splines

csaps() in matlab does a cubic spline according to a particular definition of the smoothing parameter p . Here is some matlab code and its result: % x variable age = 75:99 % y variable diffs = [-39 -2 -167 -21 -13 32 -37 -132 -143 -91 -93 -88 -62 -112 -95 -28 -90 -40 -27 -23 -28 -11 -8 -6 1] % 0.0005 is the parameter p, and the later specification of % age are the desired x for prediction csaps(age,diffs,0.0005,age) % result (column headers removed): -63.4604 -64.0474 -64.6171 -65.1397 -65.6111 -66.0165 -66.3114 -66.4123 -66.2229 -65.6726 -64.7244 -63.3582 -61.5676 -59.3568 -56.7364 -53.7382

I have used smooth.spline to estimate a cubic spline for my data. But when I calculate the 90% point-wise confidence interval using equation, the results seems to be a little bit off. Can someone please tell me if I did it wrongly? I am just wondering if there is a function that can automatically calculate a point-wise interval band associated with smooth.spline function. boneMaleSmooth = smooth.spline( bone[males,"age"], bone[males,"spnbmd"], cv=FALSE) error90_male = qnorm(.95)*sd(boneMaleSmooth$x)/sqrt(length(boneMaleSmooth$x)) plot(boneMaleSmooth, ylim=c(-0.5,0.5), col="blue", lwd=3, type=

I have a set of points which I want to smooth using B-spline curves. My question is how can I implement B-spline curves to smooth these set of points? I want to implement this using c++. Here is a function for any given number of points: void Spline(double x[N+1],double y[N+1], // input double A[N],double B[N], // output double C[N],double D[N]) // output { double w[N]; double h[N]; double ftt[N+1]; for (int i=0; i<N; i++) { w[i] = (x[i+1]-x[i]); h[i] = (y[i+1]-y[i])/w[i]; } ftt[0] = 0; for (int i=0; i<N-1; i++) ftt[i+1] = 3*(h[i+1]-h[i])/(w[i+1]+w[i]); ftt[N] = 0; for (int i=0; i<N; i++) { A

I'm trying to emulate Excel's Insert>Scatter>Scatter with smooth lines and markers command in Matplotlib The scipy function interpolate creates a similar effect, with some nice examples of how to simply implement this here: How to draw cubic spline in matplotlib However Excel's spline algorithm is also able to generate a smooth curve through just three points (e.g. x = [0,1,2] y = [4,2,1]); and it isn't possible to do this with cubic splines. I have seen discussions that suggest that the Excel algorithm uses Catmull-Rom splines; but don't really understand these, or how they could be adapted

The smooth.spline function in R allows a tradeoff between roughness (as defined by the integrated square of the second derivative) and fitting the points (as defined by summing the squares of the residuals). This tradeoff is accomplished by the spar or df parameter. At one extreme you get the least squares line, and the other you get a very wiggly curve which intersects all of the data points (or the mean if you have duplicated x values with different y values) I have looked at scipy.interpolate.UnivariateSpline and other spline variants in Python, however, they seem to only tradeoff by

I have a following model: coxph(Surv(fulength, mortality == 1) ~ pspline(predictor)) where is fulength is a duration of follow-up (including mortality), predictor is a predictor of mortality. The output of the command above is this: coef se(coef) se2 Chisq DF p pspline(predictor), line 0.174 0.0563 0.0562 9.52 1.00 0.002 pspline(predictor), nonl 4.74 3.09 0.200 How can I plot this model so that I get the nice curvy line with 95% confidence bands and hazard ratio on the y axis? What I am aiming for is something similar to this: This is when you get when you run the first example in ?cph of the

I am trying to extract the placement of the knots from a GAM model in order to delineate my predictor variable into categories for another model. My data contains a binary response variable (used) and a continuous predictor (open). data <- data.frame(Used = rep(c(1,0,0,0),1250), Open = round(runif(5000,0,50), 0)) I fit the GAM as such: mod <- gam(Used ~ s(Open), binomial, data = data) I can get the predicted values, and the model matrix etc with either type=c("response", "lpmatrix") within the predict.gam function but I am struggling with out to extract the knot locations at which which the