如何在 Pytorch 中進行三次樣條插值和積分

Question

在 Pytorch 中，是否有類似於Scipy 的三次樣條插值？ 給定一維輸入張量x和y ，我想通過這些點進行插值並在xs處評估它們以獲得ys 。 另外，我想要一個積分器 function 找到Ys ，即從x[0]到xs的樣條插值的積分。

Answer 1

這是我在Pytorch中使用Cubic Hermite Splines 高效並支持 autograd 的要點。

為方便起見，我也將代碼放在這里。

import torch as T

def h_poly_helper(tt):
  A = T.tensor([
      [1, 0, -3, 2],
      [0, 1, -2, 1],
      [0, 0, 3, -2],
      [0, 0, -1, 1]
      ], dtype=tt[-1].dtype)
  return [
    sum( A[i, j]*tt[j] for j in range(4) )
    for i in range(4) ]

def h_poly(t):
  tt = [ None for _ in range(4) ]
  tt[0] = 1
  for i in range(1, 4):
    tt[i] = tt[i-1]*t
  return h_poly_helper(tt)

def H_poly(t):
  tt = [ None for _ in range(4) ]
  tt[0] = t
  for i in range(1, 4):
    tt[i] = tt[i-1]*t*i/(i+1)
  return h_poly_helper(tt)

def interp_func(x, y):
  "Returns integral of interpolating function"
  if len(y)>1:
    m = (y[1:] - y[:-1])/(x[1:] - x[:-1])
    m = T.cat([m[[0]], (m[1:] + m[:-1])/2, m[[-1]]])
  def f(xs):
    if len(y)==1: # in the case of 1 point, treat as constant function
      return y[0] + T.zeros_like(xs)
    I = T.searchsorted(x[1:], xs)
    dx = (x[I+1]-x[I])
    hh = h_poly((xs-x[I])/dx)
    return hh[0]*y[I] + hh[1]*m[I]*dx + hh[2]*y[I+1] + hh[3]*m[I+1]*dx
  return f

def interp(x, y, xs):
  return interp_func(x,y)(xs)

def integ_func(x, y):
  "Returns interpolating function"
  if len(y)>1:
    m = (y[1:] - y[:-1])/(x[1:] - x[:-1])
    m = T.cat([m[[0]], (m[1:] + m[:-1])/2, m[[-1]]])
    Y = T.zeros_like(y)
    Y[1:] = (x[1:]-x[:-1])*(
        (y[:-1]+y[1:])/2 + (m[:-1] - m[1:])*(x[1:]-x[:-1])/12
        )
    Y = Y.cumsum(0)
  def f(xs):
    if len(y)==1:
      return y[0]*(xs - x[0])
    I = P.searchsorted(x[1:].detach(), xs)
    dx = (x[I+1]-x[I])
    hh = H_poly((xs-x[I])/dx)
    return Y[I] + dx*(
        hh[0]*y[I] + hh[1]*m[I]*dx + hh[2]*y[I+1] + hh[3]*m[I+1]*dx
        )
  return f

def integ(x, y, xs):
  return integ_func(x,y)(xs)

# Example
if __name__ == "__main__":
  import matplotlib.pylab as P # for plotting
  x = T.linspace(0, 6, 7)
  y = x.sin()
  xs = T.linspace(0, 6, 101)
  ys = interp(x, y, xs)
  Ys = integ(x, y, xs)
  P.scatter(x, y, label='Samples', color='purple')
  P.plot(xs, ys, label='Interpolated curve')
  P.plot(xs, xs.sin(), '--', label='True Curve')
  P.plot(xs, Ys, label='Spline Integral')
  P.plot(xs, 1-xs.cos(), '--', label='True Integral')
  P.legend()
  P.show()

Answer 2

這是對@chausies 答案的評論，但發布時間太長。

只是想發布他的答案的一個稍微縮小的版本，主要是為了我自己將來的參考：

import torch

def h_poly(t):
    tt = t[None, :]**torch.arange(4, device=t.device)[:, None]
    A = torch.tensor([
        [1, 0, -3, 2],
        [0, 1, -2, 1],
        [0, 0, 3, -2],
        [0, 0, -1, 1]
    ], dtype=t.dtype, device=t.device)
    return A @ tt


def interp(x, y, xs):
    m = (y[1:] - y[:-1]) / (x[1:] - x[:-1])
    m = torch.cat([m[[0]], (m[1:] + m[:-1]) / 2, m[[-1]]])
    idxs = torch.searchsorted(x[1:], xs)
    dx = (x[idxs + 1] - x[idxs])
    hh = h_poly((xs - x[idxs]) / dx)
    return hh[0] * y[idxs] + hh[1] * m[idxs] * dx + hh[2] * y[idxs + 1] + hh[3] * m[idxs + 1] * dx