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  • 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!

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  • 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!

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  • 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!

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  • Sparse linear program solver

    - by Jacob
    This great SO answer points to a good sparse solver, but I've got constraints on x (for Ax = b) such that each element in x is >=0 an <=N. The first thing which comes to mind is an LP solver for large sparse matrices. Any ideas/recommendations?

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  • O'Reilly book clarification on 2d linear system

    - by Eric
    The Oreilly book "Learning openCV" states at page 356 : Quote Before we get totally lost, let’s consider a particular realistic situation of taking measurements on a car driving in a parking lot. We might imagine that the state of the car could be summarized by two position variables, x and y, and two velocities, vx and vy. These four variables would be the elements of the state vector xk. Th is suggests that the correct form for F is: x = [ x; y; vx; vy; ]k F = [ 1, 0, dt, 0; 0, 1, 0, dt; 0, 0, 1, 0; 0, 0, 0, 1; ] It seems natural to put 'dt' just there in the F matrix but I just don't get why. What if I have a n states system, how would I spray some "dt" in the F matrix?

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  • Rendering Linear Gradients using the HTML5 Canvas

    - by dwahlin
    Related HTML5 Canvas Posts: Getting Started with the HTML5 Canvas Rendering Text with the HTML5 Canvas Creating a Line Chart using the HTML5 Canvas New Pluralsight Course: HTML5 Canvas Fundamentals Gradients are everywhere. They’re used to enhance toolbars or buttons and help add additional flare to a web page when used appropriately. In the past we’ve always had to rely on images to render gradients which works well, but isn’t necessarily the most efficient (although 1 pixel wide images do work well). CSS3 provides a great way to render gradients in modern browsers (see http://www.colorzilla.com/gradient-editor for a nice online gradient generator tool) but it’s not the only option. If you’re working with charts, games, multimedia or other HTML5 Canvas applications you can also use gradients and render them on the client-side without relying on images. In this post I’ll introduce how to use linear gradients and discuss the different functions that can be used to create them.   Creating Linear Gradients Linear gradients can be created using the 2D context’s createLinearGradient function. The function takes the starting x,y coordinates and ending x,y coordinates of the gradient:   createLinearGradient(x1, y1, x2, y2);   By changing the start and end coordinates you can control the direction that the gradient renders. For example, adding the following coordinates causes the gradient to render from left to right since the y value stays at 0 for both points while the x value changes from 0 to 200. var lgrad = ctx.createLinearGradient(0, 0, 200, 0); Here’s an example of how changing the coordinates affects the gradient direction:   Once a linear gradient object has been created you can set color stops using the addColorStop() function. It takes the location where the color should appear in the gradient with 0 being the beginning and 1 being at the end (0.5 would be in the middle) as well as the color to display in the gradient. lgrad.addColorStop(0, 'white'); lgrad.addColorStop(1, 'gray');   An example of combining createLinearGradient() with addColorStop() is shown next:   Using createLinearGradient() var canvas = document.getElementById('myCanvas'); var ctx = canvas.getContext('2d'); var lgrad = ctx.createLinearGradient(0, 0, 200, 0); lgrad.addColorStop(0, 'white'); lgrad.addColorStop(1, 'gray'); ctx.fillStyle = lgrad; ctx.fillRect(0, 0, 200, 200); ctx.strokeRect(0, 0, 200, 200); This code renders a white to gray gradient as shown next: A live example of using createLinearGradient() is shown next. Click the Result tab to see the code in action.   In the next post on the HTML5 Canvas I’ll take a look at radial gradients and how they can be used. In the meantime, if you’re interested in learning more about the HTML5 Canvas and how it can be used in your Web or Windows 8 applications, check out my HTML5 Canvas Fundamentals course from Pluralsight. It has over 4 1/2 hours of canvas goodness packed in it.

