I'm trying to make some sort of 3D editor, and I'm trying to make an "orbit" tool, much like the one in Blender.
Now I'd like to know what the angle is. I used the code provided by José Pereda that can be found here, but I need the angles to be in the range from 0 to 2π. I cannot retrieve them from the acquired angles since the output values don't range from -π to π and are different for every output.
Also, the extracted angles need to be relative to each other, much like if you put 3 different Rotate transformations where every following one is relative to the preceding ones, preferably in YXZ order, since that's the order I use everywhere else, as well as the model format the editor is going to export into.
As requested, I have uploaded the part of the code that is relevant to the question on Pastebin. As far as I could tell from the printed values, the angles are relative to each other in XYZ order. For angle calculation, I use the snippet from the linked question:
public static Vec3d getAngle(Node n) {
Transform T = n.getLocalToSceneTransform();
double roll = Math.atan2(-T.getMyx(), T.getMxx());
double pitch = Math.atan2(-T.getMzy(), T.getMzz());
double yaw = Math.atan2(T.getMzx(), Math.sqrt(T.getMzy() * T.getMzy() + T.getMzz() * T.getMzz()));
return new Vec3d(roll, pitch, yaw);
}
Just a quick notice, I have very little knowledge of matrices and their transformations, since that's something we'll be taking in our school next year, which makes it harder for me to understand.
My question addresses both mathematical and CS issues, but since I need a performant implementation I am posting it here.
Problem:
I have an estimated normal bivariate distribution, defined as a python matrix, but then I will need to transpose the same computation in Java. (dummy values here)
mean = numpy.matrix([[0],[0]])
cov = numpy.matrix([[1,0],[0,1]])
When I receive in inupt a column vector of integers values (x,y) I want to compute the probability of that given tuple.
value = numpy.matrix([[4],[3]])
probability_of_value_given_the_distribution = ???
Now, from a matematical point of view, this would be the integral for 3.5 < x < 4.5 and 2.5 < y < 3.5 over the probability density function of my normal.
What I want to know:
Is there a way to avoid the effective implementation of this, that implies dealing with expressions defined over matrices and with double integrals? Besides that it will take me a while if I had to implement it by myself, this would be computationally expensive. An approximate solution would be perfectly fine for me.
My reasonings:
In an univariate normal, one could simply use the cumulative distribution function (or even store its values for the standard one and then normalize), but unfortunately there appears not to be a closed cdf form for multivariates.
Another approach for univariate is to use the inverse of bivariate approximation (so, approximate a normal as a binomial), but extending this to the multivariate I can't figure out how to keep in count the covariances.
I really hope someone has already implemented this, I need it soon (finishing my thesis) and I couldn't find anything.
OpenTURNS provides an efficient implementation of the CDF of a multinormal distribution (see the code).
import numpy as np
mean = np.array([0.0, 0.0])
cov = np.array([[1.0, 0.0],[0.0, 1.0]])
Let us create the multinormal distribution with these parameters.
import openturns as ot
multinormal = ot.Normal(mean, ot.CovarianceMatrix(cov))
Now let us compute the probability of the square [3.5, 4.5] x |2.5, 3.5]:
prob = multinormal.computeProbability(ot.Interval([3.5,2.5], [4.5,3.5]))
print(prob)
The computed probability is
1.3701244220201715e-06
If you are looking for the probabiliy density function of a bivariate normal distribution, below are a few lines that could do the job:
import numpy as np
def multivariate_pdf(vector, mean, cov):
quadratic_form = np.dot(np.dot(vector-mean,np.linalg.inv(cov)),np.transpose(vector-mean))
return np.exp(-.5 * quadratic_form)/ (2*np.pi * np.linalg.det(cov))
mean = np.array([0,0])
cov = np.array([[1,0],[0,1]])
vector = np.array([4,3])
pdf = multivariate_pdf(vector, mean, cov)
I've started differentiating two images by counting the number of different pixels using a simple algorithm:
private int returnCountOfDifferentPixels(String pic1, String pic2)
{
Bitmap i1 = loadBitmap(pic1);
Bitmap i2 = loadBitmap(pic2);
int count=0;
for (int y = 0; y < i1.getHeight(); ++y)
for (int x = 0; x < i1.getWidth(); ++x)
if (i1.getPixel(x, y) != i2.getPixel(x, y))
{
count++;
}
return count;
}
However this approach seems to be inefficient in its initial form, as there is always a very high number of pixels which differ even in very similar photos.
