我在fortran77中编写了一个代码，它基本上是大气CO2和海洋pH值的BLAG模型的简化形式。我现在正试图在python中编写这段代码，但是我在python中找不到能够解决4个函数和4个变量初始猜测的矩阵求解器的问题。 fortran版本使用LUDCMP和LUBKSB（矩阵求解器）来获得解决方案。我环顾四周，似乎python应该有一个更简化的方法来解决这个问题，但无论如何，我都不知道。 我已经附上了下面的整个代码，注释了，供人们仔细阅读 - 我基本上只是想把这个fortran程序转换成python。 我对python比较陌生，所以如果这是一个非常明显的问题，我会提前道歉。任何指针都将非常感谢！
module coeffs
real::alpha = 3.74e-2 !! alpha value
real::K1 = 8.73e-7 !! constant
real::K2 = 5.58e-10 !! constant
real::KSP = 4.7e-7 !! constant
real::Ca = 1.03e-2 !! calcium ions
real::VOC = 3.6e19 !! volume of ocean
real::Mair = 1.7e20 !! mass of carbon in air
real::Mtot = 1.5e17 !! total mass of carbon
real::alk=2.6614e-3 !! constant alkalinity
end module
use coeffs
real,dimension(4,4)::jac
real,dimension(4)::b,x,f,dx,fp
integer::n=4
integer::np=4
integer,dimension(4)::indx
real::d,chg,pulse,pH,Mat,Moc,H2CO3,HCO3,CO3,H
real::eps=1.e-3 !! constant
real::pCO2 !! partial pressure of CO2 in atmosphere
x(1) = 3.64205e-4 !!! initial guess for pCO2
x(2) = 1.36212e-5 !!! initial guess for H2CO3
x(3) = 2.20503e-3 !!! initial guess for HCO3
x(4) = 2.28155e-4 !!! initial guess for CO3
do m = 1,1 !!! START M LOOP
print *
print *, 'Enter pulse of CO2...'
read *, pulse !! added pulse of CO2 into atmosphere (simulating something like a rapid release of carbon, or an anthropogenic input)... 0.0 is standard for no pulse
if(pulse<0.0)exit
do k = 1,10 !!! START K LOOP
call func(x,f,n,pulse)
do j = 1,4 !!! START J LOOP
xsave = x(j)
delxj = eps * x(j)
x(j) = x(j) + delxj
call func(x,fp,n,pulse)
do i = 1,4 !!! START I LOOP
jac(i,j) = (fp(i) - f(i))/delxj
end do !!! END I LOOP
x(j) = xsave
end do !!! END J LOOP
b = -f
call ludcmp(jac,n,np,indx,d)
call lubksb(jac,n,np,indx,b)
dx = b
x = x + dx
errmax = 0.
do i = 1,4 !!! START I LOOP
err = abs(dx(i)/x(i))
errmax = amax1(errmax,err)
end do !!! END I LOOP
if(errmax<1.e-5)exit
pCO2 = x(1) * 1.e6
pH = -log10((K1 * x(2))/x(3))
print *
print *, 'k = ', k, 'pH = ', pH
print *, 'pCO2 = ', pCO2
print *, 'H2CO3 = ', x(2), 'HCO3 = ', x(3), 'CO3 = ', x(4)
end do !!! END K LOOP
end do !!! END M LOOP
stop
end
!!! subroutine containing functions to be solved
subroutine func(x,f,n,pulse)
use coeffs
real,dimension(n)::x,f
f(1) = x(1) - (x(2)/alpha) !! Henry's law
f(2) = x(2) - (K2*x(3)*x(3))/(K1*x(4)) !! combination of 1st and 2nd dissociations for H2CO3 (eliminating H as a variable)
f(3) = x(3) - alk+(2.*x(4)) !! constant alkalinity
f(4) = 1.e-19 * (Mtot+pulse-(Mair*x(1))-(VOC*(x(2)+x(3)+x(4)))) !! conservation of total carbon in the atmosphere/ocean system
return
end
!!! below this point is fortran77 matrix solver subroutines
SUBROUTINE LUDCMP(A,N,NP,INDX,D)
PARAMETER (NMAX=100,TINY=1.0E-20)
DIMENSION A(NP,NP),INDX(N),VV(NMAX)
D=1.
DO 12 I=1,N
AAMAX=0.
