Search Results

Search found 386 results on 16 pages for 'sqrt'.

Page 12/16 | < Previous Page | 8 9 10 11 12 13 14 15 16  | Next Page >

• how to use units along function parameter values in Mathematica

  - by niko

  I would like to pass the parameter values in meters or kilometers (both possible) and get the result in meters/second. I've tried to do this in the following example: u = 3.986*10^14 Meter^3/Second^2; v[r_, a_] := Sqrt[u (2/r - 1/a)]; Convert[r, Meter]; Convert[a, Meter]; If I try to use the defined function and conversion: a = 24503 Kilo Meter; s = 10198.5 Meter/Second; r = 6620 Kilo Meter; Solve[v[r, x] == s, x] The function returns the following: {x -> (3310. Kilo Meter^3)/(Meter^2 - 0.000863701 Kilo Meter^2)} which is not the user-friendly format. Anyway I would like to define a and r in meters or kilometers and get the result s in meters/second (Meter/Second). I would be very thankful if anyone of you could correct the given function definition and other statements in order to get the wanted result.

  Read the article

• Fitting Gaussian KDE in numpy/scipy in Python

  - by user248237

  I am fitting a Gaussian kernel density estimator to a variable that is the difference of two vectors, called "diff", as follows: gaussian_kde_covfact(diff, smoothing_param) -- where gaussian_kde_covfact is defined as: class gaussian_kde_covfact(stats.gaussian_kde): def __init__(self, dataset, covfact = 'scotts'): self.covfact = covfact scipy.stats.gaussian_kde.__init__(self, dataset) def _compute_covariance_(self): '''not used''' self.inv_cov = np.linalg.inv(self.covariance) self._norm_factor = sqrt(np.linalg.det(2*np.pi*self.covariance)) * self.n def covariance_factor(self): if self.covfact in ['sc', 'scotts']: return self.scotts_factor() if self.covfact in ['si', 'silverman']: return self.silverman_factor() elif self.covfact: return float(self.covfact) else: raise ValueError, \ 'covariance factor has to be scotts, silverman or a number' def reset_covfact(self, covfact): self.covfact = covfact self.covariance_factor() self._compute_covariance() This works, but there is an edge case where the diff is a vector of all 0s. In that case, I get the error: File "/srv/pkg/python/python-packages/python26/scipy/scipy-0.7.1/lib/python2.6/site-packages/scipy/stats/kde.py", line 334, in _compute_covariance self.inv_cov = linalg.inv(self.covariance) File "/srv/pkg/python/python-packages/python26/scipy/scipy-0.7.1/lib/python2.6/site-packages/scipy/linalg/basic.py", line 382, in inv if info>0: raise LinAlgError, "singular matrix" numpy.linalg.linalg.LinAlgError: singular matrix What's a way to get around this? In this case, I'd like it to return a density that's essentially peaked completely at a difference of 0, with no mass everywhere else. thanks.

  Read the article

• Fastest image iteration in Python

  - by Greg

  I am creating a simple green screen app with Python 2.7.4 but am getting quite slow results. I am currently using PIL 1.1.7 to load and iterate the images and saw huge speed-ups changing from the old getpixel() to the newer load() and pixel access object indexing. However the following loop still takes around 2.5 seconds to run for an image of around 720p resolution: def colorclose(Cb_p, Cr_p, Cb_key, Cr_key, tola, tolb): temp = math.sqrt((Cb_key-Cb_p)**2+(Cr_key-Cr_p)**2) if temp < tola: return 0.0 else: if temp < tolb: return (temp-tola)/(tolb-tola) else: return 1.0 .... for x in range(width): for y in range(height): Y, cb, cr = fg_cbcr_list[x, y] mask = colorclose(cb, cr, cb_key, cr_key, tola, tolb) mask = 1 - mask bgr, bgg, bgb = bg_list[x,y] fgr, fgg, fgb = fg_list[x,y] pixels[x,y] = ( (int)(fgr - mask*key_color[0] + mask*bgr), (int)(fgg - mask*key_color[1] + mask*bgg), (int)(fgb - mask*key_color[2] + mask*bgb)) Am I doing anything hugely inefficient here which makes it run so slow? I have seen similar, simpler examples where the loop is replaced by a boolean matrix for instance, but for this case I can't see a way to replace the loop. The pixels[x,y] assignment seems to take the most amount of time but not knowing Python very well I am unsure of a more efficient way to do this. Any help would be appreciated.