I was thinking of a way of to determine if two pixels are really THAT different.
the bitmap.getpixel(x,y) from android returns a Color object.
How can I implement a proper differentiation between two Color objects, to help with my motion detection?
You are right, because of noise and other factors there is usually a lot of raw pixel change in a video stream. Here are some options you might want to consider:
Blurring the image first, ideally with a Gaussian filter or with a simple box filter. This just means that you take the (weighted) average over the neighboring pixel and the pixel itself. This should reduce the sensor noise quite a bit already.
Only adding the difference to count if it's larger than some threshold. This has the effect of only considering pixels that have really changed a lot. This is very easy to implement and might already solve your problem alone.
Thinking about it, try these two options first. If they don't work out, I can give you some more options.
EDIT: I just saw that you're not actually summing up differences but just counting different pixels. This is fine if you combine it with Option 2. Option 1 still works, but it might be an overkill.
Also, to find out the difference between two colors, use the methods of the Color class:
int p1 = i1.getPixel(x, y);
int p2 = i2.getPixel(x, y);
int totalDiff = Color.red(p1) - Color.red(p2) + Color.green(p1) - Color.green(p2) + Color.blue(p1) - Color.blue(p2);
Now you can come up with a threshold the totalDiff must exceed to contribute to count.
Of course, you can play around with these numbers in various ways. The above code for example only computes changes in pixel intensity (brightness). If you also wanted to take into account changes in hue and saturation, you would have to compute totalDifflike this:
int totalDiff = Math.abs(Color.red(p1) - Color.red(p2)) + Math.abs(Color.green(p1) - Color.green(p2)) + Math.abs(Color.blue(p1) - Color.blue(p2));
Also, have a look at the other methods of Color, for example RGBToHSV(...).
I know that this is essentially very similar another answer here but I think be restating it in a different form it might prove useful to those seeking the solution. This involves have more than two images over time. If you only literally then this will not work but an equivilent method will.
Do the history for all pixels on each frame. For example, for each pixel:
history[x, y] = (history[x, y] * (w - 1) + get_pixel(x, y)) / w
Where w might be w = 20. The higher w the larger the spike for motion but the longer motion has to be missing for it to reset.
Then to determine if something has changed you can do this for each pixel:
changed_delta = abs(history[x, y] - get_pixel(x, y))
total_delta += changed_delta
You will find that it stabilizes most of the noise and when motion happens you will get a large difference. You are essentially taking many frames and detecting motion from the many against the newest frame.
Also, for detecting positions of motion consider breaking the image into smaller pieces and doing them individually. Then you can find objects and track them across the screen by treating a single image as a grid of separate images.
Any clever ideas on how to generate random coordinates (latitude / longitude) of places on Earth? Latitude / Longitude. Precision to 5 points and avoid bodies of water.
double minLat = -90.00;
double maxLat = 90.00;
double latitude = minLat + (double)(Math.random() * ((maxLat - minLat) + 1));
double minLon = 0.00;
double maxLon = 180.00;
double longitude = minLon + (double)(Math.random() * ((maxLon - minLon) + 1));
DecimalFormat df = new DecimalFormat("#.#####");
log.info("latitude:longitude --> " + df.format(latitude) + "," + df.format(longitude));
Maybe i'm living in a dream world and the water topic is unavoidable ... but hopefully there's a nicer, cleaner and more efficient way to do this?
EDIT
Some fantastic answers/ideas -- however, at scale, let's say I need to generate 25,000 coordinates. Going to an external service provider may not be the best option due to latency, cost and a few other factors.
To deal with the body of water problem is going to be largely a data issue, e.g. do you just want to miss the oceans or do you need to also miss small streams. Either you need to use a service with the quality of data that you need, or, you need to obtain the data yourself and run it locally. From your edit, it sounds like you want to go the local data route, so I'll focus on a way to do that.
