python,interpolation,track,satellite

You shouldn't interpolate near 0, because the satellite was there at two very different points in time. You should add 360° to the second half of your data (which might correspond with the upper left curve), so that you get a continuous curve from about -30° to 360°. Then interpolate....

python,arrays,numpy,dictionary,interpolation

With a sample dictionary, I constructed the required numpy arrays, x, y, and 2d z. And then used a scipy interpolator to do the rest. import numpy as np a = {0.0 : {0.0: 0, 0.5: 3, 1.0: 6, 1.5: 9}, 0.5 : {0.0: 1, 0.5: 5, 1.0: 9, 1.5:...

So with thanks to @m.s. for providing his code, it turns out the issue is that with the second graph I was interpolating with x= -90:10:90, whereas if I interpolate with 1:19, in a similar manner to the first graph, then the problem is fixed.

javascript,algorithm,math,interpolation

The sample codes in the links you posted are all cubic interpolations either from 4 points or from 2 points and 2 derivatives. The 'mu' is the parameter at which you want to evaluate the y value. The input y0, y1,y2 and y3 are the y coordinates of the 4...

python,arrays,interpolation,indices,interpolate

Does Numpy interp do what you want?: import numpy as np x = [0,1,2] y = [2,3,4] np.interp(0.56, x, y) Out[81]: 2.56 ...

matlab,interpolation,handle,curve-fitting,piecewise

It's not all clear but why not to puts things into a class ? classdef Piecewise < handle methods % Construction function [this] = Piecewise(xmeas, ymeas) ... here compute xTrans and coeffMat... end % Interpolation function [yinterp] = Evaluate(xinterp) ... Here use previously computed xTrans and coeffMat ... end end...

c,math,2d,interpolation,linear-interpolation

You have to agree on an interpolation method first. I would suggest either bilinear or barycentric interpolation. In one of my previous posts I visualized the difference between both methods. I'll concentrate on the bilinear interpolation. We want to transform any point within a cell to its corrected point. Therefore,...

python,matplotlib,interpolation,smooth

Chaikin's corner cutting algorithm might be the ideal approach for you. For a given polygon with vertices as P0, P1, ...P(N-1), the corner cutting algorithm will generate 2 new vertices for each line segment defined by P(i) and P(i+1) as Q(i) = (3/4)P(i) + (1/4)P(i+1) R(i) = (1/4)P(i) + (3/4)P(i+1)...

python,pandas,interpolation,extrapolation

So here is a mask that ought to solve the problem. Just interpolate and then apply the mask to reset appropriate values to NaN. Honestly, this was a bit more work than I realized it would be because I had to loop through each column but then groupby didn't quite...

python,numpy,scipy,interpolation,cubic-spline

This is not possible with interp1d. You can get the spline coefficients from splrep though.

python,dictionary,interpolation

Given e.g to isolate the problem a bit: tee = { 100:2.92, 120:2.99, 140:2.97, 160:2.99, 180:3.05, 200:3.12, 240:3.25, 260:3.45, 280:3.65, 300:3.71, 320:3.79, 340:3.86, 360:3.92, 380:3.96, 400:3.99, 420:4.02, 440:4.08, 460:4.17, 480:4.28, 500:4.41, 520:4.54, 540:4.65, 560:4.74, 580:4.79, 600:4.82 } you could have...: import bisect teekeys = sorted(tee) def lookup(aval): where = bisect.bisect_left(teekeys,...

Why not using splinefun: func = splinefun(x=setx, y=population, method="fmm", ties = mean) Then you define the point to forecast you want: func(seq(1973, 2014, 0.25)) ...

python,scipy,interpolation,delaunay

After some experimenting, the solution looks simple (this post was quite helpful): # dimension of the problem (in this example I use 3D grid, # but the method works for any dimension n>=2) n = 3 # my array of grid points (array of n-dimensional coordinates) points = [[1,2,3], [2,3,4],...

c,floating-point,interpolation

In general, multiplications and divisions rarely cause a significant loss of precision. Because these are floating point numbers, with separate fields for the scale and significant digits, getting large intermediate results in itself isn't an issue. 2e100/3e100 and 2/3 are (for all intents and purposes) equally accurate. On the other...

