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datetime in microseconds

Hi I have a time in microseconds, for example 0x8C905CBA7F84A F4. I
want this to a normal view in hh:mm:ss DD:MM:YYYY. I tried with
datetime, but it only takes a max of 1000000 microseconds is there
another solution?

Aug 20 '07 #1
5 5447
On Aug 20, 6:52 am, mroelo...@gmail .com wrote:
Hi I have a time in microseconds, for example 0x8C905CBA7F84A F4. I
want this to a normal view in hh:mm:ss DD:MM:YYYY. I tried with
datetime, but it only takes a max of 1000000 microseconds is there
another solution?
Just truncate the value so that it's less than 1000000. A slight
difference of a couple seconds shouldn't matter for the output format

Mike

Aug 20 '07 #2
On Aug 20, 9:52 pm, mroelo...@gmail .com wrote:
Hi I have a time in microseconds, for example 0x8C905CBA7F84A F4. I
want this to a normal view in hh:mm:ss DD:MM:YYYY. I tried with
datetime, but it only takes a max of 1000000 microseconds is there
another solution?
Your question can be interpreted in two possible ways:

1. You have an interval or duration (independent of a calendar point)
and you want to express it in years, months, days, hours, etc. This is
not possible, due to the variable number of days in a month. The best
that you can do is express it as days, hours, etc.
>>microsecs = 0x8C905CBA7F84A F4
secs = microsecs // 1000000 # or round to nearest if you prefer
mins, secs = divmod(secs, 60)
hrs, mins = divmod(mins, 60)
days, hrs = divmod(hrs, 24)
days, hrs, mins, secs
(7326893L, 11L, 1L, 16L)
>>>
2. You want to know the (Gregorian) calendar point that is
0x8C905CBA7F84A F4 microseconds after some epoch. In this case you need
to specify what the epoch is. Then you can try something like:
>>datetime.date time.fromordina l(1) + datetime.timede lta(microsecond s=microsecs
)
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
OverflowError: date value out of range
>># Whoops!
years_appro x = days / 365.25
years_appro x
20059.939767282 682
>>>
Hmmm, one of us seems to be missing something ...

Aug 20 '07 #3
On Aug 20, 3:15 pm, John Machin <sjmac...@lexic on.netwrote:
On Aug 20, 9:52 pm, mroelo...@gmail .com wrote:
Hi I have a time in microseconds, for example 0x8C905CBA7F84A F4. I
want this to a normal view in hh:mm:ss DD:MM:YYYY. I tried with
datetime, but it only takes a max of 1000000 microseconds is there
another solution?

Your question can be interpreted in two possible ways:

1. You have an interval or duration (independent of a calendar point)
and you want to express it in years, months, days, hours, etc. This is
not possible, due to the variable number of days in a month. The best
that you can do is express it as days, hours, etc.
>microsecs = 0x8C905CBA7F84A F4
secs = microsecs // 1000000 # or round to nearest if you prefer
mins, secs = divmod(secs, 60)
hrs, mins = divmod(mins, 60)
days, hrs = divmod(hrs, 24)
days, hrs, mins, secs

(7326893L, 11L, 1L, 16L)

2. You want to know the (Gregorian) calendar point that is
0x8C905CBA7F84A F4 microseconds after some epoch. In this case you need
to specify what the epoch is. Then you can try something like:
>datetime.datet ime.fromordinal (1) + datetime.timede lta(microsecond s=microsecs

)
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
OverflowError: date value out of range
># Whoops!
years_approx = days / 365.25
years_approx
20059.939767282 682

Hmmm, one of us seems to be missing something ...
Sorry, sorry, sorry it was the wrong value, it should be
0xE0E6FAC3FF3AB 2.

Aug 20 '07 #4
On Aug 20, 4:17 pm, mroelo...@gmail .com wrote:
On Aug 20, 3:15 pm, John Machin <sjmac...@lexic on.netwrote:
On Aug 20, 9:52 pm, mroelo...@gmail .com wrote:
Hi I have a time in microseconds, for example 0x8C905CBA7F84A F4. I
want this to a normal view in hh:mm:ss DD:MM:YYYY. I tried with
datetime, but it only takes a max of 1000000 microseconds is there
another solution?
Your question can be interpreted in two possible ways:
1. You have an interval or duration (independent of a calendar point)
and you want to express it in years, months, days, hours, etc. This is
not possible, due to the variable number of days in a month. The best
that you can do is express it as days, hours, etc.
>>microsecs = 0x8C905CBA7F84A F4
>>secs = microsecs // 1000000 # or round to nearest if you prefer
>>mins, secs = divmod(secs, 60)
>>hrs, mins = divmod(mins, 60)
>>days, hrs = divmod(hrs, 24)
>>days, hrs, mins, secs
(7326893L, 11L, 1L, 16L)
2. You want to know the (Gregorian) calendar point that is
0x8C905CBA7F84A F4 microseconds after some epoch. In this case you need
to specify what the epoch is. Then you can try something like:
>>datetime.date time.fromordina l(1) + datetime.timede lta(microsecond s=microsecs
)
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
OverflowError: date value out of range
>># Whoops!
>>years_appro x = days / 365.25
>>years_appro x
20059.939767282 682
Hmmm, one of us seems to be missing something ...

Sorry, sorry, sorry it was the wrong value, it should be
0xE0E6FAC3FF3AB 2.
The solution I made, with thanks to John. Maybe someone a better one??
def DecodeDateTime( self,dateTime):
dateTime = self.Rotate(dat eTime)
microsecs = int(hexlify(dat eTime),16)
microsecs -= 31536000000000 # -1 Year
microsecs -= 1123200000000 # -13 Days (magic?)
secs = microsecs // 1000000
mins, secs = divmod(secs, 60)
hrs, mins = divmod(mins, 60)
days, hrs = divmod(hrs, 24)
timed = datetime.dateti me.fromordinal( 1) +
datetime.timede lta(days)
return "%02d-%02d-%02d %02d:%02d:%02d" %(timed.day,
timed.month, timed.year, hrs, mins, secs)

Aug 20 '07 #5
Robert Dailey wrote:
A small off topic question. Why use divmod() instead of the modulus
operator?
Because he needed both the quotient and the remainder. % only gives you the
remainder.

--
Robert Kern

"I have come to believe that the whole world is an enigma, a harmless enigma
an underlying truth."
-- Umberto Eco

Aug 20 '07 #6

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