One method is to obtain a shapefile for either land areas or water areas. You can then generate a random point and determine if it intersects a land area (or alternatively, does not intersect a water area).
To get started, you might get some low resolution data here and then get higher resolution data here for when you want to get better answers on coast lines or with lakes/rivers/etc. You mentioned that you want precision in your points to 5 decimal places, which is a little over 1m. Do be aware that if you get data to match that precision, you will have one giant data set. And, if you want really good data, be prepared to pay for it.
Once you have your shape data, you need some tools to help you determine the intersection of your random points. Geotools is a great place to start and probably will work for your needs. You will also end up looking at opengis code (docs under geotools site - not sure if they consumed them or what) and JTS for the geometry handling. Using this you can quickly open the shapefile and start doing some intersection queries.
File f = new File ( "world.shp" );
ShapefileDataStore dataStore = new ShapefileDataStore ( f.toURI ().toURL () );
FeatureSource<SimpleFeatureType, SimpleFeature> featureSource =
dataStore.getFeatureSource ();
String geomAttrName = featureSource.getSchema ()
.getGeometryDescriptor ().getLocalName ();
ResourceInfo resourceInfo = featureSource.getInfo ();
CoordinateReferenceSystem crs = resourceInfo.getCRS ();
Hints hints = GeoTools.getDefaultHints ();
hints.put ( Hints.JTS_SRID, 4326 );
hints.put ( Hints.CRS, crs );
FilterFactory2 ff = CommonFactoryFinder.getFilterFactory2 ( hints );
GeometryFactory gf = JTSFactoryFinder.getGeometryFactory ( hints );
Coordinate land = new Coordinate ( -122.0087, 47.54650 );
Point pointLand = gf.createPoint ( land );
Coordinate water = new Coordinate ( 0, 0 );
Point pointWater = gf.createPoint ( water );
Intersects filter = ff.intersects ( ff.property ( geomAttrName ),
ff.literal ( pointLand ) );
FeatureCollection<SimpleFeatureType, SimpleFeature> features = featureSource
.getFeatures ( filter );
filter = ff.intersects ( ff.property ( geomAttrName ),
ff.literal ( pointWater ) );
features = featureSource.getFeatures ( filter );
Quick explanations:
This assumes the shapefile you got is polygon data. Intersection on lines or points isn't going to give you what you want.
First section opens the shapefile - nothing interesting
you have to fetch the geometry property name for the given file
coordinate system stuff - you specified lat/long in your post but GIS can be quite a bit more complicated. In general, the data I pointed you at is geographic, wgs84, and, that is what I setup here. However, if this is not the case for you then you need to be sure you are dealing with your data in the correct coordinate system. If that all sounds like gibberish, google around for a tutorial on GIS/coordinate systems/datum/ellipsoid.
generating the coordinate geometries and the filters are pretty self-explanatory. The resulting set of features will either be empty, meaning the coordinate is in the water if your data is land cover, or not empty, meaning the opposite.
Note: if you do this with a really random set of points, you are going to hit water pretty often and it could take you a while to get to 25k points. You may want to try to scope your point generation better than truly random (like remove big chunks of the Atlantic/Pacific/Indian oceans).
Also, you may find that your intersection queries are too slow. If so, you may want to look into creating a quadtree index (qix) with a tool like GDAL. I don't recall which index types are supported by geotools, though.
This has being asked a long time ago and I now have the similar need. There are two possibilities I am looking into:
1. Define the surface ranges for the random generator.
Here it's important to identify the level of precision you are going for. The easiest way would be to have a very relaxed and approximate approach. In this case you can divide the world map into "boxes":
Each box has it's own range of lat lon. Then you first randomise to get a random box, then you randomise to get a random lat and random long within the boundaries of that box.
Precisions is of course not the best at all here... Though it depends:) If you do your homework well and define a lot of boxes covering most complex surface shapes - you might be quite ok with the precision.
2. List item
Some API to return continent name from coordinates OR address OR country OR district = something that WATER doesn't have. Google Maps API's can help here. I didn't research this one deeper, but I think it's possible, though you will have to run the check on each generated pair of coordinates and rerun IF it's wrong. So you can get a bit stuck if random generator keeps throwing you in the ocean.
