bpo-36138: Clarify docs about converting datetime.timedelta to scalars. (GH-12137)

Be explicit that timedelta division converts an overall duration to the interval
units given by the denominator.
This commit is contained in:
Yasser A 2019-03-15 23:56:58 -04:00 committed by Nick Coghlan
parent d2fdd1fedf
commit f40b4a0b62
2 changed files with 6 additions and 3 deletions

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@ -254,8 +254,9 @@ Supported operations:
| | rounded to the nearest multiple of |
| | timedelta.resolution using round-half-to-even.|
+--------------------------------+-----------------------------------------------+
| ``f = t2 / t3`` | Division (3) of *t2* by *t3*. Returns a |
| | :class:`float` object. |
| ``f = t2 / t3`` | Division (3) of overall duration *t2* by |
| | interval unit *t3*. Returns a :class:`float` |
| | object. |
+--------------------------------+-----------------------------------------------+
| ``t1 = t2 / f or t1 = t2 / i`` | Delta divided by a float or an int. The result|
| | is rounded to the nearest multiple of |
@ -351,7 +352,8 @@ Instance methods:
.. method:: timedelta.total_seconds()
Return the total number of seconds contained in the duration. Equivalent to
``td / timedelta(seconds=1)``.
``td / timedelta(seconds=1)``. For interval units other than seconds, use the
division form directly (e.g. ``td / timedelta(microseconds=1)``).
Note that for very large time intervals (greater than 270 years on
most platforms) this method will lose microsecond accuracy.

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@ -0,0 +1 @@
Improve documentation about converting datetime.timedelta to scalars.