year to Modified Julian day
in [Fortran]
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From: Fortran_follower on 6 Nov 2009 10:25 Dear Friends, Recently I've started learning Fortran 95. I want to convert the date format mentioned below to Modified Julian Day format (MJD) 04-Aug-2001 00:10:52 04-Aug-2001 00:10:53 04-Aug-2001 00:10:54 04-Aug-2001 00:10:55 04-Aug-2001 00:10:56 04-Aug-2001 00:10:57 04-Aug-2001 00:10:58 04-Aug-2001 00:10:59 04-Aug-2001 00:11:00 04-Aug-2001 00:11:01 Since, the month is in alphabetical form, I am finding difficulty to deal with char/string arrays in fortran. Can any one please help me how to come up with this issue...!! Thanks in advance. Praveen.
From: e p chandler on 6 Nov 2009 11:03 On Nov 6, 10:25 am, Fortran_follower <ezeepravee...(a)gmail.com> wrote: > Dear Friends, > > Recently I've started learning Fortran 95. > I want to convert the date format mentioned below to Modified Julian > Day format (MJD) > 04-Aug-2001 00:10:52 > 04-Aug-2001 00:10:53 > 04-Aug-2001 00:10:54 > 04-Aug-2001 00:10:55 > 04-Aug-2001 00:10:56 > 04-Aug-2001 00:10:57 > 04-Aug-2001 00:10:58 > 04-Aug-2001 00:10:59 > 04-Aug-2001 00:11:00 > 04-Aug-2001 00:11:01 > > Since, the month is in alphabetical form, I am finding difficulty to > deal with char/string arrays in fortran. > > Can any one please help me how to come up with this issue...!! > > Thanks in advance. > > Praveen. There are two simple ways that come to mind. 1. Create an array (look up table) which contains 12 entries. Each of these is a character variable of 3 characters. For each record in the input file, loop through the list of month names until one matches the month_field. Then your array index is the month number. For example 'Aug' would be in position 8. 2. Create a single character variable of 36 characters containing 'JanFeb.......'. Use INDEX to find the position of the month_field within the larger string. Adjust that number to return a number from 1 to 12. -- e
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