ruby,ruby-on-rails-4,metaprogramming,interpolation

Here is the example for you: ENTRY_HASHES = { "disbursement" => { :description => 'smth' } } ENTRY_HASHES.each do |key, value| define_method("#{ key }?") do |argument| puts "#{ key }? method called with #{ argument }" puts "you can use hash values: #{ value } of course" end end disbursement?('my_arg')...

haskell,markdown,interpolation,yesod

Well, finally I've found really simple solution. Any quasi quoter could be easily switched to file input rather than inline text with quoteFile function from Language.Haskell.TH.Quote module1. Let's describe in Foundation module following function: import Language.Haskell.TH.Quote (QuasiQuoter, quoteFile) import Text.Shakespeare.Text (st) stFile :: QuasiQuoter stFile = quoteFile st Now we...

r,ggplot2,geospatial,interpolation

The CRAN spatial view got me started on "Kriging". The code below takes ~7 minutes to run on my laptop. You could try simpler interpolations (e.g., some sort of spline). You might also remove some of the locations from the high-density regions. You don't need all of those spots to...

This syntax worked with RectBivariateSpline x2 = np.linspace(xmin,xmax,c1) y2 = np.linspace(ymin,ymax,r1) f2 = sp.interpolate.RectBivariateSpline(x2, y2, dstnc1.T,kx=1, ky=1) I can then evaluate at new points with this: out2 = f2.ev(xnew1,ynew1) For interp2d I am stuck as I am not able to bypass firewall at my office to update Anaconda (Windows). I...

jquery,angularjs,interpolation

You cannot inject HTML into Angular and have it work. At a minimum you will need to wrap the JQuery widget in a directive. <div ng-controller="blah"> <div jquery-target> </div> </div> Then define the directive. module.directive('jqueryTarget', function() { return { template: function(element, attrs) { var blah = '' // generated html....

matrix,interpolation,quaternions

I ended up solving my question through a bit more web surfing. For future reference, heres how to do it. The transformation component is stored in a 4x4 matrix like this: r r r t r r r t r r r t 0 0 0 1 where r is...

math,3d,interpolation,linear-interpolation,bicubic

It would be the same formula as in the original thread, just three dimensions would be used; the approach would be called tri-linear interpolation, as linear interpolation would be used along all three axes. More information can be found on Wikipedia, where also an illustration visualizes the approach. The underlying...

python,api,3d,interpolation,maya

Thank you joojaa! Got the function from here: http://gamedev.stackexchange.com/questions/23743/whats-the-most-efficient-way-to-find-barycentric-coordinates converted python function: import maya.OpenMaya as OpenMaya def barycentricInterp(vecA, vecB, vecC, vecP): ''' Calculates barycentricInterpolation of a point in a triangle. :param vecA - OpenMaya.MVector of a vertex point. :param vecB - OpenMaya.MVector of a vertex point. :param vecC - OpenMaya.MVector...

matlab,image-processing,interpolation,contour,points

One intuitive approach (IMO) is to create an independent variable for both x and y. Base it on arc length, and interpolate on it. % close the contour, temporarily xc = [x(:); x(1)]; yc = [y(:); y(1)]; % current spacing may not be equally spaced dx = diff(xc); dy =...

Your data contains several y values for the same x value. This violates the assumptions of most interpolation algorithms. Either discard the rows with duplicate x values, average the y values for each individual x, or obtain a better resolution for the x values such that they aren't the same...

If you just want to delete part of your data "spherically" you can do being x,y,z your geometric data and v the values: level=5000; %km (or whatever value it is in your data) V(sqrt(X.^2+Y.^+Z.^2)>level)=0; %or depending in the plotting functions. V(sqrt(X.^2+Y.^+Z.^2)>level)=NaN; % plot things ...

