects within 5-10km radius on that map?
So for example, gets given geo-location of an airport, and searching for runways within a radius of that airport, then placing vectors marking the runway for viewing purposes and specific location of the runway.
How would I go about programming something like that.
Just reading through the sample code, this is what you need:
// confine the map search area to the user's current location
MKCoordinateRegion newRegion;
newRegion.center.latitude = self.userLocation.latitude;
newRegion.center.longitude = self.userLocation.longitude;
// setup the area spanned by the map region:
// we use the delta values to indicate the desired zoom level of the map,
// (smaller delta values corresponding to a higher zoom level)
//
newRegion.span.latitudeDelta = 0.112872;
newRegion.span.longitudeDelta = 0.109863;
And the reference.
I need to read the tessellation/cgr/visualisation details from a CATIA V5R18 Part file using CATIA V5R18 API.
Visualisation details such as:
Number of Vertices
Number of Triangles
Number of Strips
Number of Fans
Number of Normal
Bounding Sphere Centre and Radius
These details I have read from .cgr files using CAT3DRep/CATRep/CATSurfacicRep, but I am not able to read the same for .CATPart files.
From .CATPart with the help of CATIVisu I got CAT3DBagRep type, when I queried from PartFeatures. But to get Visualisation details I need CATSurfacicRep.
What interface should I query and from where should I query?
I am not sure about R18 but for R22 and R23 if you have the CAA documentation there is an example located at: C:\Program Files\Dassault Systemes\B23\CAADoc\CAATessellation.edu.
This example code has everything you need to get the tessellation data except normals and bounding spheres. I used this example code when I was teaching myself how to make a tessellated nurbs surface and it was quite useful for testing.
Hi i need the flow to read the visualisation details from a CATIA V5R18 Part file.
Visualisation details lik,
1.No of Vertices
2.No of Triangles
3.No of Strips
4.No of Fans
5.No of Normal
6.Bouding Sphere Centre and Radius
These details i have red from .cgr files using CAT3DRep/CATRep/CATSurfacicRep...
But i am not able to read the same for .CATPart files.
From .CATPart with the help of CATIVisu i got CAT3DBagRep type When i queried from PartFeatures But to get Visualisation details i need CATSurfacicRep.
Can anyone help?
Wat Interface i should query and from where i should query?
Well, information about the mesh (triangle, strips, fans, etc) is only carried by leaf Reps, like CAT3DSurfacicRep.
For complex files like CATPart or CATProduct, where you have a hierarchy of geometries, there's also a hierarchy of Reps. CAT3DBagRep is the class that allows building this hierarchy, as it has children Reps (which can of course be also CAT3DBagReps).
One solution may be to recursively explore this Rep hierarchy from the root CAT3DBagRep you get. The method to get the children Reps of a CAT3DBagRep is:
list<CATRep> *GetChildren();
You can go down the Rep tree until you get Reps of the expected type, like CATSurfacicRep. You may find many of them depending on your model.
When retrieving the mesh coordinates, normals and bounding element, please take into account that they are given in local Rep coordinates. A CAT3DBagRep carries positioning and orientation information (used when you position CATProducts, for example). This is returned by the following CAT3DBagRep method:
const CAT4x4Matrix * GetMatrix() const;
Depending on your scenario/model, you may need to take this positioning information into account.
I'd like to know the local time where my user is sending his request from.
Basically, is there such a thing as a function like this
var localTime = getLocalTime( lat, long );
I'm not sure if a simple division on the lat could work, since most of the countries don't have perfect geometric shapes.
Any help would be great. Any language is accepted. I'd like to avoid calling distant APIs.
The Google Time Zone API seems to be what you're after. It, however does not have any free tier.
The Time Zone API provides time offset data for locations on the surface of the earth. Requesting the time zone information for a specific Latitude/Longitude pair will return the name of that time zone, the time offset from UTC, and the Daylight Savings offset.
The shapefile used to compute the timezone is not maintained anymore.
I just faced the same issue today, and I am not sure how relevant my answer is after all this time, but I basically just wrote a Python function that does what you want. You can find it here.
https://github.com/cstich/gpstotz
Edit:
As mentioned in the comments I should also post code. The code is based on Eric Muller's shapefile of timezones, which you can get here - http://efele.net/maps/tz/world/.
Edit 2:
As it turns out shapefiles have a somewhat archaic definition of exterior and interior rings (basically exterior rings are using the right hand rule, while interior rings are using the left hand rule). In any case fiona seems to take care of that and I updated the code accordingly.
from rtree import index # requires libspatialindex-c3.deb
from shapely.geometry import Polygon
from shapely.geometry import Point
import os
import fiona
''' Read the world timezone shapefile '''
tzshpFN = os.path.join(os.path.dirname(__file__),
'resources/world/tz_world.shp')
''' Build the geo-index '''
idx = index.Index()
with fiona.open(tzshpFN) as shapes:
for i, shape in enumerate(shapes):
assert shape['geometry']['type'] == 'Polygon'
exterior = shape['geometry']['coordinates'][0]
interior = shape['geometry']['coordinates'][1:]
record = shape['properties']['TZID']
poly = Polygon(exterior, interior)
idx.insert(i, poly.bounds, obj=(i, record, poly))
def gpsToTimezone(lat, lon):
'''
For a pair of lat, lon coordiantes returns the appropriate timezone info.
If a point is on a timezone boundary, then this point is not within the
timezone as it is on the boundary. Does not deal with maritime points.
For a discussion of those see here:
http://efele.net/maps/tz/world/
#lat: latitude
#lon: longitude
#return: Timezone info string
'''
query = [n.object for n in idx.intersection((lon, lat, lon, lat),
objects=True)]
queryPoint = Point(lon, lat)
result = [q[1] for q in query
if q[2].contains(queryPoint)]
if len(result) > 0:
return result[0]
else:
return None
if __name__ == "__main__":
''' Tests '''
assert gpsToTimezone(0, 0) is None # In the ocean somewhere
assert gpsToTimezone(51.50, 0.12) == 'Europe/London'
I was searching for the same thing couple of days ago and unfortunately I could not find an API or a simple function that does it. The reason being as you said that countries do not have perfect geometric shapes. You have to create a representation of the area of each time zone and see where your point lies. I think this will be a pain and I have no idea if it can be done at all.
The only one I found is described here: Determine timezone from latitude/longitude without using web services like Geonames.org . Basically you need a database with information about timezones and you are trying to see which one is closest to your point of interest.
However, I was looking for static solutions(without using internet), so if you can use internet connection you can use: http://www.earthtools.org/webservices.htm which provides a webservice to give you the timezone given lat/lon coordinates.
As of 2019, Google API does not have any free tier and the data source of #cstich answer is not maintained anymore.
If you want an API, timezonedb.com offers a free tier rate limited to 1 request/second.
The original maintainer of the data used by #cstich link to this project which retrieve data from OpenStreetMap. The readme contains link to look up libraries in a wide variety of languages.
Couldn't you simply use the user IP to determine which they live in ? And then you use an array of (Countries | Difference with GMT) to get the local time.