I have a point shapefile of rain gauge stations (with data) which I want to use to create a Thiessen polygons over a basin/watershed (another shapefile). All the raingauge points are located within the spatial extents of the basin/watershed.
I want my Thiessen polygon to cover the whole basin like this:
I used ArcToolbox/Analysis Tools/Proximity/Create Thiessen Polygons to create it and i got something like this:
I would like my polygons to cover the whole basin area.
You have to set your extents to another shapefile (or see other options). When you have your polygon tool open, you go into Advance (bottom RH corner) > Extents, you should find it.
Related
I'm trying to code a trivial pursuit game. I want to give an id for every space of the board in order to use them for the movements. I need to know for every space which is next to which/match with each other.
But because of the geometry of the board(extern circle + radii), I didn't find the right logic behind this.
I am thinking of an ID based on 7 numbers (for the 6 radii + the circle). For example :
//this is not my code, i'm just trying to show example of IDs
center = [0][0][0][0][0][0][-2]
one on the "2nd radius" = [0][3][0][0][0][0][-2]
one the circle and the "3rd radius" = [0][0][6][0][0][0][22]
one on the circle = [0][0][0][0][0][0][21]
I have no idea if it's gonna work or if it's optimal, i will try and see.
If some of you have any better idea for name the ID, i would be happy to listen to them.
Here is an image of the board.
enter image description here
Thank you for helping!
OK, seeems you are inventing some coordinate system for this wheel for easy addressing and easy transtions between cells. System with many indices looks too complex.
Perhaps two-index scheme would be appropriate. Resembles polar coordinates:
The first index 0..6 as distance from the center.
The second one 1..42 - angular position.
So center cell is A[0][0] (the second index is not defined, we can choose any)
It's neighbors are A[1][1], A[1][8], A[1][15] ..A[1][36] (marked with 1 at your picture)
Similar for the next cells at the rays A[2][1], A[2][8], A[2][15] ..A[2][36] and so on
Wheel cells are A[6][1], A[6][2]..A[6][42]
Now neighbor cells have coordinates where one index differs by 1 (except for central cell, extra case)
Is this scheme suitable?
I am trying to model the interior of an epithelial space and am stuck on movement around the interior edges of a cylindrical space. Basically, I'm trying to implement StickyBorders and keep agents on those borders in a cylindrical space that I am creating.
Is there a way to use cylindrical coordinates in Repast Simphony? I found this example (https://www.researchgate.net/publication/259695792_An_Agent-Based_Model_of_Vascular_Disease_Remodeling_in_Pulmonary_Arterial_Hypertension) where they seem to have done something similar, but the paper doesn't explain methods in much depth, and I don't believe this is an example in the repast simphony models.
Currently, I have a class of epithelial cells that are set up to form a cylinder and other agents start just inside that cylinder. To move, they are choosing their most desired spot (similar to the Zombie code) then pointing to a new location in the direction of that desired location within one grid square of that original location. They check that new point before moving to it and make sure that there are at least two other epithelial cells in the immediate moore neighborhood, to ensure they stay against the wall.
GridPoint intendedpt = new GridPoint((int)Math.rint(alongX),(int)Math.rint(alongY),(int)Math.rint(alongZ));
GridCellNgh<EpithelialCell> nearEpithelium = new GridCellNgh<EpithelialCell>(mac_grid, intendedpt, EpithelialCell.class, 1,1,1);
List<GridCell<EpithelialCell>> EpiCells = nearEpithelium.getNeighborhood(false);
int nearbyEpiCellsCount=0;
for (GridCell<EpithelialCell> cell: EpiCells) {
nearbyEpiCellsCount++;
}
if (nearbyEpiCellsCount<2) {
System.out.println(this + " leaving epithelial wall /r");
RunEnvironment.getInstance().pauseRun();
//TODO: where to go if false
}
I am wondering if there is a way to either set the boundaries of the space to be a cylinder or to check which side of the agent is against the wall and restrict its movement in that direction.
The sticky border code (StickyBorders.java) essentially just checks if the point that the agent moves to is beyond any of the space's dimensions, and if so the point is clamped to that dimension. So, for example, if the space is 3x4 and an agent's movement would take it to 4,2, then that point becomes 3,2 and the agent is placed there. Can you do something like that in this case? If not, can you edit your question to explain why not and maybe that will help us understand better.
