cubic-spline

Get subject-specific peak velocity and age at peak velocity values from linear mixed spline models

喜你入骨 提交于 2021-02-08 04:01:55
问题 I am fitting a linear mixed effects model with a natural spline function for age in order to describe the nonlinear trajectory for a repeated outcome y (bone mineral content in grams) across time ( age in years). How can I solve the spline derivatives to get the velocity curve and estimate for each individual their peak velocity (grams/year) and age at peak velocity (years) from this model? This is the example data dat <- structure(list(id = c(1001L, 1001L, 1001L, 1001L, 1001L, 1002L, 1003L,

Get subject-specific peak velocity and age at peak velocity values from linear mixed spline models

爷,独闯天下 提交于 2021-02-08 04:01:27
问题 I am fitting a linear mixed effects model with a natural spline function for age in order to describe the nonlinear trajectory for a repeated outcome y (bone mineral content in grams) across time ( age in years). How can I solve the spline derivatives to get the velocity curve and estimate for each individual their peak velocity (grams/year) and age at peak velocity (years) from this model? This is the example data dat <- structure(list(id = c(1001L, 1001L, 1001L, 1001L, 1001L, 1002L, 1003L,

Individual age at peak velocity from linear mixed natural spline model

耗尽温柔 提交于 2021-01-29 12:49:00
问题 This thread continues with Get subject-specific peak velocity and age at peak velocity values from linear mixed spline models. I am fitting a linear mixed effects model with a natural spline function for age. I would like to estimate age at peak velocity (apv - years) and peak velocity (pv - grams) for each person in the dataset by differentiating the spline terms. The model includes a random quadratic slope for age. How can I estimate the person-specific apv and pv? I am using the

Spline interpolation in 3D in python

谁说我不能喝 提交于 2020-04-11 08:27:28
问题 I am searching the equivalent Matlab command Vq = interp3(X,Y,Z,V,Xq,Yq,Zq) in Python. In Matlab I can use the method 'spline' interpolation, which I can not find in python for 3D data. There exists scipy.interpolate.griddata, but it doesn't have the option spline for 3D data. The data I want to interpolate is a 3D matrix (51x51x51), which is regularly distributed on a 3D grid. scipy.interpolate.Rbf may be the option, but I don't get it working: xi = yi = zi = np.linspace(1, 132651, 132651)

Spline interpolation in 3D in python

怎甘沉沦 提交于 2020-04-11 08:27:10
问题 I am searching the equivalent Matlab command Vq = interp3(X,Y,Z,V,Xq,Yq,Zq) in Python. In Matlab I can use the method 'spline' interpolation, which I can not find in python for 3D data. There exists scipy.interpolate.griddata, but it doesn't have the option spline for 3D data. The data I want to interpolate is a 3D matrix (51x51x51), which is regularly distributed on a 3D grid. scipy.interpolate.Rbf may be the option, but I don't get it working: xi = yi = zi = np.linspace(1, 132651, 132651)

Inferring/expressing the polynomial equation of a fitted smoothing spline?

夙愿已清 提交于 2020-01-25 06:52:45
问题 If I smooth a data vector with a smoothing cubic spline my understanding is that each ‘segment’ between knots should be representable as a cubic polynomial. Is it possible to infer the equation of each segment from the spline coefficients after e.g fitting by the smooth.spline function in R? This is straightforward for an interpolating spine as the array of polynomial coefficients is generated explicitly. However, I’ve not been able to find an answer as to whether this is possible with a

Inferring/expressing the polynomial equation of a fitted smoothing spline?

我的未来我决定 提交于 2020-01-25 06:52:17
问题 If I smooth a data vector with a smoothing cubic spline my understanding is that each ‘segment’ between knots should be representable as a cubic polynomial. Is it possible to infer the equation of each segment from the spline coefficients after e.g fitting by the smooth.spline function in R? This is straightforward for an interpolating spine as the array of polynomial coefficients is generated explicitly. However, I’ve not been able to find an answer as to whether this is possible with a

How to perform cubic spline interpolation in python?

折月煮酒 提交于 2019-12-20 09:18:31
问题 I have two lists to describe the function y(x): x = [0,1,2,3,4,5] y = [12,14,22,39,58,77] I would like to perform cubic spline interpolation so that given some value u in the domain of x, e.g. u = 1.25 I can find y(u). I found this in SciPy but I am not sure how to use it. 回答1: Short answer: from scipy import interpolate def f(x): x_points = [ 0, 1, 2, 3, 4, 5] y_points = [12,14,22,39,58,77] tck = interpolate.splrep(x_points, y_points) return interpolate.splev(x, tck) print(f(1.25)) Long

Python: Cubic Spline Regression for a time series data

戏子无情 提交于 2019-12-13 23:27:29
问题 I have the data as shown below. I want to find a CUBIC SPLINE curve that fits the entire data set (link to sample data). Things I've tried so far: I've gone through scipy's Cubic Spline Functions, but all of them are only able to give results at a single time only, whereas I want a single curve for the entire time range. I plotted a graph by taking an average of the spline coefficients generated by scipy.interpolate.splrep for a 4 number of knots, but the results were not good and didn't

Extracting points coordinates(x,y) from a curve c#

浪子不回头ぞ 提交于 2019-12-13 05:29:57
问题 i have a curve that i draw on a picturebox in c# using the method graphics.drawcurve(pen, points, tension) is there anyway that i can extract all points (x,y coordinates) been covered by the curve ? and save them into an array or list or any thing would be great, so i can use them in a different things. My code: void Curved() { Graphics gg = pictureBox1.CreateGraphics(); Pen pp = new Pen(Color.Green, 1); int i,j; Point[] pointss = new Point[counter]; for (i = 0; i < counter; i++) { pointss[i]