| Re: incremental time data file..... [message #26326] |
Thu, 23 August 2001 08:08  |
Manish
Messages: 20
Registered: April 2001
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Junior Member |
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Thanx William, that's great!
Manish
--
"William Thompson" <thompson@orpheus.nascom.nasa.gov> wrote in message
news:9m34cp$4sk$1@skates.gsfc.nasa.gov...
> "Manish" <mrmanish@bigfoot.com> writes:
>
>> Pavel, thanks for the help, just one more thing!!
>
>> I've altered it to zero fill the hour and minute values, but how do I
>> introduce a zero to fill the values between 0 and 9 seconds in the same
way?
>> Essentially, how do you zero fill a floating point value??
>
>> I trust this is an easy thing to fix, but I'd appreciate any help, being
>> only a mere novice....!
>
>> Cheers,
>> Manish
>
> Probably the easiest way is treat everything as integers.
>
> ss = fix(my_time-hh*3600L-mm*60L)
> fsec = round(1E5*(my_time-hh*3600L-mm*60L-ss)) ;Fractional second
> out = transpose([[hh],[mm],[ss],[fsec]])
> print, out[*, 82300:82310], format='(i2.2,":",i2.2,":",i2.2,".",i5.5)'
>
> Also, that way, everything comes out exactly the same string length, e.g.
>
> 23:58:57.64063
> 23:58:58.68750
> 23:58:59.73438
> 23:59:00.78125
> 23:59:01.83594
> 23:59:02.88281
> 23:59:03.92969
> 23:59:04.97656
> 23:59:06.03125
> 23:59:07.07813
> 23:59:08.12500
>
> William Thompson
>
>
>> "Pavel A. Romashkin" <pavel.romashkin@noaa.gov> wrote in message
>> news:3B83E468.4E0D9DE4@noaa.gov...
>>> How about
>>>
>>> my_time = (findgen(24.*60.*60./1.04906)*1.04906)
>>> hh = fix(my_time / 3600L)
>>> mm = fix((my_time - hh*3600L)/60L)
>>> ss = my_time-hh*3600L-mm*60L
>>> out = transpose([[hh],[mm],[ss]])
>>> print, out[*, 82300:82310], format='(i2,":", i2,":", F8.5)'
>>>
>>> If you need exact zero-padded field width, play with string conversion
>>> and formatted output.
>>> Cheers,
>>> Pavel
>>>
>>> Manish wrote:
>>>>
>>>> Hi,
>>>> I've only started using IDL recently, and was wondering if anyone can
>> help
>>>> me out.
>>>> I need to produce a data file which steps through increments of time
>>>> (1.04906 s) for an entire day, i.e. to produce a file which looks
like:
>>>>
>>>> 00:00:01.04906
>>>> 00:00:02.0992
>>>> ...
>>>> ...
>>>> 23:59:59....(whatever the last integer would be!)l
>
>
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| Re: incremental time data file..... [message #26328 is a reply to message #26326] |
Thu, 23 August 2001 07:36  |
thompson
Messages: 584
Registered: August 1991
|
Senior Member |
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"Manish" <mrmanish@bigfoot.com> writes:
> Pavel, thanks for the help, just one more thing!!
> I've altered it to zero fill the hour and minute values, but how do I
> introduce a zero to fill the values between 0 and 9 seconds in the same way?
> Essentially, how do you zero fill a floating point value??
> I trust this is an easy thing to fix, but I'd appreciate any help, being
> only a mere novice....!
> Cheers,
> Manish
Probably the easiest way is treat everything as integers.
ss = fix(my_time-hh*3600L-mm*60L)
fsec = round(1E5*(my_time-hh*3600L-mm*60L-ss)) ;Fractional second
out = transpose([[hh],[mm],[ss],[fsec]])
print, out[*, 82300:82310], format='(i2.2,":",i2.2,":",i2.2,".",i5.5)'
Also, that way, everything comes out exactly the same string length, e.g.
23:58:57.64063
23:58:58.68750
23:58:59.73438
23:59:00.78125
23:59:01.83594
23:59:02.88281
23:59:03.92969
23:59:04.97656
23:59:06.03125
23:59:07.07813
23:59:08.12500
William Thompson
> "Pavel A. Romashkin" <pavel.romashkin@noaa.gov> wrote in message
> news:3B83E468.4E0D9DE4@noaa.gov...
>> How about
>>
>> my_time = (findgen(24.*60.*60./1.04906)*1.04906)
>> hh = fix(my_time / 3600L)
>> mm = fix((my_time - hh*3600L)/60L)
>> ss = my_time-hh*3600L-mm*60L
>> out = transpose([[hh],[mm],[ss]])
>> print, out[*, 82300:82310], format='(i2,":", i2,":", F8.5)'
>>
>> If you need exact zero-padded field width, play with string conversion
>> and formatted output.
>> Cheers,
>> Pavel
>>
>> Manish wrote:
>>>
>>> Hi,
>>> I've only started using IDL recently, and was wondering if anyone can
> help
>>> me out.
>>> I need to produce a data file which steps through increments of time
>>> (1.04906 s) for an entire day, i.e. to produce a file which looks like:
>>>
>>> 00:00:01.04906
>>> 00:00:02.0992
>>> ...
>>> ...
