gh-93267: Fix wrong result Documented in Date Arithmetic datetime #93506
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I don't believe swapping the order fixes anything, as "removed" is commutative (at least in British English) -- e.g. your first cousin once removed could be your father's cousin or your son's cousin. We could update to "away from" as per the issue, although personally I feel that looses a little clarity of expression. Regardless, you should update line 1148 ("tetime2 is a duration of timedelta removed from datetime1, moving forward in") to the same verbiage that we decide on. cc: @ericvsmith A |
Doc/library/datetime.rst
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| @@ -589,8 +589,8 @@ Supported operations: | |||
| +-------------------------------+----------------------------------------------+ | |||
| | Operation | Result | | |||
| +===============================+==============================================+ | |||
| | ``date2 = date1 + timedelta`` | *date2* is ``timedelta.days`` days removed | | |||
| | | from *date1*. (1) | | |||
| | ``date2 = date1 + timedelta`` | *date1* is ``timedelta.days`` days removed | | |||
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The result should be what will be the value of date2 because date2 is what being assigned to. Suggestion:
Results in `timedelta.days` added to `date1`.
Even though it seems a bit repetitive to restate the operation on the left, it's already very intuitive and it would confirm to the reader that the + operator here does the obvious thing and not something weird.
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How about *date2* will be ``timedelta.days`` after *date1*. I think "after" indicates the direction correctly.
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That sounds good to me, using "after" here makes sense.
| | ``date2 = date1 + timedelta`` | *date2* is ``timedelta.days`` days removed | | ||
| | | from *date1*. (1) | | ||
| | ``date2 = date1 + timedelta`` | *date1* is ``timedelta.days`` days after | | ||
| | | *date2*. (1) | |
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Changing "removed from" to "after" is good, but swapping date1 with date2 in only one column is wrong. + here does an obvious thing, not a bizarre thing
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I think this looks good, once the noted date1/date2 bug is fixed.
| | ``date2 = date1 + timedelta`` | *date2* is ``timedelta.days`` days removed | | ||
| | | from *date1*. (1) | | ||
| | ``date2 = date1 + timedelta`` | *date1* is ``timedelta.days`` days after | | ||
| | | *date2*. (1) | |
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I think this looks good, once the noted date1/date2 bug is fixed.
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#93267