5

次の Python では、返された配列に含まれる 5 つの関数funcを統合する必要があります。このコードは、以下を使用して生成された外部 Fortran モジュールを呼び出しますf2py。

import numpy as np
from numpy import cos, sin , exp
from trapzdv import trapzdv
def func(x):
    return np.array([x**2, x**3, cos(x), sin(x), exp(x)])

if __name__ == '__main__':
    xs = np.linspace(0.,20.,100)
    ans =  trapzdv(func,xs,5)
    print 'from Fortran:', ans
    print 'exact:', np.array([20**3/3., 20**4/4., sin(20.), -cos(20.), exp(20.)])

Fortran ルーチンは次のとおりです。

      subroutine trapzdv(f,xs,nf,nxs,result)
          integer :: I
          double precision :: x1,x2
          integer, intent(in) :: nf, nxs
          double precision, dimension(nf) :: fx1,fx2
          double precision, intent(in), dimension(nxs) :: xs
          double precision, intent(out), dimension(nf) :: result
          external :: f 
          result = 0.0
          do I = 2,nxs
            x1 = xs(I-1)
            x2 = xs(I)
            fx1 = f(x1)
            fx2 = f(x2)
            result = result + (fx1+fx2)*(x2-x1)/2
          enddo
          return
      end

問題は、Fortran が最初の関数のみを に統合していることですfunc(x)。印刷結果を参照してください。

from Fortran: [ 2666.80270721  2666.80270721  2666.80270721  2666.80270721  2666.80270721]
exact: [  2.66666667e+03   4.00000000e+04   9.12945251e-01  -4.08082062e-01 4.85165195e+08]

One way to workarond that is to modify func(x) to return the value of a given position in the array of functions:

def func(x,i):
    return np.array([x**2, x**3, cos(x), sin(x), exp(x)])[i-1]

And then change the Fortran routine to call the function with two parameters:

      subroutine trapzdv(f,xs,nf,nxs,result)
          integer :: I
          double precision :: x1,x2,fx1,fx2
          integer, intent(in) :: nf, nxs
          double precision, intent(in), dimension(nxs) :: xs
          double precision, intent(out), dimension(nf) :: result
          external :: f 
          result = 0.0
          do I = 2,nxs
            x1 = xs(I-1)
            x2 = xs(I)
            do J = 1,nf
                fx1 = f(x1,J)
                fx2 = f(x2,J)
                result(J) = result(J) + (fx1+fx2)*(x2-x1)/2
            enddo
          enddo
          return
      end

Which works:

from Fortran: [  2.66680271e+03   4.00040812e+04   9.09838195e-01   5.89903440e-01 4.86814128e+08]
exact: [  2.66666667e+03   4.00000000e+04   9.12945251e-01  -4.08082062e-01 4.85165195e+08]

But here func is called 5 times more than necessary (in the real case func has above 300 functions, so it will be called 300 times more than necessary).

  • Does anyone know a better solution to make Fortran recognizes all the array returned by func(x)? In other words, make Fortran build fx1 = f(x1) as an array with 5 elements corresponding to the functions in func(x).

OBS: I am compiling using f2py -c --compiler=mingw32 -m trapzdv trapzdv.f90

4

2 に答える 2

1

この回答は質問を解決しませんが、Cython で同じことを行う回避策です。ここでは、ベクトル値関数に対して台形則と多項式積分器が実装されています。私が入れた以下のコードintegratev.pyx：

import numpy as np
from numpy.linalg import inv
cimport numpy as np
FLOAT = np.float32
ctypedef np.float_t FLOAT_t

def trapzv(f, np.ndarray xs, int nf):
    cdef int nxs = xs.shape[0]
    cdef np.ndarray ans = np.zeros(nf, dtype=FLOAT)
    cdef double x1, x2
    for i in range(1,nxs):
        x1 = xs[i-1]
        x2 = xs[i]
        ans += (f(x2)+f(x1))*(x2-x1)/2.
    return ans

def poly(f, np.ndarray xs, int nf, int order=2):
    cdef int nxs = xs.shape[0]
    cdef np.ndarray ans = np.zeros(nf, dtype=FLOAT)
    cdef np.ndarray xis = np.zeros(order+1, dtype=FLOAT)
    cdef np.ndarray ais
    if nxs % (order+1) != 0:
        raise ValueError("poly: The size of xs must be a multiple of 'order+1'")
    for i in range(order,nxs,order):
        xis = xs[i-order:i+1]
        X = np.concatenate([(xis**i)[:,None] for i in range(order+1)], axis=1)
        ais = np.dot( inv(X), f(xis).transpose() )
        for k in range(1,order+2):
            ans += ais[k-1,:]/k * (xis[-1]**k - xis[0]**k)
    return ans

次のテストが使用されました。

import numpy as np
from numpy import cos, sin , exp
import pyximport; pyximport.install()
import integratev
from subprocess import Popen
def func(x):
    return np.array([x**2, x**3, cos(x), sin(x), exp(x)])

if __name__ == '__main__':
    xs = np.linspace(0.,20.,33)
    print 'exact:', np.array([20**3/3., 20**4/4., sin(20.), -cos(20.)+1, exp(20.)-1])
    ans =  integratev.trapzv(func,xs,5)
    print 'trapzv:', ans
    ans =  integratev.poly(func,xs,5,2)
    print 'poly:', ans

与える:

exact: [  2.66666667e+03   4.00000000e+04   9.12945251e-01   5.91917938e-01 4.85165194e+08]
trapzv: [  2.66796875e+03   4.00390625e+04   8.83031547e-01   5.72522998e-01 5.00856448e+08]
poly: [  2.66666675e+03   4.00000000e+04   9.13748980e-01   5.92435718e-01 4.85562144e+08]

ポリゴンは任意の順序にすることができ、ほとんどの場合により良い結果が得られるでしょう...

于 2013-07-08T09:04:21.020 に答える