  Read the article

• correct fisheye distortion

  - by Will

  I have some points that describe positions in a picture taken with a fisheye lens. I've found this description of how to generate a fisheye effect, but not how to reverse it. How do you calculate the radial distance from the centre to go from fisheye to rectilinear? My function stub looks like this: Point correct_fisheye(const Point& p,const Size& img) { // to polar const Point centre = {img.width/2,img.height/2}; const Point rel = {p.x-centre.x,p.y-centre.y}; const double theta = atan2(rel.y,rel.x); double R = sqrt((rel.x*rel.x)+(rel.y*rel.y)); // fisheye undistortion in here please //... change R ... // back to rectangular const Point ret = Point(centre.x+R*cos(theta),centre.y+R*sin(theta)); fprintf(stderr,"(%d,%d) in (%d,%d) = %f,%f = (%d,%d)\n",p.x,p.y,img.width,img.height,theta,R,ret.x,ret.y); return ret; }

  Read the article

• What is the most platform- and Python-version-independent way to make a fast loop for use in Python?

  - by Statto

  I'm writing a scientific application in Python with a very processor-intensive loop at its core. I would like to optimise this as far as possible, at minimum inconvenience to end users, who will probably use it as an uncompiled collection of Python scripts, and will be using Windows, Mac, and (mainly Ubuntu) Linux. It is currently written in Python with a dash of NumPy, and I've included the code below. Is there a solution which would be reasonably fast which would not require compilation? This would seem to be the easiest way to maintain platform-independence. If using something like Pyrex, which does require compilation, is there an easy way to bundle many modules and have Python choose between them depending on detected OS and Python version? Is there an easy way to build the collection of modules without needing access to every system with every version of Python? Does one method lend itself particularly to multi-processor optimisation? (If you're interested, the loop is to calculate the magnetic field at a given point inside a crystal by adding together the contributions of a large number of nearby magnetic ions, treated as tiny bar magnets. Basically, a massive sum of these.) # calculate_dipole # ------------------------- # calculate_dipole works out the dipole field at a given point within the crystal unit cell # --- # INPUT # mu = position at which to calculate the dipole field # r_i = array of atomic positions # mom_i = corresponding array of magnetic moments # --- # OUTPUT # B = the B-field at this point def calculate_dipole(mu, r_i, mom_i): relative = mu - r_i r_unit = unit_vectors(relative) #4pi / mu0 (at the front of the dipole eqn) A = 1e-7 #initalise dipole field B = zeros(3,float) for i in range(len(relative)): #work out the dipole field and add it to the estimate so far B += A*(3*dot(mom_i[i],r_unit[i])*r_unit[i] - mom_i[i]) / sqrt(dot(relative[i],relative[i]))**3 return B

  Read the article

• Why does Raphael's framerate slow down on this code?

  - by Bob

  So I'm just doing a basic orbit simulator using Raphael JS, where I draw one circle as the "star" and another circle as the "planet". It seems to be working just fine, with the one snag that as the simulation continues, its framerate progressively slows down until the orbital motion no longer appears fluid. Here's the code (note: uses jQuery only to initialize the page): $(function() { var paper = Raphael(document.getElementById('canvas'), 640, 480); var star = paper.circle(320, 240, 10); var planet = paper.circle(320, 150, 5); var starVelocity = [0,0]; var planetVelocity = [20.42,0]; var starMass = 3.08e22; var planetMass = 3.303e26; var gravConstant = 1.034e-18; function calculateOrbit() { var accx = 0; var accy = 0; accx = (gravConstant * starMass * ((star.attr('cx') - planet.attr('cx')))) / (Math.pow(circleDistance(), 3)); accy = (gravConstant * starMass * ((star.attr('cy') - planet.attr('cy')))) / (Math.pow(circleDistance(), 3)); planetVelocity[0] += accx; planetVelocity[1] += accy; planet.animate({cx: planet.attr('cx') + planetVelocity[0], cy: planet.attr('cy') + planetVelocity[1]}, 150, calculateOrbit); paper.circle(planet.attr('cx'), planet.attr('cy'), 1); // added to 'trace' orbit } function circleDistance() { return (Math.sqrt(Math.pow(star.attr('cx') - planet.attr('cx'), 2) + Math.pow(star.attr('cy') - planet.attr('cy'), 2))); } calculateOrbit(); }); It doesn't appear, to me anyway, that any part of that code would cause the animation to gradually slow down to a crawl, so any help solving the problem will be appreciated!

  Read the article

• Can a GeneralPath be modified?

