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Search found 1110 results on 45 pages for 'linear interpolation'.

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  • Linear regression confidence intervals in SQL

    - by Matt Howells
    I'm using some fairly straight-forward SQL code to calculate the coefficients of regression (intercept and slope) of some (x,y) data points, using least-squares. This gives me a nice best-fit line through the data. However we would like to be able to see the 95% and 5% confidence intervals for the line of best-fit (the curves below). What these mean is that the true line has 95% probability of being below the upper curve and 95% probability of being above the lower curve. How can I calculate these curves? I have already read wikipedia etc. and done some googling but I haven't found understandable mathematical equations to be able to calculate this. Edit: here is the essence of what I have right now. --sample data create table #lr (x real not null, y real not null) insert into #lr values (0,1) insert into #lr values (4,9) insert into #lr values (2,5) insert into #lr values (3,7) declare @slope real declare @intercept real --calculate slope and intercept select @slope = ((count(*) * sum(x*y)) - (sum(x)*sum(y)))/ ((count(*) * sum(Power(x,2)))-Power(Sum(x),2)), @intercept = avg(y) - ((count(*) * sum(x*y)) - (sum(x)*sum(y)))/ ((count(*) * sum(Power(x,2)))-Power(Sum(x),2)) * avg(x) from #lr Thank you in advance.

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  • Non-graphical linearity estimation

    - by aL3xa
    In my previous post, I was looking for correlation ratio (η or η2) routines in R. I was surprised by the fact that no one uses η for linearity checking in the GLM procedures. Let's start form a simple example: how do you check linearity of bivariate correlation? Solely with scatterplot? There are several ways of doing this, one way is to compare linear and non-linear model R2, then to apply F test to seek for significant difference between them. Finally, the question is: How do you check linearity, the "non-grafical" way?

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  • Initial conditions with a non-linear ODE in Mathematica

    - by buggy
    Hi, I'm trying to use Mathematica's NDSolve[] to compute a geodesic along a sphere using the coupled ODE: x" - (x" . x) x = 0 The problem is that I can only enter initial conditions for x(0) and x'(0) and the solver is happy with the solution where x" = 0. The problem is that my geodesic on the sphere has the initial condition that x"(0) = -x(0), which I have no idea how to tell mathematica. If I add this as a condition, it says I'm adding True to the list of conditions. Here is my code: s1 = NDSolve[{x1''[t] - (x1[t] * x1''[t] + x2[t] * x2''[t] + x3[t]*x3''[t]) * x1[t] == 0, x2''[t] - (x1[t] * x1''[t] + x2[t] * x2''[t] + x3[t]*x3''[t]) * x2[t] == 0, x3''[t] - (x1[t] * x1''[t] + x2[t] * x2''[t] + x3[t]*x3''[t]) * x3[t] == 0, x1[0] == 1, x2[0] == 0, x3[0] == 0, x1'[0] == 0, x2'[0] == 0, x3'[0] == 1} , { x1, x2, x3}, {t, -1, 1}][[1]] I would like to modify this so that the initial acceleration is not zero but -x(0). Thanks

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  • Linear Recurrence for very large n

    - by Android Decoded
    I was trying to solve this problem on SPOJ (http://www.spoj.pl/problems/REC/) F(n) = a*F(n-1) + b where we have to find F(n) Mod (m) where 0 <= a, b, n <= 10^100 1 <= M <= 100000 I am trying to solve it with BigInteger in JAVA but if I run a loop from 0 to n its getting TLE. How could I solve this problem? Can anyone give some hint? Don't post the solution. I want hint on how to solve it efficiently.

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  • Can I interpolate two HEX color values without converting them to RGB?

    - by navand
    I'm trying to make a Gradient Class for a Blackberry app. At first I thought about converting the HEX values to RGB and then interpolating them before converting the result back into HEX, but since I will be doing this for every pixel line of an area, and the calculations will be made by a mobile, I thought that maybe there's a more efficient way of doing it. Maybe involving those pesky bitwise operators which I know nothing of... or something. So, is there a way of interpolating without converting to RGB and back? If so, is it faster than the original way? In any case, can you help me make the most efficient color interpolation? Thank you in advance!

