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Bugs item #1289118, was opened at 2005-09-12 16:41
Message generated for change (Comment added) made by rhettinger
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Category: Python Library
Group: None
Status: Open
Resolution: None
Priority: 5
Submitted By: Daniel Stutzbach (agthorr)
>Assigned to: Tim Peters (tim_one)
Summary: timedelta multiply and divide by floating point

Initial Comment:
In python 2.4.1, the datetime.timedelta type allows for
the multiplication and division by integers. However,
it raises a TypeError for multiplication or division by
floating point numbers. This is a counterintutive
restriction and I can't think of any good reason for it.

For example:

>>> import datetime
>>> datetime.timedelta(minutes=5)/2
datetime.timedelta(0, 150)
>>> datetime.timedelta(minutes=5)*0.5
Traceback (most recent call last):
File "<stdin>", line 1, in ?
TypeError: unsupported operand type(s) for *:
'datetime.timedelta' and 'float'


----------------------------------------------------------------------

>Comment By: Raymond Hettinger (rhettinger)
Date: 2005-09-12 23:11

Message:
Logged In: YES
user_id=80475

Tim, do you prefer the current behavior?

----------------------------------------------------------------------

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Bugs item #1289118, was opened at 2005-09-12 17:41
Message generated for change (Comment added) made by mcherm
You can respond by visiting:
https://sourceforge.net/tracker/?func=detail&atid=105470&aid=1289118&group_id=5470

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Category: Python Library
Group: None
Status: Open
Resolution: None
Priority: 5
Submitted By: Daniel Stutzbach (agthorr)
Assigned to: Tim Peters (tim_one)
Summary: timedelta multiply and divide by floating point

Initial Comment:
In python 2.4.1, the datetime.timedelta type allows for
the multiplication and division by integers. However,
it raises a TypeError for multiplication or division by
floating point numbers. This is a counterintutive
restriction and I can't think of any good reason for it.

For example:

>>> import datetime
>>> datetime.timedelta(minutes=5)/2
datetime.timedelta(0, 150)
>>> datetime.timedelta(minutes=5)*0.5
Traceback (most recent call last):
File "<stdin>", line 1, in ?
TypeError: unsupported operand type(s) for *:
'datetime.timedelta' and 'float'


----------------------------------------------------------------------

>Comment By: Michael Chermside (mcherm)
Date: 2005-09-15 12:03

Message:
Logged In: YES
user_id=99874

I, too, would like to know what Tim thinks, but for what it's
worth (not much) I find Daniel's point fairly convincing...
multiplication by floats is an operation that makes sense, has
only one possible obvious meaning, and is not particularly
likely to cause errors (the way multiplying Decimal's with
floats does). So IF it's easy to implement, I say go for it.

----------------------------------------------------------------------

Comment By: Raymond Hettinger (rhettinger)
Date: 2005-09-13 00:11

Message:
Logged In: YES
user_id=80475

Tim, do you prefer the current behavior?

----------------------------------------------------------------------

You can respond by visiting:
https://sourceforge.net/tracker/?func=detail&atid=105470&aid=1289118&group_id=5470
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Bugs item #1289118, was opened at 2005-09-12 17:41
Message generated for change (Comment added) made by tim_one
You can respond by visiting:
https://sourceforge.net/tracker/?func=detail&atid=105470&aid=1289118&group_id=5470

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not just the latest update.
Category: Python Library
Group: None
Status: Open
Resolution: None
Priority: 5
Submitted By: Daniel Stutzbach (agthorr)
>Assigned to: Nobody/Anonymous (nobody)
Summary: timedelta multiply and divide by floating point

Initial Comment:
In python 2.4.1, the datetime.timedelta type allows for
the multiplication and division by integers. However,
it raises a TypeError for multiplication or division by
floating point numbers. This is a counterintutive
restriction and I can't think of any good reason for it.

For example:

>>> import datetime
>>> datetime.timedelta(minutes=5)/2
datetime.timedelta(0, 150)
>>> datetime.timedelta(minutes=5)*0.5
Traceback (most recent call last):
File "<stdin>", line 1, in ?
TypeError: unsupported operand type(s) for *:
'datetime.timedelta' and 'float'


----------------------------------------------------------------------

>Comment By: Tim Peters (tim_one)
Date: 2005-09-15 17:04

Message:
Logged In: YES
user_id=31435

timedelta arithmetic is 100% portable now, and wholly
explainable in terms of universally understood integer
arithmetic. Throw floats into it, and that's lost.

That said, I don't have a strong objection to complicating the
implementation if there _are_ strong use cases. The OP's
example isn't "a use case": it's not worth anything to let
someone multiply a timedelta by 0.5 instead of dividing by 2.
I don't have a use case to offer in its place (never felt a need
here).

If someone wants to work on it, note that a timedelta can
contain more than 53 bits of information, so, e.g., trying to
represent a timedelta as an IEEE double-precision number of
microseconds can lose information. This makes a high-
qualty "computed as if to infinite precision with one rounding
at the end" implementation of mixed datetime/float arithmetic
tricky to do right.

