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michaesp |
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PROGRAM timeresolution
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c ***********************************************************************
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c * Change time resolution of of a trajectory file *
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c * Michael Sprenger / Winter 2010 *
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c ***********************************************************************
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implicit none
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c ----------------------------------------------------------------------
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c Declaration of variables
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c ----------------------------------------------------------------------
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c Parameters
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character*80 inpfile
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character*80 outfile
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integer ntra,otim,ncol
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real timeres
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character*80 unit
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character*80 mode
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c Trajectories
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character*80 vars(100)
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integer refdate(6)
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integer ntim
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real,allocatable, dimension (:,:,:) :: trainp
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real,allocatable, dimension (:,:,:) :: traout
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real,allocatable, dimension (:) :: timold,timnew
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real,allocatable, dimension (:) :: fldold,fldnew
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c Numerical constants
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real eps
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parameter (eps=0.001)
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c Auxiliary variables
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integer inpmode
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integer outmode
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integer stat
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integer fid
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integer i,j,k
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real hhmm,tfrac
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real range
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c ----------------------------------------------------------------------
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c Parameter handling
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c ----------------------------------------------------------------------
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c Read parameters
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open(10,file='timeres.param')
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read(10,*) inpfile
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read(10,*) outfile
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read(10,*) ntra,otim,ncol
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read(10,*) timeres
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read(10,*) unit
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read(10,*) mode
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close(10)
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c Change unit to minutes
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if ( unit.eq.'min') then
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timeres = 1./60. * timeres
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unit = 'h'
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endif
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c Determine the formats
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call mode_tra(inpmode,inpfile)
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if (inpmode.eq.-1) inpmode=1
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call mode_tra(outmode,outfile)
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if (outmode.eq.-1) outmode=1
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c ----------------------------------------------------------------------
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c Read input trajectory and allocate memory
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c ----------------------------------------------------------------------
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c Allocate memory for input trajectories
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allocate(trainp(ntra,otim,ncol),stat=stat)
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if (stat.ne.0) print*,'*** error allocating array trainp ***'
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allocate(timold(otim),stat=stat)
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if (stat.ne.0) print*,'*** error allocating array timold ***'
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allocate(fldold(otim),stat=stat)
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if (stat.ne.0) print*,'*** error allocating array fldold ***'
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c Read inpufile
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call ropen_tra(fid,inpfile,ntra,otim,ncol,refdate,vars,inpmode)
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call read_tra (fid,trainp,ntra,otim,ncol,inpmode)
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call close_tra(fid,inpmode)
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c Check that first four columns correspond to time,lon,lat,p
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if ( (vars(1).ne.'time' ).or.
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> (vars(2).ne.'xpos' ).and.(vars(2).ne.'lon' ).or.
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> (vars(3).ne.'ypos' ).and.(vars(3).ne.'lat' ).or.
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> (vars(4).ne.'ppos' ).and.(vars(4).ne.'p' ) )
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>then
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print*,' ERROR: problem with input trajectories ...'
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stop
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endif
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c Convert all times from hhmm to fractional time
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do i=1,ntra
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do j=1,otim
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hhmm = trainp(i,j,1)
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call hhmm2frac(hhmm,tfrac)
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trainp(i,j,1) = tfrac
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enddo
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enddo
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c Get the time range in minutes
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range = ( trainp(1,otim,1) - trainp(1,1,1) )
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c Determine the new number of times
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ntim = nint( abs( range ) / timeres ) + 1
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c Check that the time range and new time resolution are consistent
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if ( abs( real(ntim-1) * timeres -range ).gt.eps ) then
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print*,' ERROR: time range and resolution are not compatible'
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print*,' range = ',range
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print*,' (ntim-1) * timeres = ',real(ntim-1) * timeres
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stop
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endif
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c Allocate memory for output trajectories
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allocate(traout(ntra,ntim,ncol),stat=stat)
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if (stat.ne.0) print*,'*** error allocating array trainp ***'
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allocate(timnew(ntim),stat=stat)
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if (stat.ne.0) print*,'*** error allocating array timnew ***'
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allocate(fldnew(ntim),stat=stat)
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if (stat.ne.0) print*,'*** error allocating array fldnew ***'
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c Define the old and new times
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do i=1,otim
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timold(i) = trainp(1,i,1)
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enddo
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do i=1,ntim
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timnew(i) = timold(1) + real(i-1) * timeres
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enddo
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c ----------------------------------------------------------------------
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c Change time resolution
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c ----------------------------------------------------------------------
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do i=1,ntra
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do k=2,ncol
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c Copy old field
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do j=1,otim
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fldold(j) = trainp(i,j,k)
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enddo
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c Exception: Handle date line problem for longitude
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if ( k.eq.2 ) then
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do j=2,otim
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if ( (fldold(j-1)-fldold(j)).gt.180. ) then
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fldold(j) = fldold(j) + 360.