There is an excellent builtin matlab function for this: imresize. In this you can specify several details like interpolation method and filter kernel. Just hit F1 while imresize is selected in the matlab gui. fact = 1.5; % read image and make greyscale if necessary I = double(imread('house.png')); if size(I,...

r,csv,coordinates,interpolation

Well, based on your summaries the range of x is 0.02 and the range of y is 0.01. However, in your expand.grid call you ask for sequences from the min to the max with a step size of 3.5. So your seq calls will return 1 value each, and your...

Internally it does this (which does not involve zoo): y <- c(NA, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2.661698, 3.107128, 7.319669, 10.800864, 17.855491, 18.250267, 28.587002, 36.405397, 38.467383, 38.685956, 43.917737, 40.829615, 43.519173, 45.597497, 43.252656, 45.581646, 48.258325, 48.269969, 50.905045, 53.258165, 58.39137, 59.27844, 58.720518, 56.933438, 62.062116, 59.860849,...

python,arrays,numpy,scipy,interpolation

Instead of passing a single number to the zoom parameter, give a sequence: scipy.ndimage.zoom(x, zoom=(1.5, 2.), order=1) #array([[0, 0, 1, 1, 2, 2], # [2, 2, 3, 3, 4, 4], # [4, 4, 5, 5, 6, 6], # [6, 6, 7, 7, 8, 8]]) With the sequences (2., 2.75) and...

I'm fairly sure that the code below gives the correct answer, except for a tiny level of numerical imprecision due to the use of the approx() function. The basic idea is to use ddply to split and combine data frames, and approx to do the interpolation. library(plyr) # time_interpolate is...

python,python-2.7,refactoring,interpolation

As stated in another question (Python ConfigParser interpolation from foreign section) no, unfortunately there is not an easy way to do this. You would need to create your own parser for the returned values and iterate over them. This seems a bit kludgy though.

matlab,interpolation,interpolate,function-interposition

Your computations are correct, but you are not plotting the function the right way. The blue line in your generated plot is piecewise linear. That's because you are only evaluating your polynomial p at the interpolation points x2. The plot command then draws line segments between those points and you...

python,arrays,optimization,numpy,interpolation

scipy's interp1d can help: import numpy as np from scipy.interpolate import interp1d ntime, nheight_in, nlat, nlon = (10, 20, 30, 40) heights = np.linspace(0, 1, nheight_in) t_in = np.random.normal(size=(ntime, nheight_in, nlat, nlon)) f_out = interp1d(heights, t_in, axis=1) nheight_out = 50 new_heights = np.linspace(0, 1, nheight_out) t_out = f_out(new_heights) ...

python,matplotlib,scipy,interpolation

A standard cubic spline is not very good at reasonable looking interpolations between data points that are very unevenly spaced. Fortunately, there are plenty of other interpolation algorithms and Scipy provides a number of them. Here are a few applied to your data: import numpy as np from scipy.interpolate import...

python,optimization,numpy,scipy,interpolation

Received a tip from an external source that work, so posting the answer in case it helps anyone in the future. SciPy has an Rbf interpolation method (radial basis function) which allows better than linear interpolation at arbitrary dimensions. Taking a variable data with rows of (x1,x2,x3...,xn,v) values, the follow...

image,matlab,image-processing,interpolation,resampling

I don't know any matching image processing function, so I would do it manually: %defines the image section you want I={[2.5:1:5],[3.5:1:5.5]} %spans a grid [P{1},P{2}]=ndgrid(I{:}) %interpolates using the grid IMG2=interpn(IMG,P{:}) This code is for 2D-images (grayscale), for coloured images: %defines the image section you want I={[2.5:1:5],[3.5:1:5.5]} %We dont want to...

matlab,rgb,interpolation,grayscale,fisheye

Try changing these three lines: [Mr Nr] = size(imR); % size of rectangular image ... imP = zeros(M, N); ... imP = interp2(imR, xR, yR); %interpolate (imR, xR, yR); ...to these: [Mr Nr Pr] = size(imR); % size of rectangular image ... imP = zeros(M, N, Pr); ... for dim...

javascript,function,math,interpolation,progress

I'm going to focus on the mathematical part of your request: a function should go in "waves": faster at beginning, slowing down closer to the end to extent in case of a delayed checkpoint Translation: a monotonic function (i.e. no backward steps) with a positive initial gradient and asymptotic behaviour...