>>> 23:59:59....(whatever the last integer would be!)l
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| Re: incremental time data file..... [message #26330 is a reply to message #26328] |
Thu, 23 August 2001 05:58 |
Manish
Messages: 20
Registered: April 2001
|
Junior Member |
|
|
Pavel, thanks for the help, just one more thing!!
I've altered it to zero fill the hour and minute values, but how do I
introduce a zero to fill the values between 0 and 9 seconds in the same way?
Essentially, how do you zero fill a floating point value??
I trust this is an easy thing to fix, but I'd appreciate any help, being
only a mere novice....!
Cheers,
Manish
--
"Pavel A. Romashkin" <pavel.romashkin@noaa.gov> wrote in message
news:3B83E468.4E0D9DE4@noaa.gov...
> How about
>
> my_time = (findgen(24.*60.*60./1.04906)*1.04906)
> hh = fix(my_time / 3600L)
> mm = fix((my_time - hh*3600L)/60L)
> ss = my_time-hh*3600L-mm*60L
> out = transpose([[hh],[mm],[ss]])
> print, out[*, 82300:82310], format='(i2,":", i2,":", F8.5)'
>
> If you need exact zero-padded field width, play with string conversion
> and formatted output.
> Cheers,
> Pavel
>
> Manish wrote:
>>
>> Hi,
>> I've only started using IDL recently, and was wondering if anyone can
help
>> me out.
>> I need to produce a data file which steps through increments of time
>> (1.04906 s) for an entire day, i.e. to produce a file which looks like:
>>
>> 00:00:01.04906
>> 00:00:02.0992
>> ...
>> ...
>> 23:59:59....(whatever the last integer would be!)l
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| Re: incremental time data file..... [message #26345 is a reply to message #26331] |
Wed, 22 August 2001 09:57 |
Pavel A. Romashkin
Messages: 531
Registered: November 2000
|
Senior Member |
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|
How about
my_time = (findgen(24.*60.*60./1.04906)*1.04906)
hh = fix(my_time / 3600L)
mm = fix((my_time - hh*3600L)/60L)
ss = my_time-hh*3600L-mm*60L
out = transpose([[hh],[mm],[ss]])
print, out[*, 82300:82310], format='(i2,":", i2,":", F8.5)'
If you need exact zero-padded field width, play with string conversion
and formatted output.
Cheers,
Pavel
Manish wrote:
>
> Hi,
> I've only started using IDL recently, and was wondering if anyone can help
> me out.
> I need to produce a data file which steps through increments of time
> (1.04906 s) for an entire day, i.e. to produce a file which looks like:
>
> 00:00:01.04906
> 00:00:02.0992
> ...
> ...
> 23:59:59....(whatever the last integer would be!)l
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| Re: incremental time data file..... [message #26347 is a reply to message #26346] |
Wed, 22 August 2001 09:13 |
ugo_digirolamo[1]
Messages: 2
Registered: August 2001
|
Junior Member |
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Mah... to me this looks like a simple application of the "mod"
operator.
You can use a double time in seconds, and extract from it seconds,
minutes and hours taking seconds mod 1, (seconds/60) mod 60 and
seconds/3600.
Cheers
Ugo
"Manish" <mrmanish@bigfoot.com> wrote in message news:<9m00mj$b8e$1@yarrow.open.ac.uk>...
> Hi,
> I've only started using IDL recently, and was wondering if anyone can help
> me out.
> I need to produce a data file which steps through increments of time
> (1.04906 s) for an entire day, i.e. to produce a file which looks like:
>
> 00:00:01.04906
> 00:00:02.0992
> ...
> ...
> 23:59:59....(whatever the last integer would be!)
>
>
> Can anyone help out, but more importantly, does anyone understand what the
> hell I'm going on about!!
>
> Cheers for your time,
>
> Manish.
>
> --
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| Re: incremental time data file..... [message #26471 is a reply to message #26328] |
Thu, 23 August 2001 08:35 |
Craig Markwardt
Messages: 1869
Registered: November 1996
|
Senior Member |
|
|
thompson@orpheus.nascom.nasa.gov (William Thompson) writes:
> "Manish" <mrmanish@bigfoot.com> writes:
>
>> Pavel, thanks for the help, just one more thing!!
>
>> I've altered it to zero fill the hour and minute values, but how do I
>> introduce a zero to fill the values between 0 and 9 seconds in the same way?
>> Essentially, how do you zero fill a floating point value??
>
>> I trust this is an easy thing to fix, but I'd appreciate any help, being
>> only a mere novice....!
>
>> Cheers,
>> Manish
>
> Probably the easiest way is treat everything as integers.
>
> ss = fix(my_time-hh*3600L-mm*60L)
> fsec = round(1E5*(my_time-hh*3600L-mm*60L-ss)) ;Fractional second
> out = transpose([[hh],[mm],[ss],[fsec]])
> print, out[*, 82300:82310], format='(i2.2,":",i2.2,":",i2.2,".",i5.5)'
>
> Also, that way, everything comes out exactly the same string length, e.g.
...
I agree. I also found it was necessary to convert the number of
seconds to integers. Otherwise I was always plagued by bizarre
rounding errors which popped up at awkward moments, and were otherwise
impossible to resolve completely. For example, 04:02:60 or 04:02:-1
Craig
--
------------------------------------------------------------ --------------
Craig B. Markwardt, Ph.D. EMAIL: craigmnet@cow.physics.wisc.edu
Astrophysics, IDL, Finance, Derivatives | Remove "net" for better response
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