  - by Dov

  java2d is fairly expressive, but requires constructing lots of objects. In contrast, the older API would let you call methods to draw various shapes, but lacks all the new features like transparency, stroke, etc. Java has fairly high costs associated with object creation. For speed, I would like to create a GeneralPath whose structure does not change, but go in and change the x,y points inside. path = new GeneralPath(GeneralPath.WIND_EVEN_ODD, 10); path.moveTo(x,y); path.lineTo(x2, y2); double len = Math.sqrt((x2-x)*(x2-x) + (y2-y)*(y2-y)); double dx = (x-x2) * headLen / len; double dy = (y-y2) * headLen / len; double dx2 = -dy * (headWidth/headLen); double dy2 = dx * (headWidth/headLen); path.lineTo(x2 + dx + dx2, y2 + dy + dy2); path.moveTo(x2 + dx - dx2, y2 + dy - dy2); path.lineTo(x2,y2); This one isn't even that long. Imagine a much longer sequence of commands, and only the ones on the end are changing. I just want to be able to overwrite commands, to have an iterator effectively. Does that exist?

  Read the article

• 1D Function into 2D Function Interpolation

  - by Drazick

  Hello. I have a 1D function which I want to interpolate into 2D function. I know the function should have "Polar Symmetry". Hence I use the following code (Matlab Syntax): Assuming the 1D function is LSF of the length 15. [x, y] = meshgrid([-7:7]); r = sqrt(x.^2 + y.^2); PSF = interp1([-7:7], LSF, r(:)); % Sometimes using 'spline' option, same results. PSF = reshape(PSF, [7, 7]); I have few problems: 1. Got some overshoot at the edges (As there some Extrapolation). 2. Can't enforce some assumptions (Monotonic, Non Negative). Is there a better Interpolation method for those circumstances? I couldn't find "Lanczos" based interpolation I can use the same way as interp1 (For a certain vector of points, in "imresize" you can only set the length). Is there such function anywhere? Has anyone encountered a function which allows enforcing some assumptions (Monotonically Decreasing, Non Negative, etc..). Thanks.

  Read the article

• How to implement Excel Solver functionality in C#?

  - by Vic

  Hi, I have an application in C#, I need to do some optimization calculations, like Excel Solver Add-in does, one option is certainly to write my own solver implementation, but I'm kind of short of time, so I'm looking into libraries that already exist that can help me with this. I've been trying the Microsoft Solver Foundation, which seems pretty neat and cool, the problem is that it doesn't seem to work with the kind of calculations that I need to do. At the end of this question I'm adding the information about the calculations I need to perform and optimize. So basically my question is if any of you know of any other library that I can use for this purpose, or any tutorial that can help to do my own solver, or any idea that gives me a lead to solve this issue. Thanks. Additional Info: This is the data I need to calculate: I have 7 variables, lets call them var1, var2,...,var7 The constraints for these variables are: All of them need to be 0 <= varn <= 0.5 (where n is the number of the variable) The sum of all the variables should be equal to 1 The objective is to maximize the target formula, which in Excel looks like this: (MMULT(TRANSPOSE(L26:L32),M14:M20)) / (SQRT(MMULT(MMULT(TRANSPOSE(L26:L32),M4:S10),L26:L32))) The range that you see in this formula, L26:L32, is actually the range with the variables from above, var1, var2,..., varn. M14:M20 and M4:S10 are ranges with data that I get from different sources, there are more likely decimal values. As I said before, I was using Microsoft Solver Foundation, I modeled pretty much everything with it, I created functions that handle the operations of the target formula, but when I tried to solve the model it always fail, I think it is because of the complexity of the operations. In any case, I just wanted to show these data so you can have an idea about the kind of calculations that I need to implement.

  Read the article

• Why does Clojure hang after hacing performed my calculations?

  - by Thomas

  Hi all, I'm experimenting with filtering through elements in parallel. For each element, I need to perform a distance calculation to see if it is close enough to a target point. Never mind that data structures already exist for doing this, I'm just doing initial experiments for now. Anyway, I wanted to run some very basic experiments where I generate random vectors and filter them. Here's my implementation that does all of this (defn pfilter [pred coll] (map second (filter first (pmap (fn [item] [(pred item) item]) coll)))) (defn random-n-vector [n] (take n (repeatedly rand))) (defn distance [u v] (Math/sqrt (reduce + (map #(Math/pow (- %1 %2) 2) u v)))) (defn -main [& args] (let [[n-str vectors-str threshold-str] args n (Integer/parseInt n-str) vectors (Integer/parseInt vectors-str) threshold (Double/parseDouble threshold-str) random-vector (partial random-n-vector n) u (random-vector)] (time (println n vectors (count (pfilter (fn [v] (< (distance u v) threshold)) (take vectors (repeatedly random-vector)))))))) The code executes and returns what I expect, that is the parameter n (length of vectors), vectors (the number of vectors) and the number of vectors that are closer than a threshold to the target vector. What I don't understand is why the programs hangs for an additional minute before terminating. Here is the output of a run which demonstrates the error $ time lein run 10 100000 1.0 [null] 10 100000 12283 [null] "Elapsed time: 3300.856 msecs" real 1m6.336s user 0m7.204s sys 0m1.495s Any comments on how to filter in parallel in general are also more than welcome, as I haven't yet confirmed that pfilter actually works.