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  • What's the most effective way to interpolate between two colors? (pseudocode and bitwise ops expecte

    - by navand
    Making a Blackberry app, want a Gradient class. What's the most effective way (as in, speed and battery life) to interpolate two colors? Please be specific. // Java, of course int c1 = 0xFFAA0055 // color 1, ARGB int c2 = 0xFF00CCFF // color 2, ARGB float st = 0 // the current step in the interpolation, between 0 and 1 /* Help from here on. Should I separate each channel of each color, convert them to decimal and interpolate? Is there a simpler way? interpolatedChannel = red1+((red2-red1)*st) interpolatedChannel = interpolatedChannel.toString(16) ^ Is this the right thing to do? If speed and effectiveness is important in a mobile app, should I use bitwise operations? Help me! */

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  • Solving for the coefficent of linear equations with one known coefficent

    - by CppLearner
    clc; clear all; syms y a2 a3 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % [ 0.5 0.25 0.125 ] [ a2 ] [ y ] % [ 1 1 1 ] [ a3 ] = [ 3 ] % [ 2 4 8 ] [ 6 ] [ 2 ] %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% M = [0.5 0.25 0.125; 1 1 1; 2 4 8]; t = [a2 a3 6]; r = [y 3 2]; sol = M * t' s1 = solve(sol(1), a2) % solve for a2 s2 = solve(sol(2), a3) % solve for a3 This is what I have so far. These are my output sol = conj(a2)/2 + conj(a3)/4 + 3/4 conj(a2) + conj(a3) + 6 2*conj(a2) + 4*conj(a3) + 48 s1 = - conj(a3)/2 - 3/2 - Im(a3)*i s2 = - conj(a2) - 6 - 2*Im(a2)*i sol looks like what we would have if we put them back into equation form: 0.5 * a2 + 0.25 * a3 + 0.125 * a4 a2 + a3 + a4 = 3 2*a2 + 4*a3 + 8*a4 = 2 where a4 is known == 6. My problem is, I am stuck with how to use solve to actually solve these equations to get the values of a2 and a3. s2 solve for a3 but it doesn't match what we have on paper (not quite). a2 + a3 + 6 = 3 should yield a3 = -3 - a2. because of the imaginary. Somehow I need to equate the vector solution sol to the values [y 3 2] for each row.

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  • Data frame linear fit in R

    - by user1247384
    This is perhaps a simple question, but I am n00b.Say I have a data frame with a bunch of columns. I need to call lm function over the column 1 and 2, 1 and 3, and so on. So basically I need to loop over all columns and store the results of the fit as I build the model. The problem I am running into is that lm(df[1]~df[2], data = df) doesnt work. In this case df is the data frame object and df[1] is the first column. What is a good way to do this in a loop, as in access the columns of df in an iterative fashion. Thanks.

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  • O(N log N) Complexity - Similar to linear?

    - by gav
    Hey All, So I think I'm going to get buried for asking such a trivial but I'm a little confused about something. I have implemented quicksort in Java and C and I was doing some basic comparissons. The graph came out as two straight lines, with the C being 4ms faster than the Java counterpart over 100,000 random integers. The code for my tests can be found here; android-benchmarks I wasn't sure what an (n log n) line would look like but I didn't think it would be straight. I just wanted to check that this is the expected result and that I shouldn't try to find an error in my code. I stuck the formula into excel and for base 10 it seems to be a straight line with a kink at the start. Is this because the difference between log(n) and log(n+1) increases linearly? Thanks, Gav

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  • C# Random of cordinates is linear

    - by Shawn Mclean
    My code is to generate random cordinates of lat and long within a bound: Random lastLat = new Random(); Random lastLon = new Random(); for (int i = 0; i < 50; i++) { int lat = lastLat.Next(516400146, 630304598); //18.51640014679267 - 18.630304598192915 int lon = lastLon.Next(224464416, 341194152); //-72.34119415283203 - -72.2244644165039 SamplePostData d0 = new SamplePostData(); d0.Location = new Location(Convert.ToDouble("18." + lat), Convert.ToDouble("-72." + lon)); AddPushpin(d0); } My output looks like this: Is there something wrong with how my numbers are generated?