android,animation,interpolation

You may have set up the ObjectAnimator incorrectly. Let's assume your Point class has an instance variable xPosition that is an Integer. In order to animate the xPosition with the ObjectAnimator, you would do this: ObjectAnimator.ofInt(new Point(), // Instance of your class "xPosition", // Name of the property you want...

javascript,d3.js,interpolation,tween

Your code assumes that you have only a single element in the selection (linePath.node()) -- you're getting the length of the first element only. You could use for example .each() to make it work for every line: function transition() { d3.select(this).transition() .duration(700).attrTween("stroke-dasharray", tweenDash); } function tweenDash() { var that =...

This post that I just recently answered pretty much solves your problem. However, I'll throw you a bone and help you out. Use interp1 to do this interpolation for you. What you do is you specify the (x,y) control points, which are those points that are part of your dataset,...

OK, what you need is a regression (see: Wolfram, Wiki) or aproximation (see: Wiki). General idea of both is exactly what you need: finding a function that will match a function for which you have samples as close as possible. There are several methods, you can google them up as...

To simplify, let's begin by just considering a single intepolated value z. Assume four values z00, z01, z10, z10, and two weights w0 and w1 applied to the first and second index, giving z0 = z00 + w0 × (z10 - z00) z1 = z01 + w0 × (z11 -...

colors,mdx,interpolation,iccube

InterpolateRGBColors expect a numerical between 0 and 1 for interpolation. So we need to scale our measure to ensure we get the right colors. There is an example in our live demo , here. What we need is to scale [Measures].[NbSejours] between 0,1. There are two no documented function in...

javascript,angularjs,youtube,concatenation,interpolation

Since 1.2 you can only bind one expression to *[src], *[ng-src] or action. You can read more about it here. Try this instead: In Controller: $scope.getIframeSrc = function (videoId) { return 'https://www.youtube.com/embed/' + videoId; }; HTML: ng-src="{{getIframeSrc(video.id.videoId)}}" Note that you still need to whitelist it as you have, or you...

Well, the reason is that the parameters to the two functions are, as you have noted, different. Yes, this makes it really hard to just switch out one for the other, as I well know. Why? In general it was a clear design decision to break backward compatibility with the...

To the best of my knowledge, there isn't a direct function which you can use in MATLAB. However, you can solve the problem manually using fitlm.

opengl,glsl,webgl,interpolation,fragment-shader

Without multisampling techniques, a fragment will always be sampled at the pixel center. The behavior you see is the intended behavior. The texture sample parameters you mention do not influence interpolation at all. They have an effect only when sampling from a texture using the texture() family of GLSL functions....

So if your control points are always on the same x coordinate and linearly dispersed along whole range then you can do it like this: //--------------------------------------------------------------------------- const int N=5; // number of control points (must be >= 4) float ctrl[N]= // control points y values initiated with linear function y=x...

Using griddata was the correct approach. First, define a regular grid using meshgrid, e.g. [x,y] = meshgrid(-0.2160:-0.0001:-0.2319,0.2161:0.0001:0.2319); % replace the values by meaningful borders and then map the nonregular grid X, Y to the regular grid z = griddata(X,Y,Z,x,y,'cubic') where 'cubic' can be replaced by the preffered method....

Thanks for all those that have given their input, especially @Jaime. I have experimented a bit, and came to this conclusion: 1) Apart from rounding errors, both methods mentioned by me have the same result. 2) They both take very much the same amount of time 3) I tried the...

python,numpy,scipy,interpolation

Just add the 360° complement each time you detect there is a jump and revert back to the first 360 degrees by using the modulo operation. For example: In [1]: import numpy as np In [2]: from scipy import interpolate In [3]: data = np.array([[0, 2, 4, 6, 8], [1,...

c++,arrays,interpolation,points,spline

I had to write a Bezier spline creation routine for an "entity" that was following a path in a game I am working on. I created a base class to handle a "SplineInterface" and the created two derived classes, one based on the classic spline technique (e.g. Sedgewick/Algorithms) an a...

javascript,angularjs,interpolation

You can use the $interpolate service: var exp = $interpolate(attrs.attr) // interpolate your attriubte var interpolated = exp(scope) // evaluate it with the scope variable Check this plunker...