  Read the article

• Setting up DrJava to work through Friedman / Felleisen "A Little Java"

  - by JDelage

  All, I'm going through the Friedman & Felleisen book "A Little Java, A Few Patterns". I'm trying to type the examples in DrJava, but I'm getting some errors. I'm a beginner, so I might be making rookie mistakes. Here is what I have set-up: public class ALittleJava { //ABSTRACT CLASS POINT abstract class Point { abstract int distanceToO(); } class CartesianPt extends Point { int x; int y; int distanceToO(){ return((int)Math.sqrt(x*x+y*y)); } CartesianPt(int _x, int _y) { x=_x; y=_y; } } class ManhattanPt extends Point { int x; int y; int distanceToO(){ return(x+y); } ManhattanPt(int _x, int _y){ x=_x; y=_y; } } } And on the main's side: public class Main{ public static void main (String [] args){ Point y = new ManhattanPt(2,8); System.out.println(y.distanceToO()); } } The compiler cannot find the symbols Point and ManhattanPt in the program. If I precede each by ALittleJava., I get another error in the main, i.e., an enclosing instance that contains ALittleJava.ManhattanPt is required I've tried to find ressources on the 'net, but the book must have a pretty confidential following and I couldn't find much. Thank you all. JDelage

  Read the article

• Can you tell me why this generates time limit exceeded in spoj(Prime Number Generator)

  - by magiix

  #include<iostream> #include<string.h> #include<math.h> using namespace std; bool prime[1000000500]; void generate(long long end) { memset(prime,true,sizeof(prime)); prime[0]=false; prime[1]=false; for(long long i=0;i<=sqrt(end);i++) { if(prime[i]==true) { for(long long y=i*i;y<=end;y+=i) { prime[y]=false; } } } } int main() { int n; long long b,e; scanf("%d",&n); while(n--) { cin>>b>>e; generate(e); for(int i=b;i<e;i++) { if(prime[i]) printf("%d\n",i); } } return 0; } That's my code for spoj prime generator. Altought it generates the same output as another accepted code ..

  Read the article

• Simple in-place discrete fourier transform ( DFT )

  - by Adam

  I'm writing a very simple in-place DFT. I am using the formula shown here: http://en.wikipedia.org/wiki/Discrete_Fourier_transform#Definition along with Euler's formula to avoid having to use a complex number class just for this. So far I have this: private void fft(double[] data) { double[] real = new double[256]; double[] imag = new double[256]; double pi_div_128 = -1 * Math.PI / 128; for (int k = 0; k < 256; k++) { for (int n = 0; n < 256; n++) { real[k] += data[k] * Math.Cos(pi_div_128 * k * n); imag[k] += data[k] * Math.Sin(pi_div_128 * k * n); } data[k] = Math.Sqrt(real[k] * real[k] + imag[k] * imag[k]); } } But the Math.Cos and Math.Sin terms eventually go both positive and negative, so as I'm adding those terms multiplied with data[k], they cancel out and I just get some obscenely small value. I see how it is happening, but I can't make sense of how my code is perhaps mis-representing the mathematics. Any help is appreciated. FYI, I do have to write my own, I realize I can get off-the shelf FFT's.