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else if ( (fldold(j-1)-fldold(j)).lt.-180. ) then
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fldold(j) = fldold(j) - 360.
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endif
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enddo
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endif
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c Cubic spline fitting
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if ( mode.eq.'cubic' ) then
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call cubicfit (timold,fldold,otim,timnew,fldnew,ntim)
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else if (mode.eq.'linear' ) then
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call linearfit(timold,fldold,otim,timnew,fldnew,ntim)
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endif
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c Exception: Reverse date line handling for longitude
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if ( k.eq.2 ) then
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do j=1,ntim
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if ( fldnew(j).gt.180. ) then
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fldnew(j) = fldnew(j) -360.
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else if ( fldnew(j).lt.-180. ) then
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fldnew(j) = fldnew(j) +360.
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endif
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enddo
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endif
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c Save the new field in the output trajectory
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do j=1,ntim
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traout(i,j,1) = timnew(j)
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traout(i,j,k) = fldnew(j)
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enddo
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enddo
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enddo
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c ----------------------------------------------------------------------
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c Write output trajectory
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c ----------------------------------------------------------------------
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c Convert all times from fractional to hhmm time
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do i=1,ntra
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do j=1,ntim
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tfrac = traout(i,j,1)
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call frac2hhmm(tfrac,hhmm)
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traout(i,j,1) = hhmm
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enddo
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enddo
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c Write output file
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call wopen_tra(fid,outfile,ntra,ntim,ncol,refdate,vars,outmode)
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call write_tra(fid,traout,ntra,ntim,ncol,outmode)
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call close_tra(fid,outmode)
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end
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c ********************************************************************
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c * REPARAMETERIZATION SUBROUTINES *
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c ********************************************************************
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c -------------------------------------------------------------
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c Interpolation of the trajectory with linear interpolation
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c -------------------------------------------------------------
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SUBROUTINE linearfit (time,lon,n,sptime,splon,spn)
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c Given the curve <time,lon> with <n> data points, fit a
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c linear fit to this curve. The new curve is returned in
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c <sptime,splon,spn> with <spn> data points. The parameter
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c <spn> specifies on entry the number of interpolated points
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c along the curve.
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implicit none
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c Declaration of subroutine parameters
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integer n
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real time(n),lon(n)
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integer spn
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real sptime(spn),splon(spn)
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c Auxiliary variables
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real dt
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real s
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integer i,j,iold
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real order
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c Determine whether the input array is ascending or descending
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if (time(1).gt.time(n)) then
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order=-1.
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else
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order= 1.
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endif
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c Bring the time array into ascending order
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do i=1,n
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time(i)=order*time(i)
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enddo
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c Prepare the linear interpolation: define the new times
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dt=(time(n)-time(1))/real(spn-1)
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do i=1,spn
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sptime(i)=time(1)+real(i-1)*dt
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enddo
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c Do the interpolation
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iold = 1
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do i=1,spn
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c Decide which interval of the old time series must be taken
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do j=iold,n-1
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if ( ( sptime(i).ge.time(j ) ).and.
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> ( sptime(i).lt.time(j+1) ) )
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> then
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iold = j
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exit
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endif
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enddo
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c Do the linear interpolation
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splon(i) = lon(iold) + ( lon(iold+1) - lon(iold) ) *
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> ( sptime(i) - time(iold) ) / ( time(iold+1) - time(iold) )
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enddo
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c Change the time arrays back: original order
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do i=1,spn
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sptime(i)=order*sptime(i)
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enddo
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do i=1,n
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time(i)=order*time(i)
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enddo
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return
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end
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c -------------------------------------------------------------
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c Interpolation of the trajectory with a natural cubic spline
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c -------------------------------------------------------------
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SUBROUTINE cubicfit (time,lon,n,sptime,splon,spn)
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c Given the curve <time,lon> with <n> data points, fit a
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c cubic spline to this curve. The new curve is returned in
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c <sptime,splon,spn> with <spn> data points. The parameter
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c <spn> specifies on entry the number of spline interpolated points
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c along the curve.
