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michaesp |
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c **************************************************************************
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c * This library provides subroutines related to time and file name *
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c * handling *
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c * *
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c **************************************************************************
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c ---------------------------------------------------------------------------
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c Concatenate a date string
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c ---------------------------------------------------------------------------
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subroutine datestring(datestr,yyyy,mm,dd,hh)
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c Declaration of subroutine parameters
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integer yyyy,mm,dd,hh
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character*20 datestr
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c Auxiliary parameters
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integer f1,f2,i0
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integer yy,ce
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20 |
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i0=ichar('0')
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datestr=''
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24 |
yy=mod(yyyy,100)
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ce=int(yyyy/100)
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if ((ce.ne.19).and.(ce.ne.20)) then
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print*,'Invalid year... Stop'
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stop
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endif
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if (yy.ge.0) then
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f1=yy/10
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f2=mod(yy,10)
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if (ce.eq.19) then
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datestr=trim(datestr)//'19'//char(f1+i0)//char(f2+i0)
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else if (ce.eq.20) then
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datestr=trim(datestr)//'20'//char(f1+i0)//char(f2+i0)
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endif
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endif
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if (mm.gt.0) then
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f1=mm/10
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f2=mod(mm,10)
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datestr=trim(datestr)//char(f1+i0)//char(f2+i0)
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endif
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if (dd.gt.0) then
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f1=dd/10
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f2=mod(dd,10)
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datestr=trim(datestr)//char(f1+i0)//char(f2+i0)
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endif
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if (hh.ge.0) then
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f1=hh/10
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f2=mod(hh,10)
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datestr=trim(datestr)//'_'//char(f1+i0)//char(f2+i0)
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endif
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return
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end
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c ---------------------------------------------------------------------------
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c Calculates the new date when diff (in hours) is added to date1.
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c ---------------------------------------------------------------------------
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subroutine newdate(date1,diff,date2)
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C
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C date1 int input array contains a date in the form
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C year,month,day,hour,step
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C diff real input timestep in hours to go from date1
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C date2 int output array contains new date in the same form
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integer date1(5),date2(5)
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integer idays(12) ! array containing the days of the monthes
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real diff
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logical yearchange
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data idays/31,28,31,30,31,30,31,31,30,31,30,31/
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yearchange=.false.
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if ((mod(date1(1),4).eq.0).and.(date1(2).le.2)) idays(2)=29
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date2(1)=date1(1)
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date2(2)=date1(2)
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date2(3)=date1(3)
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date2(4)=date1(4)
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date2(5)=0
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date2(4)=date1(4)+int(diff)+date1(5)
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if (date2(4).ge.24) then
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date2(3)=date2(3)+int(date2(4)/24)
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date2(4)=date2(4)-int(date2(4)/24)*24
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endif
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if (date2(4).lt.0) then
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if (mod(date2(4),24).eq.0) then
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date2(3)=date2(3)-int(abs(date2(4))/24)
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date2(4)=date2(4)+int(abs(date2(4))/24)*24
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else
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date2(3)=date2(3)-(1+int(abs(date2(4))/24))
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date2(4)=date2(4)+(1+int(abs(date2(4))/24))*24
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endif
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endif
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100 if (date2(3).gt.idays(date2(2))) then
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if ((date2(2).eq.2).and.(mod(date2(1),4).eq.0)) idays(2)=29
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date2(3)=date2(3)-idays(date2(2))
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if (idays(2).eq.29) idays(2)=28
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date2(2)=date2(2)+1
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if (date2(2).gt.12) then
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* date2(1)=date2(1)+int(date2(2)/12)
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* date2(2)=date2(2)-int(date2(2)/12)*12
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date2(1)=date2(1)+1
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date2(2)=date2(2)-12
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endif
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if (date2(2).lt.1) then
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date2(1)=date2(1)-(1+int(abs(date2(2)/12)))
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date2(2)=date2(2)+(1+int(abs(date2(2)/12)))*12
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endif
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goto 100
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endif
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200 if (date2(3).lt.1) then
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date2(2)=date2(2)-1
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if (date2(2).gt.12) then
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date2(1)=date2(1)+int(date2(2)/12)
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date2(2)=date2(2)-int(date2(2)/12)*12
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endif
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129 |
if (date2(2).lt.1) then
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date2(1)=date2(1)-(1+int(abs(date2(2)/12)))
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date2(2)=date2(2)+(1+int(abs(date2(2)/12)))*12
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endif
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if ((date2(2).eq.2).and.(mod(date2(1),4).eq.0)) idays(2)=29
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date2(3)=date2(3)+idays(date2(2))
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if (idays(2).eq.29) idays(2)=28
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goto 200
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endif
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if (date2(2).gt.12) then
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date2(1)=date2(1)+int(date2(2)/12)
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date2(2)=date2(2)-int(date2(2)/12)*12
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endif
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if (date2(2).lt.1) then
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date2(1)=date2(1)-(1+int(abs(date2(2)/12)))
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date2(2)=date2(2)+(1+int(abs(date2(2)/12)))*12
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endif
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if (date2(1).lt.1000) then
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if (date2(1).ge.100) date2(1)=date2(1)-100
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endif
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return
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end
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c ---------------------------------------------------------------------------
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c Convert <hh.mm> -> <frac> and <frac> -> <hhmm>
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c ---------------------------------------------------------------------------
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subroutine hhmm2frac (hhmm,frac)
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real hhmm
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real frac
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frac=real(int(hhmm))+
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> 100.*(hhmm-real(int(hhmm)))/60.