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  • UIView animations on a path not linear

    - by chis54
    I have an iOS application that I want to animate a falling leaf (or several). I have my leaf image in an ImageView and I've figured out a simple animation from the documentation: [UIView animateWithDuration:4.0f delay:0 options:UIViewAnimationTransitionFlipFromLeft animations:^(void) { leaf1ImageView.frame = CGRectMake(320, 480, leaf1ImageView.frame.size.width,leaf1ImageView.frame.size.height); } completion:NULL]; This will make the leaf go from its starting position to the bottom right corner in a straight line. How would I animate this to follow a path or curve like a parabola or sinusoid and maybe even rotate the image or view? Would this be done in the animations block? Thanks in advance!

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  • Android layout issue - table/grid/linear

    - by phpmysqlguy
    I am trying to wrap my head around some basic layout issues in android. Here is what I want as my final goal: As you can see, various fields set up like that. The fields get filled in based on XML data. There could be 1 set of fields, or there could be more. I tried a tablelayout, but couldn't get it set up right even when layout_span for Field 7. It worked ok, but when I tried to change the widths of Field 1 thru 5, the spanned row below it didn't conform to the changes (not like an HTML table would). The fields in each group need to lineup if there are more than one (see red lines in image). Can someone point me in the right direction on how I should approach this? Thanks

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  • How do I improve this linear regression function?

    - by user558383
    I have the following PHP function that I'm using to draw a trend line. However, it sometimes plots the line below all the points in the scatter graph. Is there an error in my function or is there a better way to do it. I think it might be something to do with that with the line it produces, it treats all the residuals (the distances from the scatter points to the line) as positive regardless of them being above or below the line. function linear_regression($x, $y) { $n = count($x); $x_sum = array_sum($x); $y_sum = array_sum($y); $xx_sum = 0; $xy_sum = 0; for($i = 0; $i < $n; $i++) { $xy_sum+=($x[$i]*$y[$i]); $xx_sum+=($x[$i]*$x[$i]); } $m = (($n * $xy_sum) - ($x_sum * $y_sum)) / (($n * $xx_sum) - ($x_sum * $x_sum)); $b = ($y_sum - ($m * $x_sum)) / $n; return array("m"=>$m, "b"=>$b); }

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  • How can calculus and linear algebra be useful to a system programmer?

    - by Victor
    I found a website saying that calculus and linear algebra are necessary for System Programming. System Programming, as far as I know, is about osdev, drivers, utilities and so on. I just can't figure out how calculus and linear algebra can be helpful on that. I know that calculus has several applications in science, but in this particular field of programming I just can't imagine how calculus can be so important. The information was on this site: http://www.wikihow.com/Become-a-Programmer Edit: Some answers here are explaining about algorithm complexity and optimization. When I made this question I was trying to be more specific about the area of System's Programming. Algorithm complexity and optimization can be applied to any area of programming not just System's Programming. That may be why I wasn't able to came up with such thinking at the time of the question.

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  • Interpolating data points in Excel

    - by Niels Basjes
    Hi, I'm sure this is the kind of problem other have solved many times before. A group of people are going to do measurements (Home energy usage to be exact). All of them will do that at different times and in different intervals. So what I'll get from each person is a set of {date, value} pairs where there are dates missing in the set. What I need is a complete set of {date, value} pairs where for each date withing the range a value is known (either measured or calculated). I expect that a simple linear interpolation would suffice for this project. If I assume that it must be done in Excel. What is the best way to interpolate in such a dataset (so I have a value for every day) ? Thanks. NOTE: When these datasets are complete I'll determine the slope (i.e. usage per day) and from that we can start doing home-to-home comparisons. ADDITIONAL INFO After first few suggestions: I do not want to manually figure out where the holes are in my measurement set (too many incomplete measurement sets!!). I'm looking for something (existing) automatic to do that for me. So if my input is {2009-06-01, 10} {2009-06-03, 20} {2009-06-06, 110} Then I expect to automatically get {2009-06-01, 10} {2009-06-02, 15} {2009-06-03, 20} {2009-06-04, 50} {2009-06-05, 80} {2009-06-06, 110} Yes, I can write software that does this. I am just hoping that someone already has a "ready to run" software (Excel) feature for this (rather generic) problem.

