/opt/imunify360/venv/lib/python3.11/site-packages/humanize/__pycache__
Edit: /opt/imunify360/venv/lib/python3.11/site-packages/humanize/__pycache__/time.cpython-311.pyc (22579B)
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These are largely borrowed from Django's `contrib.humanize`.
) annotationsN)Enum)total_ordering)Any )_gettext) _ngettext)intcomma) naturaldelta naturaltime
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S | S )zReturn an "absolute" value for a timedelta, always representing a time distance.
Args:
delta (datetime.timedelta): Input timedelta.
Returns:
datetime.timedelta: Absolute timedelta.
r )daysr5 )r6 r4 s r! _abs_timedeltar: 1 s,
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}n# t t f$ r d| fcY S w xY w|t | fS )zTurn a value into a date and a timedelta which represents how long ago it was.
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z S ),aT Return a natural representation of a timedelta or number of seconds.
This is similar to `naturaltime`, but does not add tense to the result.
Args:
value (datetime.timedelta, int or float): A timedelta or a number of seconds.
months (bool): If `True`, then a number of months (based on 30.5 days) will be
used for fuzziness between years.
minimum_unit (str): The lowest unit that can be used.
Returns:
str (str or `value`): A natural representation of the amount of time
elapsed unless `value` is not datetime.timedelta or cannot be
converted to int. In that case, a `value` is returned unchanged.
Raises:
OverflowError: If `value` is too large to convert to datetime.timedelta.
Examples
Compare two timestamps in a custom local timezone::
import datetime as dt
from dateutil.tz import gettz
berlin = gettz("Europe/Berlin")
now = dt.datetime.now(tz=berlin)
later = now + dt.timedelta(minutes=30)
assert naturaldelta(later - now) == "30 minutes"
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527==??2 3 3E z\!!%,,, e r floatdivisorunitsuppress collections.abc.Iterable[Unit]tuple[float, float]c L ||k r| |z dfS ||v rd| fS t | | S )a Divide `value` by `divisor` returning the quotient and remainder.
If `unit` is `minimum_unit`, makes the quotient a float number and the remainder
will be zero. The rational is that if `unit` is the unit of the quotient, we cannot
represent the remainder because it would require a unit smaller than the
`minimum_unit`.
>>> from humanize.time import _quotient_and_remainder, Unit
>>> _quotient_and_remainder(36, 24, Unit.DAYS, Unit.DAYS, [])
(1.5, 0)
If unit is in `suppress`, the quotient will be zero and the remainder will be the
initial value. The idea is that if we cannot use `unit`, we are forced to use a
lower unit so we cannot do the division.
>>> _quotient_and_remainder(36, 24, Unit.DAYS, Unit.HOURS, [Unit.DAYS])
(0, 36)
In other case return quotient and remainder as `divmod` would do it.
>>> _quotient_and_remainder(36, 24, Unit.DAYS, Unit.HOURS, [])
(1, 12)
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d|| |z z fS | |fS )a Return a tuple with two values.
If the unit is in `suppress`, multiply `value1` by `ratio` and add it to `value2`
(carry to right). The idea is that if we cannot represent `value1` we need to
represent it in a lower unit.
>>> from humanize.time import _carry, Unit
>>> _carry(2, 6, 24, Unit.DAYS, Unit.SECONDS, [Unit.DAYS])
(0, 54)
If the unit is the minimum unit, `value2` is divided by `ratio` and added to
`value1` (carry to left). We assume that `value2` has a lower unit so we need to
carry it to `value1`.
>>> _carry(2, 6, 24, Unit.DAYS, Unit.DAYS, [])
(2.25, 0)
Otherwise, just return the same input:
>>> _carry(2, 6, 24, Unit.DAYS, Unit.SECONDS, [])
(2, 6)
r r/ )r r r r rn r s r! _carryr k sL < x&)) x&6E>))) 6>r# c ` | |v r)t D ]}|| k r||vr|c S d}t | | S )a Return a minimum unit suitable that is not suppressed.
If not suppressed, return the same unit:
>>> from humanize.time import _suitable_minimum_unit, Unit
>>> _suitable_minimum_unit(Unit.HOURS, []).name
'HOURS'
But if suppressed, find a unit greater than the original one that is not
suppressed:
>>> _suitable_minimum_unit(Unit.HOURS, [Unit.HOURS]).name
'DAYS'
>>> _suitable_minimum_unit(Unit.HOURS, [Unit.HOURS, Unit.DAYS]).name
'MONTHS'
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>>> from humanize.time import _suppress_lower_units, Unit
>>> [x.name for x in sorted(_suppress_lower_units(Unit.SECONDS, [Unit.DAYS]))]
['MICROSECONDS', 'MILLISECONDS', 'DAYS']
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S d |dd }"|d }#t/ d! |"|#fz S )"a% Return a precise representation of a timedelta.
```pycon
>>> import datetime as dt
>>> from humanize.time import precisedelta
>>> delta = dt.timedelta(seconds=3633, days=2, microseconds=123000)
>>> precisedelta(delta)
'2 days, 1 hour and 33.12 seconds'
```
A custom `format` can be specified to control how the fractional part
is represented:
```pycon
>>> precisedelta(delta, format="%0.4f")
'2 days, 1 hour and 33.1230 seconds'
```
Instead, the `minimum_unit` can be changed to have a better resolution;
the function will still readjust the unit to use the greatest of the
units that does not lose precision.
For example setting microseconds but still representing the date with milliseconds:
```pycon
>>> precisedelta(delta, minimum_unit="microseconds")
'2 days, 1 hour, 33 seconds and 123 milliseconds'
```
If desired, some units can be suppressed: you will not see them represented and the
time of the other units will be adjusted to keep representing the same timedelta:
```pycon
>>> precisedelta(delta, suppress=['days'])
'49 hours and 33.12 seconds'
```
Note that microseconds precision is lost if the seconds and all
the units below are suppressed:
```pycon
>>> delta = dt.timedelta(seconds=90, microseconds=100)
>>> precisedelta(delta, suppress=['seconds', 'milliseconds', 'microseconds'])
'1.50 minutes'
```
If the delta is too small to be represented with the minimum unit,
a value of zero will be returned:
```pycon
>>> delta = dt.timedelta(seconds=1)
>>> precisedelta(delta, minimum_unit="minutes")
'0.02 minutes'
>>> delta = dt.timedelta(seconds=0.1)
>>> precisedelta(delta, minimum_unit="minutes")
'0 minutes'
```
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