Also - some water does belong to countries, districts...so yeah, not very precise.
For my needs - I am going with "boxes" because I also want to control exact areas from which the random coordinates are taken and don't mind if it lands on a lake or river, just not open ocean:)
Download a truckload of KML files containing land-only locations.
Extract all coordinates from them this might help here.
Pick them at random.
Definitely you should have a map as a resource. You can take it here: http://www.naturalearthdata.com/
Then I would prepare 1bit black and white bitmap resource with 1s marking land and 0x marking water.
The size of bitmap depends on your required precision. If you need 5 degrees then your bitmap will be 360/5 x 180/5 = 72x36 pixels = 2592 bits.
Then I would load this bitmap in Java, generate random integer withing range above, read bit, and regenerate if it was zero.
P.S. Also you can dig here http://geotools.org/ for some ready made solutions.
To get a nice even distribution over latitudes and longitudes you should do something like this to get the right angles:
double longitude = Math.random() * Math.PI * 2;
double latitude = Math.acos(Math.random() * 2 - 1);
As for avoiding bodies of water, do you have the data for where water is already? Well, just resample until you get a hit! If you don't have this data already then it seems some other people have some better suggestions than I would for that...
Hope this helps, cheers.
There is another way to approach this using the Google Earth Api. I know it is javascript, but I thought it was a novel way to solve the problem.
Anyhow, I have put together a full working solution here - notice it works for rivers too: http://www.msa.mmu.ac.uk/~fraser/ge/coord/
The basic idea I have used is implement the hiTest method of the GEView object in the Google Earth Api.
Take a look at the following example of the hitest from Google.
http://earth-api-samples.googlecode.com/svn/trunk/examples/hittest.html
The hitTest method is supplied a random point on the screen in (pixel coordinates) for which it returns a GEHitTestResult object that contains information about the geographic location corresponding to the point. If one uses the GEPlugin.HIT_TEST_TERRAIN mode with the method one can limit results only to land (terrain) as long as we screen the results to points with an altitude > 1m
This is the function I use that implements the hitTest:
var hitTestTerrain = function()
{
var x = getRandomInt(0, 200); // same pixel size as the map3d div height
var y = getRandomInt(0, 200); // ditto for width
var result = ge.getView().hitTest(x, ge.UNITS_PIXELS, y, ge.UNITS_PIXELS, ge.HIT_TEST_TERRAIN);
var success = result && (result.getAltitude() > 1);
return { success: success, result: result };
};
Obviously you also want to have random results from anywhere on the globe (not just random points visible from a single viewpoint). To do this I move the earth view after each successful hitTestTerrain call. This is achieved using a small helper function.
var flyTo = function(lat, lng, rng)
{
lookAt.setLatitude(lat);
lookAt.setLongitude(lng);
lookAt.setRange(rng);
ge.getView().setAbstractView(lookAt);
};
Finally here is a stripped down version of the main code block that calls these two methods.
var getRandomLandCoordinates = function()
{
var test = hitTestTerrain();
if (test.success)
{
coords[coords.length] = { lat: test.result.getLatitude(), lng: test.result.getLongitude() };
}
if (coords.length <= number)
{
getRandomLandCoordinates();
}
else
{
displayResults();
}
};
So, the earth moves randomly to a postition
The other functions in there are just helpers to generate the random x,y and random lat,lng numbers, to output the results and also to toggle the controls etc.
I have tested the code quite a bit and the results are not 100% perfect, tweaking the altitude to something higher, like 50m solves this but obviously it is diminishing the area of possible selected coordinates.
Obviously you could adapt the idea to suit you needs. Maybe running the code multiple times to populate a database or something.
As a plan B, maybe you can pick a random country and then pick a random coordinate inside of this country. To be fair when picking a country, you can use its area as weight.
There is a library here and you can use its .random() method to get a random coordinate. Then you can use GeoNames WebServices to determine whether it is on land or not. They have a list of webservices and you'll just have to use the right one. GeoNames is free and reliable.
Go there http://wiki.openstreetmap.org/
Try to use API: http://wiki.openstreetmap.org/wiki/Databases_and_data_access_APIs
I guess you could use a world map, define a few points on it to delimit most of water bodies as you say and use a polygon.contains method to validate the coordinates.