I'm not sure if this is what you're doing already with all that diff/cumsum stuff by I would think make t start at 0: t = datenum(2013,1,1,data(:,1),data(:,2),s)*(24*60*60); t = t-t(1); sample = interp1(t,data(:,5:end), 0:0.16:t(end)); The idea here is that we know we want to sample every 0.16 seconds but only...

You cannot dynamically set local variables in this manner because what you are actually trying to do is set a String. Your code is interpreted as follows "member1" = Fabricate(:user) Which will raise a SyntaxError for unexpected = because you cannot set a String to anything. You can however perform...

Given large data frames, your best bet may be to merge df1 and df2 by their zip codes (aka get every pair of rows from the data frames where they have the same zip code), filter by the house number criteria, remove duplicates (cases where multiple rules from df1 match),...

You may want to specify xout to avoid this. For instance, if you want to always exclude the first and the last points, here's how you can do that: specify_xout <- function(x, n) { seq(from=min(x), to=max(x), length.out=n+2)[-c(1, n+2)] } plot(df$x, df$y) points(approx(df$x, df$y, xout=specify_xout(df$x, 5)), pch = "*", col =...

The 'single' argument doesn't belong there. It is interpreted as ones(1,length('single')) instead and thus yields even huger matrices. You could also generate a griddedInterpolant instead, which would save memory: F_interp = griddedInterpolant({ky_vec,kx_vec,kv_vec,ku_vec},F) ...

ruby-on-rails,ruby,model-view-controller,dictionary,interpolation

I would move all that logic into a helper, for example create a method application_helper.rb: def display_track(track) if track.artists.empty? track.name else links = track.artists.map {|artist| link_to artist.name, artist_path(artist) } sentence = links.to_sentence(two_words_connector: ' & ', last_word_connector: ' & ') "#{track.name} (feat. #{sentence})" end end Then your view simplifies to: <table>...

ruby,string,interpolation,string-interpolation

overflow is an array of integers while stack is an array of strings. There's nothing to escape in overflow. Possibly you meant to make stack an array of Fxinum as well, in which case add .map(&:to_i) at the end of the method chain: stack = 1234.to_s.chars.reverse.each_slice(3).map { |s| s.reverse.join }.reverse.map(&:to_i)...

nearest "snaps" to the nearest data point. zero is a zero order spline. It's value at any point is the last raw value seen. linear performs linear interpolation and slinear uses a first order spline. They use different code but produce the same result. quadratic uses second order spline...

Upscale your tiny images and use that instead: BufferedImage tiny = ImageIO.read(new File("...")); BufferedImage bigger = new BufferedImage(tiny.getWidth()*10, tiny.getHeight()*10, tiny.getType()); Graphics2D g = bigger.createGraphics(); g.setRenderingHint(RenderingHints.KEY_INTERPOLATION, RenderingHints.VALUE_INTERPOLATION_NEAREST_NEIGHBOR); g.drawImage(tiny, 0, 0, bigger.getWidth(), bigger.getHeight(), 0, 0, tiny.getWidth(), tiny.getHeight(), null); Replace the 10 scaling factor by the smallest value that gives acceptable results. Do...

The trick is that the interp family of functions takes the input coordinates and values in meshgrid format, so in this case the first three arguments would have to be 3D matrices of the coordinate space. You can reconstruct this using meshgrid: [X,Y,Z] = meshgrid(1:79, 1:2, 1:47]; each of X,...