----------------------------------------------------------------------

Comment By: Michael Chermside (mcherm)
Date: 2005-09-15 12:03

Message:
Logged In: YES
user_id=99874

I, too, would like to know what Tim thinks, but for what it's
worth (not much) I find Daniel's point fairly convincing...
multiplication by floats is an operation that makes sense, has
only one possible obvious meaning, and is not particularly
likely to cause errors (the way multiplying Decimal's with
floats does). So IF it's easy to implement, I say go for it.

----------------------------------------------------------------------

Comment By: Raymond Hettinger (rhettinger)
Date: 2005-09-13 00:11

Message:
Logged In: YES
user_id=80475

Tim, do you prefer the current behavior?

----------------------------------------------------------------------

You can respond by visiting:
https://sourceforge.net/tracker/?func=detail&atid=105470&aid=1289118&group_id=5470
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Bugs item #1289118, was opened at 2005-09-12 14:41
Message generated for change (Comment added) made by agthorr
You can respond by visiting:
https://sourceforge.net/tracker/?func=detail&atid=105470&aid=1289118&group_id=5470

Please note that this message will contain a full copy of the comment thread,
including the initial issue submission, for this request,
not just the latest update.
Category: Python Library
Group: None
Status: Open
Resolution: None
Priority: 5
Submitted By: Daniel Stutzbach (agthorr)
Assigned to: Nobody/Anonymous (nobody)
Summary: timedelta multiply and divide by floating point

Initial Comment:
In python 2.4.1, the datetime.timedelta type allows for
the multiplication and division by integers. However,
it raises a TypeError for multiplication or division by
floating point numbers. This is a counterintutive
restriction and I can't think of any good reason for it.

For example:

>>> import datetime
>>> datetime.timedelta(minutes=5)/2
datetime.timedelta(0, 150)
>>> datetime.timedelta(minutes=5)*0.5
Traceback (most recent call last):
File "<stdin>", line 1, in ?
TypeError: unsupported operand type(s) for *:
'datetime.timedelta' and 'float'


----------------------------------------------------------------------

>Comment By: Daniel Stutzbach (agthorr)
Date: 2005-09-15 15:00

Message:
Logged In: YES
user_id=6324

Let me elaborate on the use-case where I originally ran into
this.

I'm conducting a series of observation experiments where I
measure the duration of an event. I then want to do various
statistical analysis such as computing the mean, median,
etc. Originally, I tried using standard functions such as
lmean from the stats.py package. However, these sorts of
functions divide by a float at the end, causing them to fail
on timedelta objects. Thus, I have to either write my own
special functions, or convert the timedelta objects to
integers first (then convert them back afterwards).

Basically, I want timedelta objects to look and act like
fixed-point arithmetic objects so that I can reuse other
functions on them that were originally developed to operate
on floats. More importantly, I'd rather not have to
maintain two different versions of the functions to deal
with different types.

For implementation, why not multiply the float times .day,
.seconds, and .microseconds separately, then propagate and
fraction parts into the next smallest group (e.g., 0.5 days
becomes 24*60*60*0.5 seconds).

I agree it'd be possible to lose information with the wrong
sequence of operations, but that's always the case when
using floating point numbers. In other words, that, too, is
what I would expect from the timedelta implementation.

----------------------------------------------------------------------

Comment By: Tim Peters (tim_one)
Date: 2005-09-15 14:04

Message:
Logged In: YES
user_id=31435

timedelta arithmetic is 100% portable now, and wholly
explainable in terms of universally understood integer
arithmetic. Throw floats into it, and that's lost.

That said, I don't have a strong objection to complicating the
implementation if there _are_ strong use cases. The OP's
example isn't "a use case": it's not worth anything to let
someone multiply a timedelta by 0.5 instead of dividing by 2.
I don't have a use case to offer in its place (never felt a need
here).

If someone wants to work on it, note that a timedelta can
contain more than 53 bits of information, so, e.g., trying to
represent a timedelta as an IEEE double-precision number of
microseconds can lose information. This makes a high-
qualty "computed as if to infinite precision with one rounding
at the end" implementation of mixed datetime/float arithmetic
tricky to do right.

----------------------------------------------------------------------

Comment By: Michael Chermside (mcherm)
Date: 2005-09-15 09:03

Message:
Logged In: YES
user_id=99874

I, too, would like to know what Tim thinks, but for what it's
worth (not much) I find Daniel's point fairly convincing...
multiplication by floats is an operation that makes sense, has
only one possible obvious meaning, and is not particularly
likely to cause errors (the way multiplying Decimal's with
floats does). So IF it's easy to implement, I say go for it.

----------------------------------------------------------------------

Comment By: Raymond Hettinger (rhettinger)
Date: 2005-09-12 21:11

Message:
Logged In: YES
user_id=80475

Tim, do you prefer the current behavior?

