spline

问题 I'm trying to prepare some demographic data retrieved from Eurostat for further processing, amongst others replacing any missing data with corresponding approximated ones. First I was using data.frames only, but then I got convinced that data.tables might offer some advantages over regular data.frames, so I migrated to data.tables. One thing I've observed while doing so was getting different results when using "na.spline" in combination with "apply" versus "na.spline" as part of the data

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,

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,

问题 I'm looking to predict 'terms', especially ns splines, from an lmer model. I've replicated the problem with the mtcars dataset (technically poor example, but works to get the point across). Here is what I'm trying to do with a linear model: data(mtcars) mtcarsmodel <- lm(wt ~ ns(drat,2) + hp + as.factor(gear), data= mtcars) summary(mtcarsmodel) coef(mtcarsmodel) test <- predict(mtcarsmodel, type = "terms") Perfect. However, there is no equivalent 'terms' option for lmer predict (unresolved

问题 I'm looking to predict 'terms', especially ns splines, from an lmer model. I've replicated the problem with the mtcars dataset (technically poor example, but works to get the point across). Here is what I'm trying to do with a linear model: data(mtcars) mtcarsmodel <- lm(wt ~ ns(drat,2) + hp + as.factor(gear), data= mtcars) summary(mtcarsmodel) coef(mtcarsmodel) test <- predict(mtcarsmodel, type = "terms") Perfect. However, there is no equivalent 'terms' option for lmer predict (unresolved

问题 I have a B-Spline curve. I have all the knots, and the x,y coordinates of the Control Points. I need to convert the B-Spline curve into Bezier curves. My end goal is to be able to draw the shape on an html5 canvas element. The B-Spline is coming from a dxf file which doesn't support Beziers, while a canvas only supports Beziers. I've found several articles which attempt to explain the process, however they are quite a bit over my head and really seem to be very theory intensive. I really need

问题 I have a B-Spline curve. I have all the knots, and the x,y coordinates of the Control Points. I need to convert the B-Spline curve into Bezier curves. My end goal is to be able to draw the shape on an html5 canvas element. The B-Spline is coming from a dxf file which doesn't support Beziers, while a canvas only supports Beziers. I've found several articles which attempt to explain the process, however they are quite a bit over my head and really seem to be very theory intensive. I really need

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

问题 How can a spline be created if only the points and the coefficients are known? I'm using scipy.interpolate.BSpline here, but am open to other standard packages as well. So basically I want to be able to give someone just those short arrays of coefficients for them to be able to recreate the fit to the data. See the failed red-dashed curve below. import numpy as np import matplotlib.pyplot as plt from scipy.interpolate import BSpline, LSQUnivariateSpline x = np.linspace(0, 10, 50) # x-data y =

问题 How can a spline be created if only the points and the coefficients are known? I'm using scipy.interpolate.BSpline here, but am open to other standard packages as well. So basically I want to be able to give someone just those short arrays of coefficients for them to be able to recreate the fit to the data. See the failed red-dashed curve below. import numpy as np import matplotlib.pyplot as plt from scipy.interpolate import BSpline, LSQUnivariateSpline x = np.linspace(0, 10, 50) # x-data y =