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  • Using scipy.interpolate.splrep function

    - by Koustav Ghosal
    I am trying to fit a cubic spline to a given set of points. My points are not ordered. I CANNOT sort or reorder the points, since I need that information. But since the function scipy.interpolate.splrep works only on non-duplicate and monotonically increasing points I have defined a function that maps the x-coordinates to a monotonically increasing space. My old points are: xpoints=[4913.0, 4912.0, 4914.0, 4913.0, 4913.0, 4913.0, 4914.0, 4915.0, 4918.0, 4921.0, 4925.0, 4932.0, 4938.0, 4945.0, 4950.0, 4954.0, 4955.0, 4957.0, 4956.0, 4953.0, 4949.0, 4943.0, 4933.0, 4921.0, 4911.0, 4898.0, 4886.0, 4874.0, 4865.0, 4858.0, 4853.0, 4849.0, 4848.0, 4849.0, 4851.0, 4858.0, 4864.0, 4869.0, 4877.0, 4884.0, 4893.0, 4903.0, 4913.0, 4923.0, 4935.0, 4947.0, 4959.0, 4970.0, 4981.0, 4991.0, 5000.0, 5005.0, 5010.0, 5015.0, 5019.0, 5020.0, 5021.0, 5023.0, 5025.0, 5027.0, 5027.0, 5028.0, 5028.0, 5030.0, 5031.0, 5033.0, 5035.0, 5037.0, 5040.0, 5043.0] ypoints=[10557.0, 10563.0, 10567.0, 10571.0, 10575.0, 10577.0, 10578.0, 10581.0, 10582.0, 10582.0, 10582.0, 10581.0, 10578.0, 10576.0, 10572.0, 10567.0, 10560.0, 10550.0, 10541.0, 10531.0, 10520.0, 10511.0, 10503.0, 10496.0, 10490.0, 10487.0, 10488.0, 10488.0, 10490.0, 10495.0, 10504.0, 10513.0, 10523.0, 10533.0, 10542.0, 10550.0, 10556.0, 10559.0, 10560.0, 10559.0, 10555.0, 10550.0, 10543.0, 10533.0, 10522.0, 10514.0, 10505.0, 10496.0, 10490.0, 10486.0, 10482.0, 10481.0, 10482.0, 10486.0, 10491.0, 10497.0, 10506.0, 10516.0, 10524.0, 10534.0, 10544.0, 10552.0, 10558.0, 10564.0, 10569.0, 10573.0, 10576.0, 10578.0, 10581.0, 10582.0] Plots: The code for the mapping function and interpolation is: xnew=[] ynew=ypoints for c3,i in enumerate(xpoints): if np.isfinite(np.log(i*pow(2,c3))): xnew.append(np.log(i*pow(2,c3))) else: if c==0: xnew.append(np.random.random_sample()) else: xnew.append(xnew[c3-1]+np.random.random_sample()) xnew=np.asarray(xnew) ynew=np.asarray(ynew) constant1=10.0 nknots=len(xnew)/constant1 idx_knots = (np.arange(1,len(xnew)-1,(len(xnew)-2)/np.double(nknots))).astype('int') knots = [xnew[i] for i in idx_knots] knots = np.asarray(knots) int_range=np.linspace(min(xnew),max(xnew),len(xnew)) tck = interpolate.splrep(xnew,ynew,k=3,task=-1,t=knots) y1= interpolate.splev(int_range,tck,der=0) The code is throwing an error at the function interpolate.splrep() for some set of points like the above one. The error is: File "/home/neeraj/Desktop/koustav/res/BOS5/fit_spline3.py", line 58, in save_spline_f tck = interpolate.splrep(xnew,ynew,k=3,task=-1,t=knots) File "/usr/lib/python2.7/dist-packages/scipy/interpolate/fitpack.py", line 465, in splrep raise _iermessier(_iermess[ier][0]) ValueError: Error on input data But for other set of points it works fine. For example for the following set of points. xpoints=[1629.0, 1629.0, 1629.0, 1629.0, 1629.0, 1629.0, 1629.0, 1629.0, 1629.0, 1629.0, 1629.0, 1629.0, 1629.0, 1629.0, 1629.0, 1629.0, 1630.0, 1630.0, 1630.0, 1631.0, 1631.0, 1631.0, 1631.0, 1630.0, 1629.0, 1629.0, 1629.0, 1628.0, 1627.0, 1627.0, 1625.0, 1624.0, 1624.0, 1623.0, 1620.0, 1618.0, 1617.0, 1616.0, 1615.0, 1614.0, 1614.0, 1612.0, 1612.0, 1612.0, 1611.0, 1610.0, 1609.0, 1608.0, 1607.0, 1607.0, 1603.0, 1602.0, 1602.0, 1601.0, 1601.0, 1600.0, 1599.0, 1598.0] ypoints=[10570.0, 10572.0, 10572.0, 10573.0, 10572.0, 10572.0, 10571.0, 10570.0, 10569.0, 10565.0, 10564.0, 10563.0, 10562.0, 10560.0, 10558.0, 10556.0, 10554.0, 10551.0, 10548.0, 10547.0, 10544.0, 10542.0, 10541.0, 10538.0, 10534.0, 10532.0, 10531.0, 10528.0, 10525.0, 10522.0, 10519.0, 10517.0, 10516.0, 10512.0, 10509.0, 10509.0, 10507.0, 10504.0, 10502.0, 10500.0, 10501.0, 10499.0, 10498.0, 10496.0, 10491.0, 10492.0, 10488.0, 10488.0, 10488.0, 10486.0, 10486.0, 10485.0, 10485.0, 10486.0, 10483.0, 10483.0, 10482.0, 10480.0] Plots: Can anybody suggest what's happening ?? Thanks in advance......

