我不确定这个问题是关于这里或其他地方的主题(或者根本没有关于主题的问题)。 我继承了使用Newton Raphson插值的Fortran 90代码,其中温度的对数与压力的对数进行插值。
插值的类型为
t = a ln(p) + b
并且其中a,b定义为
a = ln(tup/tdwn)/(alogpu - alogpd)
和
b = ln T - a * ln P
这是测试程序。它仅显示一次迭代。但实际程序在k,j和i上运行三个FOR循环。实际上,pthta是一个3D数组(k,j,i),而thta是一维数组(k)
program test
implicit none
integer,parameter :: dp = SELECTED_REAL_KIND(12,307)
real(kind=dp) kappa,interc,pres,dltdlp,tup,tdwn
real(kind=dp) pthta,alogp,alogpd,alogpu,thta,f,dfdp,p1
real(kind=dp) t1,resid,potdwn,potup,pdwn,pup,epsln,thta1
integer i,j,kout,n,maxit,nmax,resmax
kappa = 2./7.
epsln = 1.
potdwn = 259.39996337890625
potup = 268.41687198359159
pdwn = 100000.00000000000
pup = 92500.000000000000
alogpu = 11.43496392350051
alogpd = 11.512925464970229
thta = 260.00000000000000
alogp = 11.512925464970229
! known temperature at lower level
tdwn = potdwn * (pdwn/100000.)** kappa
! known temperature at upper level
tup = potup *(pup/100000.)** kappa
! linear change of temperature wrt lnP between different levels
dltdlp = dlog(tup/tdwn)/(alogpu-alogpd)
! ln(T) value(intercept) where Pressure is 1 Pa and assume a linear
! relationship between P and T
interc = dlog(tup) - dltdlp*alogpu
! Initial guess value for pressure
pthta = exp((dlog(thta)-interc-kappa*alogp)/(dltdlp-kappa))
n=0
1900 continue
!First guess of temperature at intermediate level
t1 = exp(dltdlp * dlog(pthta)+interc)
!Residual error when calculating Newton Raphson iteration(Pascal)
resid = pthta - 100000.*(t1/thta)**(1./kappa)
print *, dltdlp,interc,t1,resid,pthta
if (abs(resid) .gt. epsln) then
n=n+1
if (n .le. nmax) then
! First guess of potential temperature given T1 and
! pressure level guess
thta1 = t1 * (100000./pthta)**kappa
f= thta - thta1
dfdp = (kappa-dltdlp)*(100000./pthta)**kappa*exp(interc + (dltdlp -1.)*dlog(pthta))
p1 = pthta - f/dfdp
if (p1 .le. pdwn) then
if (p1 .ge. pup) then
pthta = p1
goto 1900
else
n = nmax
end if
end if
else
if (resid .gt. resmax) resmax = resid
maxit = maxit+1
goto 2100
end if
end if
2100 continue
end program test
当您使用数据文件中的实际数据运行此程序时,resid的值为以下
2.7648638933897018E-010
并且整个执行没有太大差别。大多数值都在
范围内1E-10 and 1E-12
所以给这些值以下IF条件
IF (abs(resid) .gt. epsln)
永远不会被调用,Newton Raphson迭代永远不会被执行。所以我看了两种让它发挥作用的方法。一种是在这两个步骤中删除指数调用
pthta = exp((dlog(thta)-interc-kappa*alogp)/(dltdlp-kappa))
t1 = exp(dltdlp * dlog(pthta)+interc)
即。将所有内容保存在对数空间中,并在Newton Raphson迭代完成后取指数。那部分确实收敛而没有问题。
我尝试做这项工作的另一种方法是截断
t1 = exp(dltdlp * dlog(pthta)+interc)
当我将其截断为整数时,resid的值会发生显着变化 1E-10到813.我不明白截断该函数调用会导致如此大的值变化。截断该结果确实导致成功完成。 所以我不确定哪个是更好的方式继续前进。
我如何决定哪种方法更好?