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end
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subroutine frac2hhmm (frac,hhmm)
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real hhmm
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real frac
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real hh,mm
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hh = real(int(frac))
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mm = 60. * (frac-real(int(frac)))
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hhmm = hh + 0.01 * mm
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end
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c ---------------------------------------------------------------------------
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c Time difference between two dates
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c ---------------------------------------------------------------------------
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subroutine timediff(date1,date2,diff)
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C New version with hour and minutes! (for hour and step [in hours]
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C use the routine oldtimediff!)
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C
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C Calculates the time difference in hours (and minutes) for the two
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C dates specified by the two arrays date1 and date2.
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C They are expected to contain the following date information:
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C year month day hour minute.
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C
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C date1 array specifying the first date
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C date2 array specifying the second date
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C diff time differenc between date1 and date2 in hours
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C
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202 |
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integer date1(5),date2(5)
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integer idays(12) ! array containing the days of the monthes
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real diff
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integer ixday,imdiff,ihdiff,iddiff,j
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integer yy,yy1,yy2
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209 |
idays(1)=31
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idays(2)=28
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idays(3)=31
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idays(4)=30
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idays(5)=31
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idays(6)=30
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idays(7)=31
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idays(8)=31
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idays(9)=30
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idays(10)=31
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idays(11)=30
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idays(12)=31
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C Check format of year (YYYY or YY - in case of YY assume 19YY)
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if (date1(1).lt.100) date1(1)=1900+date1(1)
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if (date2(1).lt.100) date2(1)=1900+date2(1)
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C Determine if the period between date1 and date2 contains a Feb.29
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ixday=0 ! extra day flag
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yy1=min(date1(1),date2(1))
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yy2=max(date1(1),date2(1))
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if (yy1.eq.yy2) then
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if (mod(yy1,4).eq.0) then
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idays(2)=29
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endif
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else
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if (mod(yy1,4).eq.0) then
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if (((yy1.eq.date1(1)).and.(date1(2).le.2)).or.
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> ((yy1.eq.date2(1)).and.(date2(2).le.2))) then
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ixday=ixday+1
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endif
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endif
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if (mod(yy2,4).eq.0) then
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if (((yy2.eq.date1(1)).and.(date1(2).gt.2)).or.
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> ((yy2.eq.date2(1)).and.(date2(2).gt.2))) then
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ixday=ixday+1
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endif
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endif
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if (yy2-yy1.gt.1) then
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do yy=yy1+1,yy2-1
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if (mod(yy,4).eq.0) then
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ixday=ixday+1
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endif
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enddo
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endif
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endif
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ihdiff=0 ! diff. in hours between date1/date2
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iddiff=0 ! diff. in days between date1/date2
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if (date1(1).gt.date2(1)) then ! compare years
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do j=date2(1),date1(1)-1
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iddiff=iddiff+365
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enddo
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iddiff=iddiff+ixday
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else if (date1(1).lt.date2(1)) then
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do j=date1(1),date2(1)-1
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iddiff=iddiff-365
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enddo
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iddiff=iddiff-ixday
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endif
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if (date1(2).gt.date2(2)) then ! compare monthes
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do j=date2(2),date1(2)-1
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iddiff=iddiff+idays(j)
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enddo
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else if (date1(2).lt.date2(2)) then
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do j=date1(2),date2(2)-1
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280 |
iddiff=iddiff-idays(j)
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enddo
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endif
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283 |
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284 |
iddiff=iddiff+date1(3)-date2(3)
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285 |
ihdiff=iddiff*24+date1(4)-date2(4)
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imdiff=ihdiff*60+date1(5)-date2(5)
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287 |
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288 |
ihdiff=imdiff/60
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imdiff=mod(imdiff,60)
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290 |
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diff=real(ihdiff)+real(imdiff)/100.
