How can I find days and minutes difference between two datetime
objects?
For example If I have
b=datetime.datetime(2006, 8, 2, 8, 57, 28, 687000)
a=datetime.datetime(2006, 8, 1, 18, 19, 45, 765000)
Thank you for help
L.
4  1901 
Lad wrote:
How can I find days and minutes difference between two datetime
objects?
For example If I have
b=datetime.datetime(2006, 8, 2, 8, 57, 28, 687000)
a=datetime.datetime(2006, 8, 1, 18, 19, 45, 765000)
a - b
Lookup datetime.timedelta - all of this is neatly documented.
Diez
Sybren Stuvel wrote:
Lad enlightened us with:
How can I find days and minutes difference between two datetime
objects?
For example If I have
b=datetime.datetime(2006, 8, 2, 8, 57, 28, 687000)
a=datetime.datetime(2006, 8, 1, 18, 19, 45, 765000)
diff = b - a
Ok, I tried
>>diff=b-a
diff
datetime.timedelta(0, 52662, 922000)
>>diff.min
datetime.timedelta(-999999999)
which is not good for me.
So I tried to use toordinal like this
diff=b.toordinal()-a.toordinal()
but I get
diff=1
Why?
where do I make a mistake?
Thank you for help
Lad wrote:
Sybren Stuvel wrote:
Lad enlightened us with:
How can I find days and minutes difference between two datetime
objects?
For example If I have
b=datetime.datetime(2006, 8, 2, 8, 57, 28, 687000)
a=datetime.datetime(2006, 8, 1, 18, 19, 45, 765000)
diff = b - a
Ok, I tried
>diff=b-a
diff
datetime.timedelta(0, 52662, 922000)
>diff.min
datetime.timedelta(-999999999)
Reread the manual:
1. "min" is minIMUM, not minUTES
2. You need:
>>diff.days
0
>>diff.seconds
52662
>>diff.microseconds
922000
>>minutes = (diff.seconds + diff.microseconds / 1000000.0) / 60.0
minutes
877.71536666666668
>>>
which is not good for me.
So I tried to use toordinal like this
diff=b.toordinal()-a.toordinal()
but I get
diff=1
Why?
because toordinal() works only on the date part, ignoring the time
part.
HTH,
John
John Machin wrote:
Lad wrote:
Sybren Stuvel wrote:
Lad enlightened us with:
How can I find days and minutes difference between two datetime
objects?
For example If I have
b=datetime.datetime(2006, 8, 2, 8, 57, 28, 687000)
a=datetime.datetime(2006, 8, 1, 18, 19, 45, 765000)
>
diff = b - a
Ok, I tried
>>diff=b-a
>>diff
datetime.timedelta(0, 52662, 922000)
>>diff.min
datetime.timedelta(-999999999)
Reread the manual:
1. "min" is minIMUM, not minUTES
2. You need:
>diff.days
0
>diff.seconds
52662
>diff.microseconds
922000
>minutes = (diff.seconds + diff.microseconds / 1000000.0) / 60.0
minutes
877.71536666666668
>>
which is not good for me.
So I tried to use toordinal like this
diff=b.toordinal()-a.toordinal()
but I get
diff=1
Why?
because toordinal() works only on the date part, ignoring the time
part.
HTH,
John
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