A faster algorithm would be to use this map, take some random point and check the color beneath, if it's blue, then water... when you have the coordinates, you convert them to lat/long.
You might also do the blue green thing , and then store all the green points for later look up. This has the benifit of being "step wise" refinable. As you figure out a better way to make your list of points you can just point your random graber at a more and more acurate group of points.
Maybe a service provider has an answer to your question already: e.g. https://www.google.com/enterprise/marketplace/viewListing?productListingId=3030+17310026046429031496&pli=1
Elevation api? http://code.google.com/apis/maps/documentation/elevation/ above sea level or below? (no dutch points for you!)
Generating is easy, the Problem is that they should not be on water. I would import the "Open Streetmap" for example here http://ftp.ecki-netz.de/osm/ and import it to an Database (verry easy data Structure). I would suggest PostgreSQL, it comes with some geo functions http://www.postgresql.org/docs/8.2/static/functions-geometry.html . For that you have to save the points in a "polygon"-column, then you can check with the "&&" operator if it is in an Water polygon. For the attributes of an OpenStreetmap Way-Entry you should have a look at http://wiki.openstreetmap.org/wiki/Category:En:Keys
Supplementary to what bsimic said about digging into GeoNames' Webservices, here is a shortcut:
they have a dedicated WebService for requesting an ocean name.
(I am aware the of OP's constraint to not using public web services due to the amount of requests. Nevertheless I stumbled upon this with the same basic question and consider this helpful.)
Go to http://www.geonames.org/export/web-services.html#astergdem and have a look at "Ocean / reverse geocoding". It is available as XML and JSON. Create a free user account to prevent daily limits on the demo account.
Request example on ocean area (Baltic Sea, JSON-URL):
http://api.geonames.org/oceanJSON?lat=54.049889&lng=10.851388&username=demo
results in
{
"ocean": {
"distance": "0",
"name": "Baltic Sea"
}
}
while some coordinates on land result in
{
"status": {
"message": "we are afraid we could not find an ocean for latitude and longitude :53.0,9.0",
"value": 15
}
}
Do the random points have to be uniformly distributed all over the world? If you could settle for a seemingly uniform distribution, you can do this:
Open your favorite map service, draw a rectangle inside the United States, Russia, China, Western Europe and definitely the northern part of Africa - making sure there are no big lakes or Caspian seas inside the rectangles. Take the corner coordinates of each rectangle, and then select coordinates at random inside those rectangles.
You are guaranteed non of these points will be on any sea or lake. You might find an occasional river, but I'm not sure how many geoservices are going to be accurate enough for that anyway.
This is an extremely interesting question, from both a theoretical and practical perspective. The most suitable solution will largely depend on your exact requirements. Do you need to account for every body of water, or just the major seas and oceans? How critical are accuracy and correctness; Will identifying sea as land or vice-versa be a catastrophic failure?
I think machine learning techniques would be an excellent solution to this problem, provided that you don't mind the (hopefully small) probability that a point of water is incorrectly classified as land. If that's not an issue, then this approach should have a number of advantages against other techniques.
Using a bitmap is a nice solution, simple and elegant. It can be produced to a specified accuracy and the classification is guaranteed to be correct (Or a least as correct as you made the bitmap). But its practicality is dependent on how accurate you need the solution to be. You mention that you want the coordinate accuracy to 5 decimal places (which would be equivalent to mapping the whole surface of the planet to about the nearest metre). Using 1 bit per element, the bitmap would weigh in at ~73.6 terabytes!
We don't need to store all of this data though; We only need to know where the coastlines are. Just by knowing where a point is in relation to the coast, we can determine whether it is on land or sea. As a rough estimate, the CIA world factbook reports that there are 22498km of coastline on Earth. If we were to store coordiates for every metre of coastline, using a 32 bit word for each latitude and longitude, this would take less than 1.35GB to store. It's still a lot if this is for a trivial application, but a few orders of magnitude less than using a bitmap. If having such a high degree of accuracy isn't neccessary though, these numbers would drop considerably. Reducing the mapping to only the nearest kilometre would make the bitmap just ~75GB and the coordinates for the world's coastline could fit on a floppy disk.
