spline

问题 Is there any method in scipy for spline interpolation in which I can use constraint on derivative at each data points? I found one "scipy.interpolate.PiecewisePolynomial" but PiecewisePolynomial class has been deprecated. 回答1: Yes. The BPoly class in scipy.interpolate has a method that constructs a piecewise polynomial in the Bernstein basis, compatible with the specified values and derivatives at breakpoints. as stated in the scipy reference, here. Basic usage on Python3 can go as following:

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

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

问题 I am trying to create a line, spline chart using JSON data. I want multiple series but I am confused on how to do that. Right now I am able to create the multiple series when the date is in 2019-07-06 format. I also have a JSON that has a column for the month and a column for the year Please help on how I can fix this. Right now I only have the code for group by day. JSON Data: [ { "month": 6, "year": 2019, "starts": 21278998, "completes": 9309458 }, { "month": 7, "year": 2019, "starts":

问题 My C++ program requires access to 3D splines that have been constructed in 3ds Max (2011). I have found a simple maxscript that exports these splines as XML files - an example as follows: <spline name='Line001' knots='5' closed='true'> <knot x='-4.67297e-005' y='0.0' z='0.0'> <invec x='-0.000144482' y='-600.0' z='-1.52588e-005' /> <outvec x='5.10227e-005' y='600.0' z='1.52588e-005' /> </knot> <knot x='6.17511e-005' y='800.0' z='500.0'> <invec x='7.92357e-005' y='800.0' z='100.0' /> <outvec x=

问题 I have data that are strictly increasing and would like to fit a smoothing spline that is monotonically increasing as well with the smooth.spline() function if possible, due to the ease of use of this function. For example, my data can be effectively reproduced with the example: testx <- 1:100 testy <- abs(rnorm(length(testx)))^3 testy <- cumsum(testy) plot(testx,testy) sspl <- smooth.spline(testx,testy) lines(sspl,col="blue") which is not necessarily increasing everywhere. Any suggestions?

问题 I want to use the natural cubic smoothing splines smooth.spline from R in Python (like som many others want as well (Python natural smoothing splines, Is there a Python equivalent to the smooth.spline function in R, Python SciPy UnivariateSpline vs R smooth.spline, ...)) Therefore I am using rpy2 like described in https://morioh.com/p/eb4151821dc4, but I want to set directly lambda instead of spar : import rpy2.robjects as robjects r_y = robjects.FloatVector(y_train) r_x = robjects

问题 When calling the derivative method on the CatmullRomSpline, what exactly does this value represent? It’s been a while since calculus, but if my memory serves me correctly taking the derivative of the position with respect to time gives you the velocity at that point in time. But with CatmullRomSpline the “time” value is a percentage, so is the resulting derivative in pixels/percent? I printed out the derivative values (vector length) along my path and the values go as high as “989.6049”,

问题 I am porting a script written in R over to Python. In R I am using smooth.spline and in Python I am using SciPy UnivariateSpline. They don't produce the same results (even though they are both based on a cubic spline method). Is there a way, or an alternative to UnivariateSpline, to make the Python spline return the same spline as R? I'm a mathematician. I understand the general idea of splines. But not the fine details of their implementation in Python or R. Here is the code in R and then

问题 Basically I want to draw a polygon, but I want the edges to appear soft rather than hard. Since the shape of the polygon is important, the edges have to go over the points. I've found monotone cubic splines to be accurate for open curves (i.e., curves that don't wrap around on themselves), but the algorithms I've found precalculate points 0 and N. Can I somehow change them to work with a closed curve? I am implementing this in JavaScript, but pseudo-code would just as well. 回答1: There is an