The approach we took in that model was to use a 3D grid space with custom borders and query methods. The space itself was still Cartesian - we just visualized it as a cylinder using custom display code. Using the Cartesian grid was an reasonable approximation for this application since the cell dimensions were significantly smaller that the vessel radius, so curvature effects were neglected. The boundary conditions on the vessel space were wrap around in the angular dimension, so that cells could move continuously around the circumference of the vessel, and the axial boundary conditions were also wrapped, as we assumed a long enough vessel length that this would be reasonable. The wall thickness dimension had hard boundaries at the basement membrane (y=0) and at the fluid interface (y=wall thickness).
Depending on which type of space you are using, you will need to implement a PointTranslator or GridPointTranslator that performs the border functions. If you want specific examples of the code I suggest you reach out to the author's directly.
For an edge in jgraphx which has its geometry set to relative, there are no points provided as they are derived from the source and target for the edge.
However, the points found there are the top left corner of the objects: what I would like to know is if there is a way to get the exact points in which the edge connects to the source and target vertex rather than just the position or center point of the object.
Try graph.getView().getPerimeterPoint(mxCellState, mxPoint...)
Provide as cell state your local cell (mxGraphView.getState(myLocalCell)) and as point the distant center point of your other cell linked by the edge (you may compute it from its geometry X, Y, Width and Height).
An jGraphx can be exported to svg format. Maybe the drawCell() in the com.mxgraph.canvas.mxSvgCanvas can help you out. It is the case when shape.equals(mxConstants.SHAPE_LINE) and how the M and L commands are being calculated.
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.
I have several boxes (x,y,width,height) randomly scattered around, and some of them need to be linked from point (x1,y1) in box1 to point (x2,y2) in box2 by drawing a line. I am trying to figure a way to make such line avoid passing through any other boxes (other than box1 and box2) by drawing several straight interconnected lines to go around any box in the way (if it is not possible to go with one straight line). The problem is that I don't know an algorithm for such thing (let alone having a technical/common name for it). Would appreciate any help in the form of algorithm or expressed ideas.
Thanks
Assuming that the lines can't be diagonal, here's one simple way. It's based on BFS and will also find the shortest line connecting the points:
Just create a graph, containing one vertex for each point (x, y) and for each point the edges:
((x,y),(x+1,y)) ((x,y),(x-1,y)) ((x,y),(x,y+1)) ((x,y),(x,y-1))
But each of this edges must be present only if it doesn't overlap a box.
Now just do a plain BFS from point (x1,y1) to (x2,y2)
It's really easy to obtain also diagonal lines the same way but you will need 8 edges for each vertex, that are, in addition to the previouses 4:
((x,y),(x-1,y+1)) ((x,y),(x-1,y-1)) ((x,y),(x+1,y-1)) ((x,y),(x+1,y+1))
Still, each edge must be present only if it doesn't overlap a box.
EDIT
If you can't consider space divided into a grid, here's another possibility, it won't give you the very shortest path, though.
Create a graph, in which each box is a vertex and has an edge to any other box that can be reached without the line to overlap a third box. Now find the shortet path using dijkstra between box1 and box2 containing the two points.
Now consider each box to have a small countour that doesn't overlap any other box. This way you can link the entering and the exiting point of each box in the path found through dijistra, passing through the countour.
Put all (x,y) coords of the corners of the boxes in a set V
Add the start- and end coordinates to V.
Create a set of edges E connecting each corner that does not cross any box-side (except for the diagonals in the boxes).
How to check if a line crosses a box side can be done with this algorithm
Now use a path-finding algorithm of your choice, to find a path in the graph (V, E).
If you need a simple algorithm that finds the shortest path, just go with a BFS.
(This will produce a path that goes along the sides of some boxes. If this is undesirable, you could in step 1 put the points at some distance delta from the actual corners.)
If the edges may not be diagonal:
Create a large grid of lines that goes between the boxes.
Throw away the grid-edges that cross a box-side.
Find a path in the grid using a path-finding algorithm of your choice.