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  • Alpha interpolation in a pixel shader

    - by c4sh
    How does the interpolation in a fragment shader work when it comes to the alpha parameter? I'm programming a shader with SharpDX, DirectX11. My idea is to interpolate 2 3d points of a segment, so that I'll have the position interpolated in between in the pixel shader. But I want to know what happens with the alpha parameter when that position is blocked by another polygon. For instance, if alpha is 1.0 at the left end of my segment and 0.0 at the other one. What is the value of alpha in the middle, 0.5? Or does it depend on the visibility at that point (meaning it could be, for instance, 1.0 OR 0.0 depending on if that part of the segment is hidden by a poolygon?

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  • How to set position for a linear-gradient background in css3

    - by Virender Sehwag
    I am trying to set the position (that is, margin or padding from top) of body tag's linear background with image. My code is background-image: linear-gradient(to bottom, rgba(255, 255, 255, 0), rgba(255, 255, 255, 0), rgba(0, 0, 0, 0.9), rgb(0, 0, 0)), url("g2hd.jpg"); background-repeat: repeat, no-repeat; background-attachment: fixed; background-position: 0% 30px, center center; but 30px is not working but it works for normal for background-image:url("g2hd.jpg"); any idea

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  • Fixed timestep with interpolation in AS3

    - by Jim Sreven
    I'm trying to implement Glenn Fiedler's popular fixed timestep system as documented here: http://gafferongames.com/game-physics/fix-your-timestep/ In Flash. I'm fairly sure that I've got it set up correctly, along with state interpolation. The result is that if my character is supposed to move at 6 pixels per frame, 35 frames per second = 210 pixels a second, it does exactly that, even if the framerate climbs or falls. The problem is it looks awful. The movement is very stuttery and just doesn't look good. I find that the amount of time in between ENTER_FRAME events, which I'm adding on to my accumulator, averages out to 28.5ms (1000/35) just as it should, but individual frame times vary wildly, sometimes an ENTER_FRAME event will come 16ms after the last, sometimes 42ms. This means that at each graphical redraw the character graphic moves by a different amount, because a different amount of time has passed since the last draw. In theory it should look smooth, but it doesn't at all. In contrast, if I just use the ultra simple system of moving the character 6px every frame, it looks completely smooth, even with these large variances in frame times. How can this be possible? I'm using getTimer() to measure these time differences, are they even reliable?

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  • How can I match end-of-line multiple times in a regex without interpolation?

    - by harschware
    Hi, if I have a input with new lines in it like: [INFO] xyz [INFO] How can I pull out the xyz part? I tried a pattern like /^\[INFO\]$(.*?)$\[INFO\]/ms, but perl gives me: Use of uninitialized value $\ in regexp compilation at scripts\t.pl line 6. I've been trying things to get interpolation to stop like using qr// but alas, no love. EDIT: The key is that the end-of-line anchor is a dollar sign but at times it may be necessary to intersperse the end-of-line anchor through the pattern. If the pattern is interpolating then you might get problems such as uninitialized $\. For instance an acceptable solution here is /^\[INFO\]\s*^(.*?)\s*^\[INFO\]/ms but that does not solve the crux of the first problem. I've changed the anchors to be ^ so there is no interpolation going on, and with this input I'm free to do that. But what about when I really do want to reference EOL with $ in my pattern? How do I get the regex to compile?

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  • How can match end-of-line multiple times in a regex without interpolation?

    - by harschware
    Hi, if I have a input with new lines in it like: [INFO] xyz [INFO] How can I pull out the xyz part? I tried a pattern like /^\[INFO\]$(.*?)$\[INFO\]/ms, but perl gives me: Use of uninitialized value $\ in regexp compilation at scripts\t.pl line 6. I've been trying things to get interpolation to stop like using qr// but alas, no love. EDIT: The key is that the end-of-line anchor is a dollar sign but at times it may be necessary to intersperse the end-of-line anchor through the pattern. If the pattern is interpolating then you might get problems such as uninitialized $\. For instance an acceptable solution here is /^\[INFO\]\s*^(.*?)\s*^\[INFO\]/ms but that does not solve the crux of the first problem. I've changed the anchors to be ^ so there is no interpolation going on, and with this input I'm free to do that. But what about when I really do want to reference EOL with $ in my pattern? How do I get the regex to compile?

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  • Speed of interpolation algorithms, C# and C++ working together.