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  • Graphing perpendicular offsets in a least squares regression plot in R

    - by D W
    I'm interested in making a plot with a least squares regression line and line segments connecting the datapoints to the regression line as illustrated here in the graphic called perpendicular offsets: http://mathworld.wolfram.com/LeastSquaresFitting.html I have the plot and regression line done here: ## Dataset from http://www.apsnet.org/education/advancedplantpath/topics/RModules/doc1/04_Linear_regression.html ## Disease severity as a function of temperature # Response variable, disease severity diseasesev<-c(1.9,3.1,3.3,4.8,5.3,6.1,6.4,7.6,9.8,12.4) # Predictor variable, (Centigrade) temperature<-c(2,1,5,5,20,20,23,10,30,25) ## Fit a linear model for the data and summarize the output from function lm() severity.lm <- lm(diseasesev~temperature,data=severity) # Take a look at the data plot( diseasesev~temperature, data=severity, xlab="Temperature", ylab="% Disease Severity", pch=16 ) abline(severity.lm,lty=1) title(main="Graph of % Disease Severity vs Temperature") Should I use some kind of for loop and segments http://www.iiap.res.in/astrostat/School07/R/html/graphics/html/segments.html to do the perpendicular offsets? Is there a more efficient way? Please provide an example if possible.

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  • How to use Vendor Properties in Multiple Backgrounds?

    - by barraponto
    I want to use multiple backgrounds in css, which are currently supported by Firefox 3.61, Chrome/Safari, supposedly Opera10.5 (doesn't run on gnu/linux). It is working fine, however i would like to use linear-gradients as a background. it works ok for Firefox, doesn't work at all with Chrome, yet i can't figure out how to make it work for both at the same time. any clues? http://snook.ca/archives/html_and_css/multiple-bg-css-gradients came the closest to match what i need, but i couldn't get it to work with chrome yet.

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  • In R draw two lines, with slopes double and half the value of the best fit line

    - by D W
    I have data with a best fit line draw. I need to draw two other lines. One needs to have double the slope and the other need to have half the slope. Later I will use the region to differentially color points outside it as per: http://stackoverflow.com/questions/2687212/conditionally-colour-data-points-outside-of-confidence-bands-in-r Example dataset: ## Dataset from http://www.apsnet.org/education/advancedplantpath/topics/RModules/doc1/04_Linear_regression.html ## Disease severity as a function of temperature # Response variable, disease severity diseasesev<-c(1.9,3.1,3.3,4.8,5.3,6.1,6.4,7.6,9.8,12.4) # Predictor variable, (Centigrade) temperature<-c(2,1,5,5,20,20,23,10,30,25) ## For convenience, the data may be formatted into a dataframe severity <- as.data.frame(cbind(diseasesev,temperature)) ## Fit a linear model for the data and summarize the output from function lm() severity.lm <- lm(diseasesev~temperature,data=severity) # Take a look at the data plot( diseasesev~temperature, data=severity, xlab="Temperature", ylab="% Disease Severity", pch=16, pty="s", xlim=c(0,30), ylim=c(0,30) ) title(main="Graph of % Disease Severity vs Temperature") par(new=TRUE) # don't start a new plot abline(severity.lm, col="blue")