You may consider using inpaint_nans which is designed to interpolate NaN elements in a 1-d or 2-d array using non-NaN elements. It can also extrapolate, as it does not use a triangulation of the data. It also allows different approaches to the interpolation.

matlab,interpolation,triangulation,numerical-integration

You could assume your function to be piecewise linear on your integration area and then integrate it using a midpoint quadrature rule: For every triangle you compute the midpoint value as the mean of the nodal values and multiply by the triangle's area. You sum it all up to get...

r,interpolation,spatial,grid-layout,netcdf

This is the solution I think you're looking for, which uses bilinear resampling. However this is not the only way to do such interpolation and you'd likely need to justify not using a more sophisticated approach (e.g. geostatistical, splines, etc.): library(raster) set.seed(2002) ## Create an extent boundary: ex <- extent(c(0,...

c,arrays,recursion,signal-processing,interpolation

You can get rid of memory allocation completely. Here is a simplified version of the imf function: void imf(long double *hilbert, long double *upper, long double *lower) { for (int i = 0; i < ARRAYSIZE; i++) { hilbert[i] -= (upper[i] + lower[i]) / 2.0; } sift can be simplified...

python,image,image-processing,python-imaging-library,interpolation

These are listed in order of lowest to higest complexity. There will be visual differences between them. The main difference will be how long the algorithm takes to execute. You'll have to decide what matters more to you, speed, or quality. If you're only doing 5 images, go for quality....

matlab,matrix,interpolation,polynomial-math

I would use diag and cumprod to help you accomplish this. First use diag to extract the diagonals of your matrix Q. After, use cumprod to generate a vector of cumulative products. How cumprod works on a vector is that for each element in the vector, the i'th element collects...

python,matlab,numpy,scipy,interpolation

Looking at the documentation you should note a few things: If x and y represent a regular grid, consider using RectBivariateSpline. x and y should be 1D vectors unlike in Matlab and they match with Matlab's x and y and NOT with xi and yi. Those come later. SciPy will...

java,math,image-processing,geometry,interpolation

I had sent this Stack OVerflow to a friend for some advice. He didn't have time to look at it then and there but he did send me this: The X & Y Problem. It may be possible to solve this math question, but I found a much better way...

c++,graphics,syntax,static-libraries,interpolation

I figured out what's happening. It turns out the real_2d_array will not take a string, but it will take characters, or in my case an array of characters. That's it. My implementation is xyz_data.replace(xyz_data.size()-1,1,"]"); //Delete the extra comma as a result of the loop above. char *DATA=new char[xyz_data.size()]; //Declare pointer...

ruby,loops,interpolation,quotes

The problem you are seeing is because you are outputting the string using p. p is the same as calling: puts object.inspect The simple fix is to use puts instead of p....

r,interpolation,linear-interpolation

Try approx: (xp <- unique(c(dat$x.low, dat$x.high))) ## [1] 0 1 2 3 (yp <- unique(c(dat$y.low, dat$y.high))) ## [1] 0 2 3 10 x <- c(1.75, 2.5) approx(xp, yp, x) ## $x ## [1] 1.75 2.50 ## ## $y ## [1] 2.75 6.50 or approxfun (which returns a new function): f...

No, Lagrangian interpolation with 200 equally spaced points would be an incredibly bad idea. (See: Runge's phenomenon). The plot command simply connects the given (x,y) points by straight lines, in the order given. To see this for yourself, use fewer points: x = 0:pi/4:2*pi; y = sin(x); plot(x,y) ...

c++,c,math,interpolation,easing

As I said in comment, exponent fits quite well: double mBase = 5; // higher = more "curvy" double minY = pow(mBase, mMin - mBase); double maxY = pow(mBase, mMax - mBase); double scale = (mMax - mMin) / (maxY - minY); double shift = mMin - minY; return pow(mBase,...

r,row,interpolation,na,missing-data

You could also use the na.approx function from the zoo package. Note that this has a slightly different behavior (than the solution by @flodel) when you have two consecutive NA values. For the first and last row you could then use na.locf. y <- na.approx(x) y[nrow(y), ] <- na.locf(y[(nrow(y)-1):nrow(y), ])[2,...

python,pandas,plot,interpolation

Got it. With help from this question, here's what I did: Resample my tsgroup from minutes to seconds. \>>> tsres = tsgroup.resample('S') \>>> tsres 2014-11-08 10:30:00 3 2014-11-08 10:30:01 NaN 2014-11-08 10:30:02 NaN 2014-11-08 10:30:03 NaN ... 2014-11-08 10:54:58 NaN 2014-11-08 10:54:59 NaN 2014-11-08 10:55:00 2 Freq: S, Length: 1501...