    - by Kaminari
    Hello. I need fast implementation of popular interpolation algorithms. I figured it out that C# in such simple algorithms will be much slower than C++ so i think of writing some native code and using it in my C# GUI. First of all i run some tests and few operations on 1024x1024x3 matrix took 32ms in C# and 4ms in C++ and that's what i basicly need. Interpolation however is not a good word because i need them only for downscaling. But the question is: Will it be faster than C# methods in Drawing2D Image outputImage = new Bitmap(destWidth, destHeight, PixelFormat.Format24bppRgb); Graphics grPhoto = Graphics.FromImage(outputImage); grPhoto.InterpolationMode = InterpolationMode.*; //all of them grPhoto.DrawImage(bmp, new Rectangle(0, 0, destWidth, destHeight), Rectangle(0, 0, sourceWidth, sourceHeight), GraphicsUnit.Pixel); grPhoto.Dispose(); Some of these method run in 20ms and some in 80. Is there a way to do it faster?

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  • Sparse quadratic program solver

    - by Jacob
    This great SO answer points to a good sparse solver, but I've got constraints on x (for Ax = b) such that each element in x is >=0 an <=N. The first thing which comes to mind is an QP solver for large sparse matrices. Also, A is huge (around 2e6x2e6) but very sparse with <=4 elements per row. Any ideas/recommendations?

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  • Windows media scaling/interpolation method

    - by MichaelH
    Usually in Windows, if videos or other media is upscaled from a certain resolution to a higher resolution (e.g. "monitor size"), a bilinear filtering algorithm or similar is used, such that the upscaled material doesn't look blocky. On my system however, the used interpolation algorithm changed from 'bilinear' to 'nearest neighbor' at some point, with the effect that upscaled videos (e.g. viewed in MPC or WMP, and also Skype video streams) and games (e.g. from PopCap) appear rather blocky. Not sure what the common factor between those is, could be DirectShow(?). I am not aware of having changed any setting that could have affected this state, in fact I am not even aware such a setting exists. I'm guesing that some installed software must have changed something on my computer. My computer is running Windows 7, but I had already experienced the same effect on an XP machine some while ago, where it changed back again to the more pleasing bilinear interpolation after a while, as magically as the first time. What could be wrong with this installation, and how can I change this upscaling interpolation behavior?

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  • Create a linear trend line in Excel graphs with logarithmic scale

    - by Redsoft7
    I I have an Excel scatter chart with x and y values. I set the logarithmic scale in x-axis and y-axis. When I add a linear trend line to the graph, the line is not linear but appears like a curve. How can I make a linear trend line on a logarithmic-scaled chart? Sample data: x: 18449 22829 25395 36869 101419 125498 208144 2001508 14359478 17301785 y: 269,09 273,89 239,50 239,50 175,13 176,73 151,94 135,15 131,55 121,55

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  • Increasing speed of circle over time as linear with Box2d

    - by Whispered
    Assume that there is a circle and it can be moved by using keyboard arrows.Is required that increasing speed over time like increasing car speed. For example; max speed is 25 and time to reach max speed shall be 5 sec. Over 5 sec the speed will reach to max speed. Does Box2d handle that situation?. I tried setting linear valocity but it seems to make the circle have constant speed instead of increased speed over time. Thank You! Note: I'm using Box2DWeb Javascript port of Box2D.

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  • How to solve linear recurrences involving two functions?

    - by Aditya Bahuguna
    Actually I came across a question in Dynamic Programming where we need to find the number of ways to tile a 2 X N area with tiles of given dimensions.. Here is the problem statement Now after a bit of recurrence solving I came out with these. F(n) = F(n-1) + F(n-2) + 2G(n-1), and G(n) = G(n-1) + F(n-1) I know how to solve LR model where one function is there.For large N as is the case in the above problem we can do the matrix exponentiation and achieve O(k^3log(N)) time where k is the minimum number such that for all km F(n) does not depend on F(n-k). The method of solving linear recurrence with matrix exponentiation as it is given in that blog. Now for the LR involving two functions can anyone suggest an approach feasible enough for large N.