  Read the article

• Draw LINE_STRIP with Unity

  - by Boozzz

  For a new project I am thinking about whether to use OpenGL or Unity3d. I have a bit of experience with OpenGL, but I am completely new to Unity. I already read through the Unity documentation and tutorials on the Unity Website. However, I could not find a way to draw a simple Line-Strip with Unity. In the following example (C#, OpenGL/SharpGL) I draw a round trajectory from a predifined point to an obstacle, which can be imagined as a divided circle with midpoint [cx,cy] and radius r. The position (x-y coordinates) of the obstacle is given by obst_x and obst_y. Question 1: How could I do the same with Unity? Question 2: In my new project, I will have to draw quite a lot of such geometric primitives. Does it make any sense to use Unity for those things? void drawCircle(float cx, float cy, float r, const float obst_x, const float obst_y) { float theta = 0.0f, pos_x, pos_y, dist; const float delta = 0.1; glBegin(GL_LINE_STRIP); while (theta < 180) { theta += delta; //get the current angle float x = r * cosf(theta); //calculate the x component float y = r * sinf(theta); //calculate the y component pos_x = x + cx; //calculate current x position pos_y = y + cy; //calculate current y position //calculate distance from current vertex to obstacle dist = sqrt(pow(pos_x - obst_x) + pow(pos_y - obst_y)); //check if current vertex intersects with obstacle if dist <= 0 { break; //stop drawing circle } else { glVertex2f(pos_x, pos_y); //draw vertex } } glEnd(); }

  Read the article

• correcting fisheye distortion programmatically

  - by Will

  I have some points that describe positions in a picture taken with a fisheye lens. I've found this description of how to generate a fisheye effect, but not how to reverse it. How do you calculate the radial distance from the centre to go from fisheye to rectilinear? My function stub looks like this: Point correct_fisheye(const Point& p,const Size& img) { // to polar const Point centre = {img.width/2,img.height/2}; const Point rel = {p.x-centre.x,p.y-centre.y}; const double theta = atan2(rel.y,rel.x); double R = sqrt((rel.x*rel.x)+(rel.y*rel.y)); // fisheye undistortion in here please //... change R ... // back to rectangular const Point ret = Point(centre.x+R*cos(theta),centre.y+R*sin(theta)); fprintf(stderr,"(%d,%d) in (%d,%d) = %f,%f = (%d,%d)\n",p.x,p.y,img.width,img.height,theta,R,ret.x,ret.y); return ret; } Alternatively, I could somehow convert the image from fisheye to rectilinear before finding the points, but I'm completely befuddled by the OpenCV documentation. Is there a straightforward way to do it in OpenCV, and does it perform well enough to do it to a live video feed?

  Read the article

• Ruby on Rails script/console printing more than expected

  - by Lloyd

  I have a simple model setup in my Ruby on Rails app. (User {name, username, lat, lon}) and am writing a basic extension to the model. I would like the method to return users within a certain distance. It all works just fine in the page view, but as I am debugging I would like to work through some testing using the script/console. My question: It seems to be printing to the screen the entire result set when I run it from the command line and script/console. My model: class User < ActiveRecord::Base def distance_from(aLat, aLon) Math.sqrt((69.1*(aLat - self.lat))**2 + (49*(aLon - self.lon))**2 ) end def distance_from_user(aUser) distance_from(aUser.lat, aUser.lon) end def users_within(distance) close_users = [] users = User.find(:all) users.each do |u| close_users << u if u.distance_from_user(self) < distance end return close_users end end and from the command line I am running >> u = User.find_by_username("someuser") >> print u.users_within(1) So, I guess I would like to know why it's printing the whole result set, and if there is a way to suppress it so as to only print what I want?

  Read the article

• ruby comma operator and step question

  - by ryan_m

  so, i'm trying to learn ruby by doing some project euler questions, and i've run into a couple things i can't explain, and the comma ?operator? is in the middle of both. i haven't been able to find good documentation for this, maybe i'm just not using the google as I should, but good ruby documentation seems a little sparse . . . 1: how do you describe how this is working? the first snippet is the ruby code i don't understand, the second is the code i wrote that does the same thing only after painstakingly tracing the first: #what is this doing? cur, nxt = nxt, cur + nxt #this, apparently, but how to describe the above? nxt = cur + nxt cur = nxt - cur 2: in the following example, how do you describe what the line with 'step' is doing? from what i can gather, the step command works like (range).step(step_size), but this seems to be doing (starting_point).step(ending_point, step_size). Am i right with this assumption? where do i find good doc of this? #/usr/share/doc/ruby1.9.1-examples/examples/sieve.rb # sieve of Eratosthenes max = Integer(ARGV.shift || 100) sieve = [] for i in 2 .. max sieve[i] = i end for i in 2 .. Math.sqrt(max) next unless sieve[i] (i*i).step(max, i) do |j| sieve[j] = nil end end puts sieve.compact.join(", ")

  Read the article

• Optimal two variable linear regression SQL statement (censoring outliers)