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implicit none
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c Declaration of subroutine parameters
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integer n
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real time(n),lon(n)
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integer spn
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real sptime(spn),splon(spn)
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c Auxiliary variables
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real y2ax(n)
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real dt
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real s
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integer i
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real order
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c Determine whether the input array is ascending or descending
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if (time(1).gt.time(n)) then
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order=-1.
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else
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order= 1.
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endif
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c Bring the time array into ascending order
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do i=1,n
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time(i)=order*time(i)
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enddo
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c Prepare the (natural) cubic spline interpolation
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call spline (time,lon,n,1.e30,1.e30,y2ax)
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dt=(time(n)-time(1))/real(spn-1)
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do i=1,spn
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sptime(i)=time(1)+real(i-1)*dt
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enddo
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c Do the spline interpolation
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do i=1,spn
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call splint(time,lon,y2ax,n,sptime(i),s)
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splon(i)=s
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enddo
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c Change the time arrays back
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do i=1,spn
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sptime(i)=order*sptime(i)
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enddo
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do i=1,n
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time(i)=order*time(i)
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enddo
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return
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end
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c -------------------------------------------------------------
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c Basic routines for spline interpolation (Numerical Recipes)
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c -------------------------------------------------------------
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SUBROUTINE spline(x,y,n,yp1,ypn,y2)
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INTEGER n,NMAX
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REAL yp1,ypn,x(n),y(n),y2(n)
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PARAMETER (NMAX=500)
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INTEGER i,k
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REAL p,qn,sig,un,u(NMAX)
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if (yp1.gt..99e30) then
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y2(1)=0.
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u(1)=0.
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else
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y2(1)=-0.5
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u(1)=(3./(x(2)-x(1)))*((y(2)-y(1))/(x(2)-x(1))-yp1)
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endif
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do 11 i=2,n-1
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sig=(x(i)-x(i-1))/(x(i+1)-x(i-1))
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p=sig*y2(i-1)+2.
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y2(i)=(sig-1.)/p
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u(i)=(6.*((y(i+1)-y(i))/(x(i+
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372 |
*1)-x(i))-(y(i)-y(i-1))/(x(i)-x(i-1)))/(x(i+1)-x(i-1))-sig*
|
|
|
373 |
*u(i-1))/p
|
|
|
374 |
11 continue
|
|
|
375 |
if (ypn.gt..99e30) then
|
|
|
376 |
qn=0.
|
|
|
377 |
un=0.
|
|
|
378 |
else
|
|
|
379 |
qn=0.5
|
|
|
380 |
un=(3./(x(n)-x(n-1)))*(ypn-(y(n)-y(n-1))/(x(n)-x(n-1)))
|
|
|
381 |
endif
|
|
|
382 |
y2(n)=(un-qn*u(n-1))/(qn*y2(n-1)+1.)
|
|
|
383 |
do 12 k=n-1,1,-1
|
|
|
384 |
y2(k)=y2(k)*y2(k+1)+u(k)
|
|
|
385 |
12 continue
|
|
|
386 |
return
|
|
|
387 |
END
|
|
|
388 |
|
|
|
389 |
SUBROUTINE splint(xa,ya,y2a,n,x,y)
|
|
|
390 |
INTEGER n
|
|
|
391 |
REAL x,y,xa(n),y2a(n),ya(n)
|
|
|
392 |
INTEGER k,khi,klo
|
|
|
393 |
REAL a,b,h
|
|
|
394 |
klo=1
|
|
|
395 |
khi=n
|
|
|
396 |
1 if (khi-klo.gt.1) then
|
|
|
397 |
k=(khi+klo)/2
|
|
|
398 |
if(xa(k).gt.x)then
|
|
|
399 |
khi=k
|
|
|
400 |
else
|
|
|
401 |
klo=k
|
|
|
402 |
endif
|
|
|
403 |
goto 1
|
|
|
404 |
endif
|
|
|
405 |
h=xa(khi)-xa(klo)
|
| 11 |
michaesp |
406 |
if (h.eq.0.) then
|
|
|
407 |
print*, 'bad xa input in splint'
|
|
|
408 |
stop
|
|
|
409 |
endif
|
| 3 |
michaesp |
410 |
a=(xa(khi)-x)/h
|
|
|
411 |
b=(x-xa(klo))/h
|
|
|
412 |
y=a*ya(klo)+b*ya(khi)+((a**3-a)*y2a(klo)+(b**3-b)*y2a(khi))*(h**
|
|
|
413 |
*2)/6.
|
|
|
414 |
return
|
|
|
415 |
END
|
|
|
416 |
|
|
|
417 |
|
|
|
418 |
|