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  • C Programming arrays, I dont understand how I would go about making this program, If anyone can just guide me through the basic outline please :) [on hold]

    - by Rashmi Kohli
    Problem The temperature of a car engine has been measured, from real-world experiments, as shown in the table and graph below: Time (min) Temperature (oC) 0 20 1 36 2 61 3 68 4 77 5 110 Use linear regression to find the engine’s temperature at 1.5 minutes, 4.3 minutes, and any other time specified by the user. Background In engineering, many times we measure several data points in an experiment, but then we need to predict a value that we have not measured which lies between two measured values, such as the problem statement above. If the relation between the measured parameters seems to be roughly linear, then we can use linear regression to find the relationship between those parameters. In the graph of the problem statement above, the relation seems to be roughly linear. Hence, we can apply linear regression to the above problem. Assuming y {y0, y1, …yn-1} has a linear relation with x {x0, x1, … xn-1}, we can say that: y = mx+b where m and b can be found with linear regression as follows: For the problem in this lab, using linear regression gives us the following line (in blue) compared to the measured curve (in red). As you can see, there is usually a difference between the measured values and the estimated (predicted) values. What linear regression does is to minimize those differences and still give us a straight line (blue). Other methods, such as non-linear regression, are also possible to achieve higher accuracy and better curve fitting. Requirements Your program should first print the table of the temperatures similar to the way it’s printed in the problem statement. It should then calculate the temperature at minute 1.5 and 4.3 and show the answers to the user. Next, it should prompt the user to enter a time in minutes (or -1 to quit), and after reading the user’s specified time it should give the value of the engine’s temperature at that time. It should then go back to the prompt. Hints •Use a one dimensional array to store the temperature values given in the problem statement. •Use functions to separate tasks such as calculating m, calculating b, calculating the temperature at a given time, printing the prompt, etc. You can then give your algorithm as well as you pseudo code per function, as opposed to one large algorithm diagram or one large sequence of pseudo code.

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  • Conditionally colour data points outside of confidence bands in R

    - by D W
    I need to colour datapoints that are outside of the the confidence bands on the plot below differently from those within the bands. Should I add a separate column to my dataset to record whether the data points are within the confidence bands? Can you provide an example please? Example dataset: ## Dataset from http://www.apsnet.org/education/advancedplantpath/topics/RModules/doc1/04_Linear_regression.html ## Disease severity as a function of temperature # Response variable, disease severity diseasesev<-c(1.9,3.1,3.3,4.8,5.3,6.1,6.4,7.6,9.8,12.4) # Predictor variable, (Centigrade) temperature<-c(2,1,5,5,20,20,23,10,30,25) ## For convenience, the data may be formatted into a dataframe severity <- as.data.frame(cbind(diseasesev,temperature)) ## Fit a linear model for the data and summarize the output from function lm() severity.lm <- lm(diseasesev~temperature,data=severity) jpeg('~/Desktop/test1.jpg') # Take a look at the data plot( diseasesev~temperature, data=severity, xlab="Temperature", ylab="% Disease Severity", pch=16, pty="s", xlim=c(0,30), ylim=c(0,30) ) title(main="Graph of % Disease Severity vs Temperature") par(new=TRUE) # don't start a new plot ## Get datapoints predicted by best fit line and confidence bands ## at every 0.01 interval xRange=data.frame(temperature=seq(min(temperature),max(temperature),0.01)) pred4plot <- predict( lm(diseasesev~temperature), xRange, level=0.95, interval="confidence" ) ## Plot lines derrived from best fit line and confidence band datapoints matplot( xRange, pred4plot, lty=c(1,2,2), #vector of line types and widths type="l", #type of plot for each column of y xlim=c(0,30), ylim=c(0,30), xlab="", ylab="" )

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