What I propose is to use a clustering algorithm to decide whether a point is on land or not. We would first need a suitably large number of coordinates that we already know to be on either land or sea. Existing GIS databases would be suitable for this. Then we can analyse the points to determine clusters of land and sea. The decision boundary between the clusters should fall on the coastlines, and all points not determining the decision boundary can be removed. This process can be iterated to give a progressively more accurate boundary.
Only the points determining the decision boundary/the coastline need to be stored, and by using a simple distance metric we can quickly and easily decide if a set of coordinates are on land or sea. A large amount of resources would be required to train the system, but once complete the classifier would require very little space or time.
Assuming Atlantis isn't in the database, you could randomly select cities. This also provides a more realistic distribution of points if you intend to mimic human activity:
https://simplemaps.com/data/world-cities
There's only 7,300 cities in the free version.
what i am trying to do: the user selects start and destination on a map and then from their coordinates i want to show the closest point location from a list of locations on map. i have a simple Sqlite database containing the longitude,latitude and name of the possible locations.
i did some research and this is what i found:
http://www.scribd.com/doc/2569355/Geo-Distance-Search-with-MySQL
but this is meant for using it with mySql and some kind of spatial search extension.
is there a possibility i can do something similar using android api or external libs?
public Point dialogFindClosestLocationToPoint(geometry.Point aStartPoint){
List<PointWithDistance> helperList=new ArrayList<PointWithDistance>();
try {
openDataBase();
Cursor c=getCursorQueryWithAllTheData();
if(c.moveToFirst())
do{
PointWithDistance helper=new PointWithDistance(c.getDouble(1),c.getDouble(2),c.getString(3));
int distance=returnDistanceBetween2Points(aStartPoint, helper);
if(distance<MAX_SEARCH_DISTANCE){
helper.setDistance(distance);
Log.i("values", helper.name);
helperList.add(helper);
}
}while (c.moveToNext());
Collections.sort(helperList,new PointComparator());
if(helperList!=null)
return helperList.get(0);
else return null;
}catch(SQLException sqle){
throw sqle;
}
finally{
close();
}
this is the code in the PointComparator() class:
public int compare(PointWithDistance o1, PointWithDistance o2) {
return (o1.getDistance()<o2.getDistance() ? -1 : (o1.getDistance()==o2.getDistance() ? 0 : 1));
}
where PointWithDistance is a object that contains: lat, long , distance, name
however this solution doesn't provide the right return info... and i realize that is it not scalable at all and very slow. i need a solution that will execute fast with a database with max of 1000 rows.
edit: my there was a mistake in this code in the sorting now i have it changed( should be < instead of >)
This kind of thing is done most efficiently using an R-Tree. The JSI library provides a Java implementation that I have used successfully with an index of 80.000 locations, processing thousands of lookups per second. However, it may not run on Android.
I was looking for something very similar some time ago:
Android sqlite sort on calculated column (co-ordinates distance)
I was using a MySQL lookup on my server, MySQL allows you to create a virtual column, performs the calculation and sorts by distance, and then you can set the max results returned or the max distance - it works very well:
Select Lat, Lon, acos(sin($lat)*sin(radians(Lat)) + cos($lat)*cos(radians(Lat))cos(radians(Lon)-$lon))$R As dist From MyTable ORDER BY dist DESC
I wanted to perform the same operation in my app - pull all the points in order to distance from the users location allowing me to show the closest ones. I ended up going with the a solution along the lines of the one suggested on the link above but realise its probably not the optimal solution but works for the purpose I wanted.
i haven't tried running your code, but it seems like it would work, it's just that it's not efficient. like you don't actually need to sort, you need the extract the minimum.
you can restrict your query to just the square that is of size (2*MAX_SEARCH_DISTANCE)^2 (with your point in the middle.
This way you are localizing your query and that will return you less results to compute distance for.
Of course this will not help if all your locations are in the localized square (maybe unlikely?).
Also, I suppose you could use hamiltonian distance instead of euclidean.
euclidean distance = sqrt((lat0 - lat1)^2 + (lon0 - lon1)^2)
hamitonian distance = (lat0 - lat1) + (lon0 - lon1)