----------------------------------------------------------------------

You can respond by visiting:
https://sourceforge.net/tracker/?func=detail&atid=105470&aid=1289118&group_id=5470
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Bugs item #1289118, was opened at 2005-09-12 16:41
Message generated for change (Comment added) made by montanaro
You can respond by visiting:
https://sourceforge.net/tracker/?func=detail&atid=105470&aid=1289118&group_id=5470

Please note that this message will contain a full copy of the comment thread,
including the initial issue submission, for this request,
not just the latest update.
Category: Python Library
Group: None
Status: Open
Resolution: None
Priority: 5
Submitted By: Daniel Stutzbach (agthorr)
Assigned to: Nobody/Anonymous (nobody)
Summary: timedelta multiply and divide by floating point

Initial Comment:
In python 2.4.1, the datetime.timedelta type allows for
the multiplication and division by integers. However,
it raises a TypeError for multiplication or division by
floating point numbers. This is a counterintutive
restriction and I can't think of any good reason for it.

For example:

>>> import datetime
>>> datetime.timedelta(minutes=5)/2
datetime.timedelta(0, 150)
>>> datetime.timedelta(minutes=5)*0.5
Traceback (most recent call last):
File "<stdin>", line 1, in ?
TypeError: unsupported operand type(s) for *:
'datetime.timedelta' and 'float'


----------------------------------------------------------------------

>Comment By: Skip Montanaro (montanaro)
Date: 2005-09-16 20:48

Message:
Logged In: YES
user_id=44345

>> Thus, I have to either write my own special functions, or convert
>> the timedelta objects to integers first (then convert them back
>> afterwards).

How about adding tolong() that returns the number of microseconds
in the timedelta and fromlong() that accepts a long representing
microseconds and returns a timedelta object? That way the timedelta
object does a reasonably simple thing and the user is still responsible
for overflow the normal arithmetic stuff. You can do any sort of
arithmetic operations on the long (including converting to other
numeric types) with all the attendant caveats, then convert back to
a timedelta object at the end.


----------------------------------------------------------------------

Comment By: Daniel Stutzbach (agthorr)
Date: 2005-09-15 17:00

Message:
Logged In: YES
user_id=6324

Let me elaborate on the use-case where I originally ran into
this.

I'm conducting a series of observation experiments where I
measure the duration of an event. I then want to do various
statistical analysis such as computing the mean, median,
etc. Originally, I tried using standard functions such as
lmean from the stats.py package. However, these sorts of
functions divide by a float at the end, causing them to fail
on timedelta objects. Thus, I have to either write my own
special functions, or convert the timedelta objects to
integers first (then convert them back afterwards).

Basically, I want timedelta objects to look and act like
fixed-point arithmetic objects so that I can reuse other
functions on them that were originally developed to operate
on floats. More importantly, I'd rather not have to
maintain two different versions of the functions to deal
with different types.

For implementation, why not multiply the float times .day,
.seconds, and .microseconds separately, then propagate and
fraction parts into the next smallest group (e.g., 0.5 days
becomes 24*60*60*0.5 seconds).

I agree it'd be possible to lose information with the wrong
sequence of operations, but that's always the case when
using floating point numbers. In other words, that, too, is
what I would expect from the timedelta implementation.

----------------------------------------------------------------------

Comment By: Tim Peters (tim_one)
Date: 2005-09-15 16:04

Message:
Logged In: YES
user_id=31435

timedelta arithmetic is 100% portable now, and wholly
explainable in terms of universally understood integer
arithmetic. Throw floats into it, and that's lost.

That said, I don't have a strong objection to complicating the
implementation if there _are_ strong use cases. The OP's
example isn't "a use case": it's not worth anything to let
someone multiply a timedelta by 0.5 instead of dividing by 2.
I don't have a use case to offer in its place (never felt a need
here).

If someone wants to work on it, note that a timedelta can
contain more than 53 bits of information, so, e.g., trying to
represent a timedelta as an IEEE double-precision number of
microseconds can lose information. This makes a high-
qualty "computed as if to infinite precision with one rounding
at the end" implementation of mixed datetime/float arithmetic
tricky to do right.

----------------------------------------------------------------------

Comment By: Michael Chermside (mcherm)
Date: 2005-09-15 11:03

Message:
Logged In: YES
user_id=99874

I, too, would like to know what Tim thinks, but for what it's
worth (not much) I find Daniel's point fairly convincing...
multiplication by floats is an operation that makes sense, has
only one possible obvious meaning, and is not particularly
likely to cause errors (the way multiplying Decimal's with
floats does). So IF it's easy to implement, I say go for it.

----------------------------------------------------------------------

Comment By: Raymond Hettinger (rhettinger)
Date: 2005-09-12 23:11

Message:
Logged In: YES
user_id=80475

Tim, do you prefer the current behavior?

----------------------------------------------------------------------

You can respond by visiting:
https://sourceforge.net/tracker/?func=detail&atid=105470&aid=1289118&group_id=5470
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