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  • Show path of a body of where it should go after linear impulse is applied

    - by Farooq Arshed
    I am making a game with Andengine and Box2D. I have a dynamic body and I apply linear impulse on the body to move it around when the user have touched the screen. Now I want to show the path where the body will go when the user have touched. If you have played Angry Birds or Basket Ball Shoot or any other which have projectile motion with a path shown you will get my point. I want to show the white dots which are shown in those games.

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  • -webkit-linear-gradient Not working in Dreamweaver CS6

    - by Ken
    I've tried multiple times to apply the following piece of code in a CSS document: display: block; width: 500px; margin: 500px auto; padding: 15px; text-align: center; border: 4px solid blue; background: -webkit-linear-gradient(top,black,white); outline: 7px solid red; Everything appears correctly, except the gradient. I have CS6 Live View turned on, and it still refuses to appear. All I get in my box is a white background, as opposed to the black to white gradient. However, when I type the same line of code into the trial of Coda 2 I downloaded, it works perfectly. Is there anything I can do to resolve the issue?

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  • Interpolation between two 3D points?

    - by meds
    I'm working with some splines which define a path a character follows (you can see a gameplay video here to get a better understanding of what's going on: http://www.youtube.com/watch?v=BndobjOiZ6g). Basically the characters 'forward' look direction is set to the 'forward' direction of the spline and when players tilt their phone left and right the character is strafed along its 'right' coordinate. The issue with this is (rather obviously) in performance, interpolating over a spline to find the nearest position and tangent relative to the player is an incredibly costly operation. To get by this I cache a finite number of positions in what I call 'SplineDetails', the class is as follows: public class SplineDetails { public SplineDetails() { Forward = Vector3.forward; Position = Vector3.one * float.MaxValue; Alpha = -1; } public float Alpha; // [0,1] measured along length of spline where 0 is the initial point and 1 is the end point of the spline public Vector3 Position; // the point of the spline at this alpha public Vector3 Forward; // the forward tangent of the spline at this alpha } I populate this with say 30 coordinates and I can give a rough estimate of a coordinate and 'forward' based on a position past in. It's not as accurate but it's much faster. But now I'd like to make the system work better by estimating positions and 'forward' directions by interpolating between two of the cached points though I'm stuck trying to figure out some logic. My first problem is, how can I determine between which two points the object is? Given each point can be placed at different intervals along the spline it could mean that two points in front or behind the object can be closer to the object. The other problem is to figure out the proportion between the two paths it's between, i.e. if there is a point a at coordinate (0,0,0) and point b at coordinate (1,0,0) if the object is at position (0.5,0,0) then the result it should give is '0.5' (as it is equal distance away from point a and point b). That's a simple example, but what if the object is at coordinate (0.5,3,0) for example?

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  • How can I teleport seamlessly, without using interpolation?

    - by modchan
    I've been implementing Bukkit plugin for creating toggleable in-game warping areas that will teleport any catched entity to other similar area. I was going to implement concept of non-Euclidean maze using this plugin, but, unfortunately, I've discovered that doing Entity.teleport() causes client to interpolate movement while teleporting, so player slides towards target like Enderman and receives screen updates, so for a split second all underground stuff is visible. While for "just teleport me where I want" usage this is just fine, it ruins whole idea of seamless teleporting, as player can clearly see when transfer happened even without need to look at debug screen. Is there possibility to somehow disable interpolating while teleporting without modifying client, or maybe prevent client from updating screen while it's being teleported?

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  • Spherical harmonics lighting interpolation

    - by TravisG
    I want to use hardware filtering to smooth out colors in texels of a texture when I'm accessing texels at coordinates that are not directly at the center of the texel, the catch being that the texels store 2 bands of spherical harmonics coefficients (=4 coefficients), not RGBA intensity values. Can I just use hardware filtering like that (GL_LINEAR with and without mip mapping) without any considerations? In other terms: If I were to first convert the coefficients back to intensity representations, than manually interpolate between two intensities, would the resulting intensity be the same as if I interpolated between the coefficient vectors directly and then converted the interpolated result to intensities?

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