  - by Dave Jarvis

  Problem Am looking to apply the y = mx + b equation (where m is SLOPE, b is INTERCEPT) to a data set, which is retrieved as shown in the SQL code. The values from the (MySQL) query are: SLOPE = 0.0276653965651912 INTERCEPT = -57.2338357550468 SQL Code SELECT ((sum(t.YEAR) * sum(t.AMOUNT)) - (count(1) * sum(t.YEAR * t.AMOUNT))) / (power(sum(t.YEAR), 2) - count(1) * sum(power(t.YEAR, 2))) as SLOPE, ((sum( t.YEAR ) * sum( t.YEAR * t.AMOUNT )) - (sum( t.AMOUNT ) * sum(power(t.YEAR, 2)))) / (power(sum(t.YEAR), 2) - count(1) * sum(power(t.YEAR, 2))) as INTERCEPT FROM (SELECT D.AMOUNT, Y.YEAR FROM CITY C, STATION S, YEAR_REF Y, MONTH_REF M, DAILY D WHERE -- For a specific city ... -- C.ID = 8590 AND -- Find all the stations within a 15 unit radius ... -- SQRT( POW( C.LATITUDE - S.LATITUDE, 2 ) + POW( C.LONGITUDE - S.LONGITUDE, 2 ) ) <15 AND -- Gather all known years for that station ... -- S.STATION_DISTRICT_ID = Y.STATION_DISTRICT_ID AND -- The data before 1900 is shaky; insufficient after 2009. -- Y.YEAR BETWEEN 1900 AND 2009 AND -- Filtered by all known months ... -- M.YEAR_REF_ID = Y.ID AND -- Whittled down by category ... -- M.CATEGORY_ID = '001' AND -- Into the valid daily climate data. -- M.ID = D.MONTH_REF_ID AND D.DAILY_FLAG_ID <> 'M' GROUP BY Y.YEAR ORDER BY Y.YEAR ) t Data The data is visualized here (with five outliers highlighted): Questions How do I return the y value against all rows without repeating the same query to collect and collate the data? That is, how do I "reuse" the list of t values? How would you change the query to eliminate outliers (at an 85% confidence interval)? The following results (to calculate the start and end points of the line) appear incorrect. Why are the results off by ~10 degrees (e.g., outliers skewing the data)? (1900 * 0.0276653965651912) + (-57.2338357550468) = -4.66958228 (2009 * 0.0276653965651912) + (-57.2338357550468) = -1.65405406 I would have expected the 1900 result to be around 10 (not -4.67) and the 2009 result to be around 11.50 (not -1.65). Thank you!

  Read the article

• Precision of cos(atan2(y,x)) versus using complex <double>, C++

  - by Ivan

  Hi all, I'm writing some coordinate transformations (more specifically the Joukoswky Transform, Wikipedia Joukowsky Transform), and I'm interested in performance, but of course precision. I'm trying to do the coordinate transformations in two ways: 1) Calculating the real and complex parts in separate, using double precision, as below: double r2 = chi.x*chi.x + chi.y*chi.y; //double sq = pow(r2,-0.5*n) + pow(r2,0.5*n); //slow!!! double sq = sqrt(r2); //way faster! double co = cos(atan2(chi.y,chi.x)); double si = sin(atan2(chi.y,chi.x)); Z.x = 0.5*(co*sq + co/sq); Z.y = 0.5*si*sq; where chi and Z are simple structures with double x and y as members. 2) Using complex : Z = 0.5 * (chi + (1.0 / chi)); Where Z and chi are complex . There interesting part is that indeed the case 1) is faster (about 20%), but the precision is bad, giving error in the third decimal number after the comma after the inverse transform, while the complex gives back the exact number. So, the problem is on the cos(atan2), sin(atan2)? But if it is, how the complex handles that? Thanks!

