r,delphi,delphi-xe5,linear-interpolation

Linear interpolation of 1D-data is very simple: Find such index in X-array, that X[i] <= Xb < X[i+1] (binary search for case of random access, linear search for case of step-by step Xb changing) Calculate Yb = Y[i] + (Y[i+1] - Y[i]) * (Xb - X[i]) / (X[i+1] - X[i])...

ios,math,curve,linear-interpolation,beizer

It seems you want an approximate parametrization by arc length. For the quadratic case, there is a closed-form expression for the arc length of a Bézier curve, but it is complicated and you still need table lookup. These papers discuss general techniques: Approximate Arc Length Parametrization, in SIBGRAPI 1996. Adaptive...

matlab,computer-vision,linear-interpolation

newframes=11 %number of frames you want interpolate_data=interpn(1:size(Data,1),1:size(Data,2),Data,linspace(1,size(Data,1),newframes),1:size(Data,2)) For the second dimensions, indices are untouched to have no interpolation. For the first dimension linspace is used to define a interpolation grid with newframes values. ...

python,algorithm,list,linear-interpolation

Interpolating on the fly to create a linear interpolation: def interpolate(l): last = None count = 0 for num in l: if num is None: # count the number of gaps count += 1 continue if count: # fill a gap based on the last seen value and the current...

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...

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...

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,...

java,arrays,linear-interpolation

If you want to interpolate intervals to different count of numbers, you can just add the count of output numbers to function parameter. Example: /*** * Interpolating method * @param start start of the interval * @param end end of the interval * @param count count of output interpolated numbers...

You can create a linear spline with the points in a as control points. After that you can specify as many points as you want from a beginning interval to an ending interval. As what Raab70 said, you can use interp1. interp1 can be called in the following way (using...