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return
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end
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295 |
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c ---------------------------------------------------------------------------
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298 |
c Time check based on date strings from input files
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299 |
c ---------------------------------------------------------------------------
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300 |
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301 |
subroutine do_timecheck (datestr,numdat,timeinc,fbflag,refdate)
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c Perform a simple timecheck based on the names of the input files;
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c the following tests have to be passed:
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c (1) the order of <datestr> must match with <fbflag>
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306 |
c (2) the step between <datestr> must match with <timeinc>
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307 |
c (3) <refdate> must match with <datestr(1)>
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309 |
implicit none
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310 |
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311 |
c Declaration of subroutine parameters
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312 |
integer numdat
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313 |
character*13 datestr(numdat)
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314 |
real timeinc
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315 |
integer fbflag
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316 |
integer refdate(6)
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317 |
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318 |
c Numerical epsilon
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319 |
real eps
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320 |
parameter (eps=0.0001)
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321 |
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322 |
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323 |
c Auxilary variables
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324 |
integer yyyy(numdat)
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325 |
integer mm (numdat)
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326 |
integer dd (numdat)
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327 |
integer hh (numdat)
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328 |
integer mi (numdat)
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329 |
integer i
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330 |
character*13 str
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331 |
integer date1(5),date2(5)
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332 |
real diff,ok
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333 |
integer hires
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334 |
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335 |
c Get date mode (including or without minutes)
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336 |
if ( len_trim(datestr(1)).eq.11 ) then
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hires = 0
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338 |
elseif ( len_trim(datestr(1)).eq.13 ) then
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hires = 1
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340 |
else
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341 |
hires = -1
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342 |
endif
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343 |
do i=2,numdat
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344 |
if ((len_trim(datestr(i)).eq.11).and.(hires.eq.1)) hires = -1
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345 |
if ((len_trim(datestr(i)).eq.13).and.(hires.eq.0)) hires = -1
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346 |
enddo
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347 |
if ( hires.eq.-1 ) then
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348 |
print*,' ERROR: string format is wrong:'
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349 |
print*,' either {YYYYMMDD_HH} or {YYYYMMDD_HHMM}'
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350 |
stop
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351 |
endif
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352 |
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353 |
c Convert date strings into integers
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354 |
do i=1,numdat
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355 |
if (hires.eq.0) then
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356 |
read(datestr(i),'(i4,i2,i2,1x,i2)')
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357 |
> yyyy(i),mm(i),dd(i),hh(i)
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358 |
mi(i) = 0
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|
359 |
else
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|
360 |
read(datestr(i),'(i4,i2,i2,1x,i2,i2)')
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|
361 |
> yyyy(i),mm(i),dd(i),hh(i),mi(i)
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|
362 |
endif
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|
363 |
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|
364 |
enddo
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|
365 |
|
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|
366 |
|
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|
367 |
c Check time increment and consistence with time increment
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|
|
368 |
do i=1,numdat-1
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|
369 |
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|
370 |
date1(1) = yyyy(i)
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|
371 |
date1(2) = mm (i)
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|
|
372 |
date1(3) = dd (i)
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|
373 |
date1(4) = hh (i)
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|
374 |
date1(5) = mi (i)
|
|
|
375 |
date2(1) = yyyy(i+1)
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|
|
376 |
date2(2) = mm (i+1)
|
|
|
377 |
date2(3) = dd (i+1)
|
|
|
378 |
date2(4) = hh (i+1)
|
|
|
379 |
date2(5) = mi (i+1)
|
|
|
380 |
|
|
|
381 |
call timediff (date2,date1,diff)
|
|
|
382 |
|
|
|
383 |
if (abs(diff).lt.(1.-eps)) diff = 100. * diff/60
|
|
|
384 |
|
|
|
385 |
if (fbflag.eq.1) then
|
|
|
386 |
ok = diff - timeinc
|
|
|
387 |
elseif (fbflag.eq.-1) then
|
|
|
388 |
ok = -diff - timeinc
|
|
|
389 |
endif
|
|
|
390 |
|
|
|
391 |
if ( abs(ok).gt.eps ) then
|
|
|
392 |
print*,' ERROR: timecheck failed !!!! Stop'
|
|
|
393 |
print*,' date 1 : ',date1
|
|
|
394 |
print*,' date 2 : ',date2
|
|
|
395 |
print*,' diff,timeinc : ',diff,timeinc
|
|
|
396 |
stop
|
|
|
397 |
endif
|
|
|
398 |
|
|
|
399 |
enddo
|
|
|
400 |
|
|
|
401 |
c Check the reference date
|
|
|
402 |
if ( ( refdate(1).ne.yyyy(1) ).or.
|
|
|
403 |
> ( refdate(2).ne.mm (1) ).or.
|
|
|
404 |
> ( refdate(3).ne.dd (1) ).or.
|
|
|
405 |
> ( refdate(4).ne.hh (1) ) )
|
|
|
406 |
>then
|
|
|
407 |
print*,' ERROR: timecheck failed (refdate) !!!! Stop'
|
|
|
408 |
print*,'ref : ',refdate
|
|
|
409 |
print*,'P : ',yyyy(1),mm(1),dd(1),hh(1),mi(1)
|
|
|
410 |
stop
|
|
|
411 |
endif
|
|
|
412 |
|
|
|
413 |
end
|
|
|
414 |
|
|
|
415 |
|
|
|
416 |
|
|
|
417 |
|
|
|
418 |
|
|
|
419 |
|
|
|
420 |
|