  Read the article

• Optimal two variable linear regression SQL statement

  - by Dave Jarvis

  Problem Am looking to apply the y = mx + b equation (where m is SLOPE, b is INTERCEPT) to a data set, which is retrieved as shown in the SQL code. The values from the (MySQL) query are: SLOPE = 0.0276653965651912 INTERCEPT = -57.2338357550468 SQL Code SELECT ((sum(t.YEAR) * sum(t.AMOUNT)) - (count(1) * sum(t.YEAR * t.AMOUNT))) / (power(sum(t.YEAR), 2) - count(1) * sum(power(t.YEAR, 2))) as SLOPE, ((sum( t.YEAR ) * sum( t.YEAR * t.AMOUNT )) - (sum( t.AMOUNT ) * sum(power(t.YEAR, 2)))) / (power(sum(t.YEAR), 2) - count(1) * sum(power(t.YEAR, 2))) as INTERCEPT FROM (SELECT D.AMOUNT, Y.YEAR FROM CITY C, STATION S, YEAR_REF Y, MONTH_REF M, DAILY D WHERE -- For a specific city ... -- C.ID = 8590 AND -- Find all the stations within a 5 unit radius ... -- SQRT( POW( C.LATITUDE - S.LATITUDE, 2 ) + POW( C.LONGITUDE - S.LONGITUDE, 2 ) ) <15 AND -- Gather all known years for that station ... -- S.STATION_DISTRICT_ID = Y.STATION_DISTRICT_ID AND -- The data before 1900 is shaky; and insufficient after 2009. -- Y.YEAR BETWEEN 1900 AND 2009 AND -- Filtered by all known months ... -- M.YEAR_REF_ID = Y.ID AND -- Whittled down by category ... -- M.CATEGORY_ID = '001' AND -- Into the valid daily climate data. -- M.ID = D.MONTH_REF_ID AND D.DAILY_FLAG_ID <> 'M' GROUP BY Y.YEAR ORDER BY Y.YEAR ) t Data The data is visualized here: Questions How do I return the y value against all rows without repeating the same query to collect and collate the data? That is, how do I "reuse" the list of t values? How would you change the query to eliminate outliers (at an 85% confidence interval)? The following results (to calculate the start and end points of the line) appear incorrect. Why are the results off by ~10 degrees (e.g., outliers skewing the data)? (1900 * 0.0276653965651912) + (-57.2338357550468) = -4.66958228 (2009 * 0.0276653965651912) + (-57.2338357550468) = -1.65405406 I would have expected the 1900 result to be around 10 (not -4.67) and the 2009 result to be around 11.50 (not -1.65). Thank you!

  Read the article

• Optimal two variable linear regression calculation

  - by Dave Jarvis

  Problem Am looking to apply the y = mx + b equation (where m is SLOPE, b is INTERCEPT) to a data set, which is retrieved as shown in the SQL code. The values from the (MySQL) query are: SLOPE = 0.0276653965651912 INTERCEPT = -57.2338357550468 SQL Code SELECT ((sum(t.YEAR) * sum(t.AMOUNT)) - (count(1) * sum(t.YEAR * t.AMOUNT))) / (power(sum(t.YEAR), 2) - count(1) * sum(power(t.YEAR, 2))) as SLOPE, ((sum( t.YEAR ) * sum( t.YEAR * t.AMOUNT )) - (sum( t.AMOUNT ) * sum(power(t.YEAR, 2)))) / (power(sum(t.YEAR), 2) - count(1) * sum(power(t.YEAR, 2))) as INTERCEPT, FROM (SELECT D.AMOUNT, Y.YEAR FROM CITY C, STATION S, YEAR_REF Y, MONTH_REF M, DAILY D WHERE -- For a specific city ... -- C.ID = 8590 AND -- Find all the stations within a 15 unit radius ... -- SQRT( POW( C.LATITUDE - S.LATITUDE, 2 ) + POW( C.LONGITUDE - S.LONGITUDE, 2 ) ) < 15 AND -- Gather all known years for that station ... -- S.STATION_DISTRICT_ID = Y.STATION_DISTRICT_ID AND -- The data before 1900 is shaky; insufficient after 2009. -- Y.YEAR BETWEEN 1900 AND 2009 AND -- Filtered by all known months ... -- M.YEAR_REF_ID = Y.ID AND -- Whittled down by category ... -- M.CATEGORY_ID = '001' AND -- Into the valid daily climate data. -- M.ID = D.MONTH_REF_ID AND D.DAILY_FLAG_ID <> 'M' GROUP BY Y.YEAR ORDER BY Y.YEAR ) t Data The data is visualized here: Question The following results (to calculate the start and end points of the line) appear incorrect. Why are the results off by ~10 degrees (e.g., outliers skewing the data)? (1900 * 0.0276653965651912) + (-57.2338357550468) = -4.66958228 (2009 * 0.0276653965651912) + (-57.2338357550468) = -1.65405406 I would have expected the 1900 result to be around 10 (not -4.67) and the 2009 result to be around 11.50 (not -1.65). Related Sites Least absolute deviations Robust regression Thank you!

  Read the article

• Rotate MapView in Android

  - by Matthew B.