答案 0 :(得分:1)
从研究的角度来看,我认为你的第一个解决方案可能是更合适的方法。在物理模拟中,应始终使用属性的对数,这些属性的定义始终为正。在上面的代码中,这些将是温度和压力。无论您使用Fortran还是任何其他编程语言,或任何可能的变量类型,严格肯定的物理变量通常都会导致计算中的溢出和下溢。如果发生了某些事情,它就会发生。
其他物理量也是如此,例如,能量(伽马射线爆发的典型能量约为10 ^ 54尔格),任意维度的物体体积(100维的体积)半径10米的球是~10 ^ 100),甚至是概率(许多统计问题中的似然函数可以取约10 ^ { - 1000}或更小的值)。使用正定变量的对数变换将使您的代码能够处理大到~10 ^ 10 ^ 307(对于双精度变量)的数字。
关于代码中使用的Fortran语法的一些注释:
变量RESMAX
在您的代码中使用,无需初始化。
为变量赋值时,必须适当指定文字常量的类型,否则可能会影响程序结果。例如,以下是在调试模式下使用英特尔Fortran编译器2018编译的原始代码的输出:
-0.152581477302743 7.31503025786548 259.608693509165
-3.152934473473579E-002 99474.1999921620
这是相同代码的输出,但所有文字常量都以kind参数_dp
为后缀(请参阅下面代码的修订版本):
-0.152580456940175 7.31501855886952 259.608692604963
-8.731149137020111E-011 99474.2302854451
此答案中修订代码的输出与上述问题中原始代码的输出略有不同。
无需使用.gt.
,.ge.
,.le.
,.lt.
,...进行比较。据我所知,这些是遗留的FORTRAN语法。请使用更具吸引力的符号(<
,>
,<=
,>=
,==
)进行比较。
没有必要在Fortran程序中使用GOTO
语句。这又是传统的FORTRAN。通常,简单优雅的do循环和if-blocks可以替换GOTO
语句,就像下面的修订代码一样。
不再需要在Fortran中使用特定于类的内部函数(例如dexp
,dlog
,...用于双精度)。在当前的Fortran标准中,几乎所有(也许全部)Fortran内部函数都具有通用名称(exp
,log
,...)。
以下是该问题中程序的修订版,它解决了上述所有过时的语法,以及处理极大或极小的正定变量的问题(我可能在日志转换中走得太远了)一些永远不会导致溢出或下溢的变量,但我的目的只是展示正定变量的对数变换背后的逻辑以及如何处理它们的算术而不会导致溢出/下溢/ error_in_results。
program test
implicit none
integer,parameter :: dp = SELECTED_REAL_KIND(12,307)
real(kind=dp) kappa,interc,pres,dltdlp,tup,tdwn
real(kind=dp) pthta,alogp,alogpd,alogpu,thta,f,dfdp,p1
real(kind=dp) t1,resid,potdwn,potup,pdwn,pup,epsln,thta1
integer i,j,kout,n,maxit,nmax,resmax
real(kind=dp) :: log_resmax, log_pthta, log_t1, log_dummy, log_residAbsolute, sign_of_f
real(kind=dp) :: log_epsln, log_pdwn, log_pup, log_thta, log_thta1, log_p1, log_dfdp, log_f
logical :: residIsPositive, resmaxIsPositive, residIsBigger
log_resmax = log(log_resmax)
resmaxIsPositive = .true.
kappa = 2._dp/7._dp
epsln = 1._dp
potdwn = 259.39996337890625_dp
potup = 268.41687198359159_dp
pdwn = 100000.00000000000_dp
pup = 92500.000000000000_dp
alogpu = 11.43496392350051_dp
alogpd = 11.512925464970229_dp
thta = 260.00000000000000_dp
alogp = 11.512925464970229_dp
log_epsln = log(epsln)
log_pup = log(pup)
log_pdwn = log(pdwn)
log_thta = log(thta)
! known temperature at lower level
tdwn = potdwn * (pdwn/1.e5_dp)**kappa
! known temperature at upper level
tup = potup *(pup/1.e5_dp)** kappa
! linear change of temperature wrt lnP between different levels
dltdlp = log(tup/tdwn)/(alogpu-alogpd)
! ln(T) value(intercept) where Pressure is 1 Pa and assume a linear
! relationship between P and T
interc = log(tup) - dltdlp*alogpu
! Initial guess value for pressure
!pthta = exp( (log(thta)-interc-kappa*alogp) / (dltdlp-kappa) )
log_pthta = ( log_thta - interc - kappa*alogp ) / ( dltdlp - kappa )
n=0
MyDoLoop: do
!First guess of temperature at intermediate level
!t1 = exp(dltdlp * log(pthta)+interc)
log_t1 = dltdlp * log_pthta + interc
!Residual error when calculating Newton Raphson iteration(Pascal)
!resid = pthta - 1.e5_dp*(t1/thta)**(1._dp/kappa)
log_dummy = log(1.e5_dp) + ( log_t1 - log_thta ) / kappa
if (log_pthta>=log_dummy) then
residIsPositive = .true.