DO 11 J=1,N
IF (ABS(A(I,J)).GT.AAMAX) AAMAX=ABS(A(I,J))
11 CONTINUE
IF (AAMAX.EQ.0.) PAUSE 'Singular matrix.'
VV(I)=1./AAMAX
12 CONTINUE
DO 19 J=1,N
IF (J.GT.1) THEN
DO 14 I=1,J-1
SUM=A(I,J)
IF (I.GT.1)THEN
DO 13 K=1,I-1
SUM=SUM-A(I,K)*A(K,J)
13 CONTINUE
A(I,J)=SUM
ENDIF
14 CONTINUE
ENDIF
AAMAX=0.
DO 16 I=J,N
SUM=A(I,J)
IF (J.GT.1)THEN
DO 15 K=1,J-1
SUM=SUM-A(I,K)*A(K,J)
15 CONTINUE
A(I,J)=SUM
ENDIF
DUM=VV(I)*ABS(SUM)
IF (DUM.GE.AAMAX) THEN
IMAX=I
AAMAX=DUM
ENDIF
16 CONTINUE
IF (J.NE.IMAX)THEN
DO 17 K=1,N
DUM=A(IMAX,K)
A(IMAX,K)=A(J,K)
A(J,K)=DUM
17 CONTINUE
D=-D
VV(IMAX)=VV(J)
ENDIF
INDX(J)=IMAX
IF(J.NE.N)THEN
IF(A(J,J).EQ.0.)A(J,J)=TINY
DUM=1./A(J,J)
DO 18 I=J+1,N
A(I,J)=A(I,J)*DUM
18 CONTINUE
ENDIF
19 CONTINUE
IF(A(N,N).EQ.0.)A(N,N)=TINY
RETURN
END
SUBROUTINE LUBKSB(A,N,NP,INDX,B)
DIMENSION A(NP,NP),INDX(N),B(N)
II=0
DO 12 I=1,N
LL=INDX(I)
SUM=B(LL)
B(LL)=B(I)
IF (II.NE.0)THEN
DO 11 J=II,I-1
SUM=SUM-A(I,J)*B(J)
11 CONTINUE
ELSE IF (SUM.NE.0.) THEN
II=I
ENDIF
B(I)=SUM
12 CONTINUE
DO 14 I=N,1,-1
SUM=B(I)
IF(I.LT.N)THEN
DO 13 J=I+1,N
SUM=SUM-A(I,J)*B(J)
13 CONTINUE
ENDIF
B(I)=SUM/A(I,I)
14 CONTINUE
RETURN
END
答案 0 :(得分：1)
如果我正确理解您的Fortran代码，您应该能够使用scipy.optimize.fsolve来解决您的方程组。对于非线性方程组，fsolve是两个强大的Fortran准牛顿求解器（hybrd和hybrdj）的包装器。要在python中解决您的系统，您可以执行以下操作：
from scipy.optimize import fsolve
import numpy as np
alpha = 3.74e-2 # alpha value
K1 = 8.73e-7 # constant
K2 = 5.58e-10 # constant
KSP = 4.7e-7 # constant
Ca = 1.03e-2 # calcium ions
VOC = 3.6e19 # volume of ocean
Mair = 1.7e20 # mass of carbon in air
Mtot = 1.5e17 # total mass of carbon
alk=2.6614e-3 # constant alkalinity
x0 = np.array([
3.64205e-4, # initial guess for pCO2
1.36212e-5, # initial guess for H2CO3
2.20503e-3, # initial guess for HCO3
2.28155e-4 # initial guess for CO3
])
def objective_func(x, pulse):
return np.array([
x[0] - x[1] / alpha,
x[1] - K2 * x[2] * x[2] / (K1 * x[3]),
x[2] - alk + 2 * x[3],
1e-19 * (Mtot + pulse - Mair * x[0] - VOC * (x[1] + x[2] + x[3]))
])
pulse = 0
sol, info, conv_flag, conv_msg = fsolve(objective_func, x0, args=(pulse,), full_output=True)
print(conv_msg)
print('Solution is: ', sol)
<强>输出：
The solution converged.
Solution is: [ 3.64195924e-04 1.36209276e-05 2.20506331e-03 2.28168347e-04]
如果您只对解决方案感兴趣，可以省略full_output=True或将其设置为False。