I see several issue, and I'm not sure of the direction you have choose (at least never seen anything similar before). Preamble: Vector3.Distance(lerpStart, DesiredPosition) this is a constant and is the radius around DesiredPosition. Here's some problems I noticed: Problem 1 lerpT is always >=1 for every point more distant...

c++,opengl,interpolation,vbo,vertex-shader

Add another set of attributes and setup the glVertexAttribPointer to point into the vertex position buffer as well, but with an offset. Keep in mind, to add a bit of dummy padding to the end, so that when reaching the end of the array you don't access out of bounds....

java,opengl,lwjgl,interpolation,heightmap

mix() is really just a convenience function for something you can easily write yourself. The definition is: mix(v1, v2, a) = v1 * (1 - a) + v2 * a Or putting it differently, it calculates a weighted average of v1 and v2, with two weights w1 and w2 that...

python,numpy,interpolation,curve-fitting,data-fitting

Well to perform fitting the answer provided in the link you have given is good enough. But since you say you find it difficult I have an example code with data in the form a sine curve and a user defined function that fits the data. Here is the code:...

I think the scipy filter functions do what you want, in particular lfilter. Going from this HOWTO: import numpy, scipy.signal taps = numpy.repeat(1.0/9, 9) smoothed_x = scipy.signal.lfilter(taps, 1.0, x) ...

matlab,time-series,interpolation

You're essentially trying to do a zero-order hold as you've seen from the link you provided. Let's call the first set of times t_hr (a cell array of strings) and the second set t_bp (also a cell array of strings). Then call the heart rates hr and the blood pressures...

Make a master dataset then use approx to linearly interpolate: newdat <- merge(data.frame(age=seq(0,100,10)),dat,all=TRUE) newdat$surv[is.na(newdat$surv)] <- with(newdat, approx(age, surv, xout=age)$y)[is.na(newdat$surv)] newdat # age surv #1 0 1.0 #2 10 0.9 #3 20 0.8 #4 30 0.7 #5 40 0.6 #6 50 0.5 #7 60 0.4 #8 70 0.3 #9 80 0.2...

python,numpy,scipy,interpolation

If you want to interpolate the result, rather than just find the z value for the nearest neighbour, I would consider doing something like the following: Use a k-d tree to partition your data points according to their (x, y) coordinates For a given (xi, yi) point to interpolate, find...

wolfram-mathematica,interpolation

Sounds like you want something like this in order to "plus" your lines. it1 = Interpolation[myPoints1]; it2 = Interpolation[myPoints2]; t1 = Table[it1[i], {i, 1.467, 1.497, 0.001}]; t2 = Table[it2[i], {i, 1.467, 1.497, 0.001}]; t3 = Transpose[{Range[1.467, 1.497, 0.001], t1 + t2}]; ListPlot[{myPoints1, myPoints2, t3}, Joined -> True, PlotRange -> {{1.467,...

c,arrays,math,matrix,interpolation

Apparently you don't really interpolate as expected. rotatedX and rotatedY are of type int; if you cast them to double, they remain integral values and the denominators ((double)(rotatedX+1.0)-(double)rotatedX)) ((double)(rotatedX+1.0)-(double)rotatedX)) cancel out to be 1.0, which means that no real interpolation as desired takes place, but f1 and f2 actually get...

This is likely due to floating point round-off. Simplifying gives: In [10]: sympy.simplify(Lagrange(Lx,Ly)) Out[10]: X*(1.85037170770859e-17*X**2 + 1.0*X - 1.11022302462516e-16) Which is basically X**2. Try getting rid of those float casts: def Lagrange (Lx, Ly): X=sympy.symbols('X') if len(Lx)!= len(Ly): print "ERROR" return 1 y=0 for k in range ( len(Lx) ):...

python,math,numpy,scipy,interpolation

I feel pretty stupid now! The answer was very simple - this answer on MathOverflow clued me in. There is no problem with discontinuity provided that I convert from a polar coordinate space to a Cartesian one, then interpolate the x and y components of the vector independently: x, y...