log_residAbsolute = log_pthta + log( 1._dp - exp(log_dummy-log_pthta) )
else
residIsPositive = .false.
log_residAbsolute = log_dummy + log( 1._dp - exp(log_pthta-log_dummy) )
end if
print *, "log-transformed values:"
print *, dltdlp,interc,log_t1,log_residAbsolute,log_pthta
print *, "non-log-transformed values:"
if (residIsPositive) print *, dltdlp,interc,exp(log_t1),exp(log_residAbsolute),exp(log_pthta)
if (.not.residIsPositive) print *, dltdlp,interc,exp(log_t1),-exp(log_residAbsolute),exp(log_pthta)
!if (abs(resid) > epsln) then
if ( log_residAbsolute > log_epsln ) then
n=n+1
if (n <= nmax) then
! First guess of potential temperature given T1 and
! pressure level guess
!thta1 = t1 * (1.e5_dp/pthta)**kappa
log_thta1 = log_t1 + ( log(1.e5_dp)-log_pthta ) * kappa
!f = thta - thta1
if ( log_thta>=thta1 ) then
log_f = log_thta + log( 1._dp - exp( log_thta1 - log_thta ) )
sign_of_f = 1._dp
else
log_f = log_thta + log( 1._dp - exp( log_thta - log_thta1 ) )
sign_of_f = 1._dp
end if
!dfdp = (kappa-dltdlp)*(1.e5_dp/pthta)**kappa*exp(interc + (dltdlp -1._dp)*log(pthta))
! assuming kappa-dltdlp>0 is TRUE always:
log_dfdp = log(kappa-dltdlp) + kappa*(log(1.e5_dp)-log_pthta) + interc + (dltdlp -1._dp)*log_pthta
!p1 = pthta - f/dfdp
! p1 should be, by definition, positive. Therefore:
log_dummy = log_f - log_dfdp
if (log_pthta>=log_dummy) then
log_p1 = log_pthta + log( 1._dp - sign_of_f*exp(log_dummy-log_pthta) )
else
log_p1 = log_dummy + log( 1._dp - sign_of_f*exp(log_pthta-log_dummy) )
end if
!if (p1 <= pdwn) then
if (log_p1 <= log_pdwn) then
!if (p1 >= pup) then
if (log_p1 >= log_pup) then
log_pthta = log_p1
cycle MyDoLoop
else
n = nmax
end if
end if
else
!if (resid > resmax) resmax = resid
residIsBigger = ( residIsPositive .and. resmaxIsPositive .and. log_residAbsolute>log_resmax ) .or. &
( .not.residIsPositive .and. .not.resmaxIsPositive .and. log_residAbsolute<log_resmax ) .or. &
( residIsPositive .and. .not. resmaxIsPositive )
if ( residIsBigger ) then
log_resmax = log_residAbsolute
resmaxIsPositive = residIsPositive
end if
maxit = maxit+1
end if
end if
exit MyDoLoop
end do MyDoLoop
end program test
以下是此程序的示例输出,它与原始代码的输出很吻合:
log-transformed values:
-0.152580456940175 7.31501855886952 5.55917546888014
-22.4565579499410 11.5076538974964
non-log-transformed values:
-0.152580456940175 7.31501855886952 259.608692604963
-1.767017293116268E-010 99474.2302854451
为了比较,这是原始代码的输出:
-0.152580456940175 7.31501855886952 259.608692604963
-8.731149137020111E-011 99474.2302854451