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PROGRAM timeresolution
c ***********************************************************************
c * Change time resolution of of a trajectory file *
c * Michael Sprenger / Winter 2010 *
c ***********************************************************************
implicit none
c ----------------------------------------------------------------------
c Declaration of variables
c ----------------------------------------------------------------------
c Parameters
character*80 inpfile
character*80 outfile
integer ntra,otim,ncol
real timeres
character*80 unit
character*80 mode
c Trajectories
character*80 vars(100)
integer refdate(6)
integer ntim
real,allocatable, dimension (:,:,:) :: trainp
real,allocatable, dimension (:,:,:) :: traout
real,allocatable, dimension (:) :: timold,timnew
real,allocatable, dimension (:) :: fldold,fldnew
c Numerical constants
real eps
parameter (eps=0.001)
c Auxiliary variables
integer inpmode
integer outmode
integer stat
integer fid
integer i,j,k
real hhmm,tfrac
real range
c ----------------------------------------------------------------------
c Parameter handling
c ----------------------------------------------------------------------
c Read parameters
open(10,file='timeres.param')
read(10,*) inpfile
read(10,*) outfile
read(10,*) ntra,otim,ncol
read(10,*) timeres
read(10,*) unit
read(10,*) mode
close(10)
c Change unit to minutes
if ( unit.eq.'min') then
timeres = 1./60. * timeres
unit = 'h'
endif
c Determine the formats
call mode_tra(inpmode,inpfile)
if (inpmode.eq.-1) inpmode=1
call mode_tra(outmode,outfile)
if (outmode.eq.-1) outmode=1
c ----------------------------------------------------------------------
c Read input trajectory and allocate memory
c ----------------------------------------------------------------------
c Allocate memory for input trajectories
allocate(trainp(ntra,otim,ncol),stat=stat)
if (stat.ne.0) print*,'*** error allocating array trainp ***'
allocate(timold(otim),stat=stat)
if (stat.ne.0) print*,'*** error allocating array timold ***'
allocate(fldold(otim),stat=stat)
if (stat.ne.0) print*,'*** error allocating array fldold ***'
c Read inpufile
call ropen_tra(fid,inpfile,ntra,otim,ncol,refdate,vars,inpmode)
call read_tra (fid,trainp,ntra,otim,ncol,inpmode)
call close_tra(fid,inpmode)
c Check that first four columns correspond to time,lon,lat,p
if ( (vars(1).ne.'time' ).or.
> (vars(2).ne.'xpos' ).and.(vars(2).ne.'lon' ).or.
> (vars(3).ne.'ypos' ).and.(vars(3).ne.'lat' ).or.
> (vars(4).ne.'ppos' ).and.(vars(4).ne.'p' ) )
>then
print*,' ERROR: problem with input trajectories ...'
stop
endif
c Convert all times from hhmm to fractional time
do i=1,ntra
do j=1,otim
hhmm = trainp(i,j,1)
call hhmm2frac(hhmm,tfrac)
trainp(i,j,1) = tfrac
enddo
enddo
c Get the time range in minutes
range = ( trainp(1,otim,1) - trainp(1,1,1) )
c Determine the new number of times
ntim = nint( abs( range ) / timeres ) + 1
c Check that the time range and new time resolution are consistent
if ( abs( real(ntim-1) * timeres -range ).gt.eps ) then
print*,' ERROR: time range and resolution are not compatible'
print*,' range = ',range
print*,' (ntim-1) * timeres = ',real(ntim-1) * timeres
stop
endif
c Allocate memory for output trajectories
allocate(traout(ntra,ntim,ncol),stat=stat)
if (stat.ne.0) print*,'*** error allocating array trainp ***'
allocate(timnew(ntim),stat=stat)
if (stat.ne.0) print*,'*** error allocating array timnew ***'
allocate(fldnew(ntim),stat=stat)
if (stat.ne.0) print*,'*** error allocating array fldnew ***'
c Define the old and new times
do i=1,otim
timold(i) = trainp(1,i,1)
enddo
do i=1,ntim
timnew(i) = timold(1) + real(i-1) * timeres
enddo
c ----------------------------------------------------------------------
c Change time resolution
c ----------------------------------------------------------------------
do i=1,ntra
do k=2,ncol
c Copy old field
do j=1,otim
fldold(j) = trainp(i,j,k)
enddo
c Exception: Handle date line problem for longitude
if ( k.eq.2 ) then
do j=2,otim
if ( (fldold(j-1)-fldold(j)).gt.180. ) then
fldold(j) = fldold(j) + 360.
else if ( (fldold(j-1)-fldold(j)).lt.-180. ) then
fldold(j) = fldold(j) - 360.
endif
enddo
endif
c Cubic spline fitting
if ( mode.eq.'cubic' ) then
call cubicfit (timold,fldold,otim,timnew,fldnew,ntim)
else if (mode.eq.'linear' ) then
call linearfit(timold,fldold,otim,timnew,fldnew,ntim)
endif
c Exception: Reverse date line handling for longitude
if ( k.eq.2 ) then
do j=1,ntim
if ( fldnew(j).gt.180. ) then
fldnew(j) = fldnew(j) -360.
else if ( fldnew(j).lt.-180. ) then
fldnew(j) = fldnew(j) +360.