  I am writing an Android app where one of the features is that the map will rotate according to the compass (i.e. if the phone is pointing east, the map will be oriented so that the east side of the map is on top). Previous answers that I have found suggested over writing the onDraw() method in mapView, however, the api changed the method to final so it cannot be overwritten. As a result I have tried to overwrite the dispatchDraw() method like so: Note: -compass is a boolean that if true, rotate the view -bearing is a float variable that has the degrees that the view should rotate protected void dispatchDraw(Canvas canvas) { canvas.save(); if (compass) { final float w = this.getWidth(); final float h = this.getHeight(); final float scaleFactor = (float)(Math.sqrt(h * h + w * w) / Math.min(w, h)); final float centerX = w / 2.0f; final float centerY = h / 2.0f; canvas.rotate(bearing, centerX, centerY); canvas.scale(scaleFactor, scaleFactor, centerX, centerY); } super.dispatchDraw(canvas); canvas.restore(); }

  Read the article

• MySQL 1064 error, works in command line and phpMyAdmin; not in app

  - by hundleyj

  Here is my query: select * from (select *, 3956 * 2 * ASIN(SQRT(POWER(SIN(RADIANS(45.5200077 - lat)/ 2), 2) + COS(RADIANS(45.5200077)) * COS(RADIANS(lat)) * POWER(SIN(RADIANS(-122.6942014 - lng)/2),2))) AS distance from stops order by distance, route asc) as p group by route, dir order by distance asc limit 10 This works fine at the command line and in PHPMyAdmin. I'm using Dbslayer to connect to MySQL via my JavaScript backend, and the request is returning a 1064 error. Here is the encoded DBSlayer request string: http://localhost:9090/db?{%22SQL%22:%22select%20*%20from%20%28select%20*,%203956%20*%202%20*%20ASIN%28SQRT%28POWER%28SIN%28RADIANS%2845.5200077%20-%20lat%29/%202%29,%202%29%20+%20COS%28RADIANS%2845.5200077%29%29%20*%20COS%28RADIANS%28lat%29%29%20*%20POWER%28SIN%28RADIANS%28-122.6942014%20-%20lng%29/2%29,2%29%29%29%20AS%20distance%20from%20%60stops%60%20order%20by%20%60distance%60,%20%60route%60%20asc%29%20as%20p%20group%20by%20%60route%60,%20%60dir%60%20order%20by%20%60distance%60%20asc%20limit%2010%22} And the response: {"MYSQL_ERRNO" : 1064 , "MYSQL_ERROR" : "You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '(RADIANS(45.5200077)) * COS(RADIANS(lat)) * POWER(SIN(RADIANS(-122.6942014 - lng' at line 1" , "SERVER" : "trimet"} Thanks!

  Read the article

• Choosing between instance methods and separate functions?

  - by StackedCrooked

  Adding functionality to a class can be done by adding a method or by defining a function that takes an object as its first parameter. Most programmers that I know would choose for the solution of adding a instance method. However, I sometimes prefer to create a separate function. For example, in the example code below Area and Diagonal are defined as separate functions instead of methods. I find it better this way because I think these functions provide enhancements rather than core functionality. Is this considered a good/bad practice? If the answer is "it depends", then what are the rules for deciding between adding method or defining a separate function? class Rect { public: Rect(int x, int y, int w, int h) : mX(x), mY(y), mWidth(w), mHeight(h) { } int x() const { return mX; } int y() const { return mY; } int width() const { return mWidth; } int height() const { return mHeight; } private: int mX, mY, mWidth, mHeight; }; int Area(const Rect & inRect) { return inRect.width() * inRect.height(); } float Diagonal(const Rect & inRect) { return std::sqrt(std::pow(static_cast<float>(inRect.width()), 2) + std::pow(static_cast<float>(inRect.height()), 2)); }

  Read the article

• Designing small comparable objects

  - by Thomas Ahle

  Intro Consider you have a list of key/value pairs: (0,a) (1,b) (2,c) You have a function, that inserts a new value between two current pairs, and you need to give it a key that keeps the order: (0,a) (0.5,z) (1,b) (2,c) Here the new key was chosen as the average between the average of keys of the bounding pairs. The problem is, that you list may have milions of inserts. If these inserts are all put close to each other, you may end up with keys such to 2^(-1000000), which are not easily storagable in any standard nor special number class. The problem How can you design a system for generating keys that: Gives the correct result (larger/smaller than) when compared to all the rest of the keys. Takes up only O(logn) memory (where n is the number of items in the list). My tries First I tried different number classes. Like fractions and even polynomium, but I could always find examples where the key size would grow linear with the number of inserts. Then I thought about saving pointers to a number of other keys, and saving the lower/greater than relationship, but that would always require at least O(sqrt) memory and time for comparison. Extra info: Ideally the algorithm shouldn't break when pairs are deleted from the list.

  Read the article

< Previous Page | 8 9 10 11 12 13 14 15 16  | Next Page >