endif
enddo
endif
c Save the new field in the output trajectory
do j=1,ntim
traout(i,j,1) = timnew(j)
traout(i,j,k) = fldnew(j)
enddo
enddo
enddo
c ----------------------------------------------------------------------
c Write output trajectory
c ----------------------------------------------------------------------
c Convert all times from fractional to hhmm time
do i=1,ntra
do j=1,ntim
tfrac = traout(i,j,1)
call frac2hhmm(tfrac,hhmm)
traout(i,j,1) = hhmm
enddo
enddo
c Write output file
call wopen_tra(fid,outfile,ntra,ntim,ncol,refdate,vars,outmode)
call write_tra(fid,traout,ntra,ntim,ncol,outmode)
call close_tra(fid,outmode)
end
c ********************************************************************
c * REPARAMETERIZATION SUBROUTINES *
c ********************************************************************
c -------------------------------------------------------------
c Interpolation of the trajectory with linear interpolation
c -------------------------------------------------------------
SUBROUTINE linearfit (time,lon,n,sptime,splon,spn)
c Given the curve <time,lon> with <n> data points, fit a
c linear fit to this curve. The new curve is returned in
c <sptime,splon,spn> with <spn> data points. The parameter
c <spn> specifies on entry the number of interpolated points
c along the curve.
implicit none
c Declaration of subroutine parameters
integer n
real time(n),lon(n)
integer spn
real sptime(spn),splon(spn)
c Auxiliary variables
real dt
real s
integer i,j,iold
real order
c Determine whether the input array is ascending or descending
if (time(1).gt.time(n)) then
order=-1.
else
order= 1.
endif
c Bring the time array into ascending order
do i=1,n
time(i)=order*time(i)
enddo
c Prepare the linear interpolation: define the new times
dt=(time(n)-time(1))/real(spn-1)
do i=1,spn
sptime(i)=time(1)+real(i-1)*dt
enddo
c Do the interpolation
iold = 1
do i=1,spn
c Decide which interval of the old time series must be taken
do j=iold,n-1
if ( ( sptime(i).ge.time(j ) ).and.
> ( sptime(i).lt.time(j+1) ) )
> then
iold = j
exit
endif
enddo
c Do the linear interpolation
splon(i) = lon(iold) + ( lon(iold+1) - lon(iold) ) *
> ( sptime(i) - time(iold) ) / ( time(iold+1) - time(iold) )
enddo
c Change the time arrays back: original order
do i=1,spn
sptime(i)=order*sptime(i)
enddo
do i=1,n
time(i)=order*time(i)
enddo
return
end
c -------------------------------------------------------------
c Interpolation of the trajectory with a natural cubic spline
c -------------------------------------------------------------
SUBROUTINE cubicfit (time,lon,n,sptime,splon,spn)
c Given the curve <time,lon> with <n> data points, fit a
c cubic spline to this curve. The new curve is returned in
c <sptime,splon,spn> with <spn> data points. The parameter
c <spn> specifies on entry the number of spline interpolated points
c along the curve.
implicit none
c Declaration of subroutine parameters
integer n
real time(n),lon(n)
integer spn
real sptime(spn),splon(spn)
c Auxiliary variables
real y2ax(n)
real dt
real s
integer i
real order
c Determine whether the input array is ascending or descending
if (time(1).gt.time(n)) then
order=-1.
else
order= 1.
endif
c Bring the time array into ascending order
do i=1,n
time(i)=order*time(i)
enddo
c Prepare the (natural) cubic spline interpolation
call spline (time,lon,n,1.e30,1.e30,y2ax)
dt=(time(n)-time(1))/real(spn-1)
do i=1,spn
sptime(i)=time(1)+real(i-1)*dt
enddo
c Do the spline interpolation
do i=1,spn
call splint(time,lon,y2ax,n,sptime(i),s)
splon(i)=s
enddo
c Change the time arrays back
do i=1,spn
sptime(i)=order*sptime(i)
enddo
do i=1,n
time(i)=order*time(i)
enddo
return
end
c -------------------------------------------------------------
c Basic routines for spline interpolation (Numerical Recipes)
c -------------------------------------------------------------
SUBROUTINE spline(x,y,n,yp1,ypn,y2)
INTEGER n,NMAX
REAL yp1,ypn,x(n),y(n),y2(n)
PARAMETER (NMAX=500)
INTEGER i,k
REAL p,qn,sig,un,u(NMAX)
if (yp1.gt..99e30) then
y2(1)=0.
u(1)=0.
else
y2(1)=-0.5
u(1)=(3./(x(2)-x(1)))*((y(2)-y(1))/(x(2)-x(1))-yp1)
endif
do 11 i=2,n-1
sig=(x(i)-x(i-1))/(x(i+1)-x(i-1))
p=sig*y2(i-1)+2.
y2(i)=(sig-1.)/p
u(i)=(6.*((y(i+1)-y(i))/(x(i+
*1)-x(i))-(y(i)-y(i-1))/(x(i)-x(i-1)))/(x(i+1)-x(i-1))-sig*
*u(i-1))/p
11 continue
if (ypn.gt..99e30) then
qn=0.
un=0.
else
qn=0.5
un=(3./(x(n)-x(n-1)))*(ypn-(y(n)-y(n-1))/(x(n)-x(n-1)))
endif
y2(n)=(un-qn*u(n-1))/(qn*y2(n-1)+1.)
do 12 k=n-1,1,-1
y2(k)=y2(k)*y2(k+1)+u(k)
12 continue
return
END
SUBROUTINE splint(xa,ya,y2a,n,x,y)
INTEGER n
REAL x,y,xa(n),y2a(n),ya(n)
INTEGER k,khi,klo
REAL a,b,h
klo=1
khi=n
1 if (khi-klo.gt.1) then
k=(khi+klo)/2
if(xa(k).gt.x)then
khi=k
else
klo=k
endif
goto 1
endif
h=xa(khi)-xa(klo)
if (h.eq.0.) then
print*, 'bad xa input in splint'
stop
endif
a=(xa(khi)-x)/h
b=(x-xa(klo))/h
y=a*ya(klo)+b*ya(khi)+((a**3-a)*y2a(klo)+(b**3-b)*y2a(khi))*(h**
*2)/6.
return
END