MiPSrSSKJr SSKJr SSKJr SSKJr SSKJ r SSK J r S r \ (aSS KrSS KJr SS KJr /S Qr\"SS\55rSSjrSSjrS S S.S SjjrS!S"SjjrS#S$SjjrS%SjrS&S'SjjrS(SjrS)SjrS*SjrS+Sjr S,S-Sjjr!S.Sjr"g )/zYTime humanizing functions. These are largely borrowed from Django's `contrib.humanize`. ) annotations)Enum)total_ordering)_gettext) _ngettext)intcommaFN)Iterable)Any) naturaldate naturalday naturaldelta naturaltime precisedeltac>\rSrSrSrSrSrSrSrSr Sr S r S S jr S r g )UnitrrcrURURLaURUR:$[$)N) __class__valueNotImplemented)selfothers g/var/www/eduai.edurigo.com/storigo/production/storigo_env/lib/python3.13/site-packages/humanize/time.py__lt__ Unit.__lt__)s+ >>U__ ,:: + +N)rr returnr )__name__ __module__ __qualname____firstlineno__ MICROSECONDS MILLISECONDSSECONDSMINUTESHOURSDAYSMONTHSYEARSr!__static_attributes__r$r#r rrs/LLGG E D F Er#rc>SSKnURR5$)Nr)datetimenow)dts r _nowr7/s ;;?? r#cHURS:a[5nXU-- $U$)zReturn an "absolute" value for a timedelta, always representing a time distance. Args: delta (datetime.timedelta): Input timedelta. Returns: datetime.timedelta: Absolute timedelta. r)daysr7)deltar5s r _abs_timedeltar;5s( zzA~fEk"" Lr#)r5precisecPSSKnU(d [5n[XR5(aUnX- nOJ[XR5(aX- nUnO)U(aUO [ U5nUR"US9nX- nU[U54$![ [ 4a SU4s$f=f)zTurn a value into a date and a timedelta which represents how long ago it was. If that's not possible, return `(None, value)`. rNseconds)r4r7 isinstance timedeltaround ValueError TypeErrorr;)rr5r<r6dater:s r _date_and_deltarFDs f%%%  E<< ( ({ $E%,ELL/E;D & &&I& ;  s(BB%$B%cSSKn[UR5nU[R[R[R 4;aSUS3n[ U5eUn[XR5(aUnO'[U5 [U5nUR"US9nUn[U5nURS-n URS-n [U S- 5n U S:XGaU S:GaUR S:XaU[R :Xa4UR"S :a$[%S S UR"5UR"-$U[R:Xd.U[R :XaBS UR"s=::aS :a+O O(UR"S - n [%S S[U 55U -$['S5$UR S:Xa ['S5$UR S:a$[%SSUR 5UR -$SUR s=::aS:aMO OJ[UR S- 5n U S:Xa ['S5$U S:Xa ['S5$[%SSU 5U -$SUR ::aJ[UR S- 5nUS:Xa ['S5$US:Xa ['S5$[%SSU5U-$OU S:XapU S:Xa ['S5$U(d[%SSU 5U -$U S:Xa[%SSU 5U -$U S:Xa ['S5$U S :Xa ['S!5$[%S"S#U 5U -$U S:XaU S:XaU S:Xa ['S!5$U S:Xa[%S$S%U 5U -$U(a<U S:Xa ['S&5$U S :XaU S- n [%S'S(U 5U -$[%S)S*U 5U -$[%S$S%U 5U -$[%S'S(U 5R)S+S,5[+U 5-$![ [4a [U5s$f=f)-aReturn a natural representation of a timedelta or number of seconds. This is similar to `naturaltime`, but does not add tense to the result. The timedelta will be rounded to the nearest unit that makes sense. 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 (cannot be float due to 'inf' or 'nan'). 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:: ```pycon >>> 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" True ``` rNzMinimum unit 'z' not supportedr>m>@r%d microsecond%d microsecondsi@B%d millisecond%d millisecondsa momentza second< %d second %d secondsza minutezan hour %d minute %d minutesza day%d hour%d hours%d day%d daysza month za year%d month %d monthsz1 year, %d dayz1 year, %d daysz1 year, 1 month%d year%d yearsz1 year, %d monthz1 year, %d months%d%s)r4rupperr,r+r*rCr@rAintfloatrDstrabsr9rBr? microsecondsr_replacer )rmonths minimum_unitr6tmpmsgmin_unitr: use_monthsyearsr9 num_months millisecondsminuteshourss r rr_sT |!!# $C 4<<!2!2D4E4EFF|nO<oH%&&  J%LELL/EJ JE JJ# E :: Dtd{#J zdQh ==A 4,,,1C1Cd1J.0A5CUCUV(() 4,,,D---$%:L:L2Xy2X$11D8 .0A3|CTU"#Z= ==A Z= ==2 [, FV V  % %EMMB./G!|}$"}|#[,@7J J 5== %--$./Ez|#{z!Y E:UB B ! ! 19W: Xy$7$> > ?Xy$7$> > ?Y<   X; [*= JJ ! ?tqyX;  ?-/@$G$N N Q*++R  J>FF,.A:N  )+>& !!# 5z : &5zs2B66C* C*)C*cvSSKnUR"URURUR5n[XRR5- 5nURS:a [US5$[U5$![ a [ U5s$[[4a [ U5s$f=f)zKLike `naturalday`, but append a year for dates more than ~five months away.rNgc@z%b %d %Y) r4rErrrrrerrCr;rr9r )rr6r:s r r r Ws EKK; 577==?2 3E zz\!%,, e  5z : &5zs2BB8B87B8crX#:Xa[XPU- 5S4$X$;aSU4$[X5upgU[U54$)aDivide `value` by `divisor`, returning the quotient and remainder. If `unit` is `minimum_unit`, the quotient will be the rounding of `value / divisor` according to the `format` string and the remainder will be zero. The rationale 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, [], "%0.2f") (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], "%0.2f") (0, 36) In other cases, return the quotient and remainder as `divmod` would do it. >>> _quotient_and_remainder(36, 24, Unit.DAYS, Unit.HOURS, [], "%0.2f") (1, 12) r)_rounding_by_fmtdivmodrc)rdivisorunitrksuppressrqrs r _quotient_and_remainderrisJ@ 8!;; %x % !DA c!f9r#cdX;a*[HnX :dM X!;dMUs $ Sn[U5eU$)aReturn a minimum unit suitable that is not suppressed. If not suppressed, return the same unit: >>> from humanize.time import _suitable_minimum_unit, Unit >>> _suitable_minimum_unit(Unit.HOURS, []).name 'HOURS' But if suppressed, find a unit greater than the original one that is not suppressed: >>> _suitable_minimum_unit(Unit.HOURS, [Unit.HOURS]).name 'DAYS' >>> _suitable_minimum_unit(Unit.HOURS, [Unit.HOURS, Unit.DAYS]).name 'MONTHS' z@Minimum unit is suppressed and no suitable replacement was found)rrC)rnrrrms r _suitable_minimum_unitrs:$D4#7 Qo Or#ch[U5n[HnX :Xa U$URU5 M U$)aExtend suppressed units (if any) with all units lower than the minimum unit. >>> from humanize.time import _suppress_lower_units, Unit >>> [x.name for x in sorted(_suppress_lower_units(Unit.SECONDS, [Unit.DAYS]))] ['MICROSECONDS', 'MILLISECONDS', 'DAYS'] )setradd)rnrrs r _suppress_lower_unitsrs<8}H    O  T Or#c ^[USS9upEUc [U5$UVs1sHn[UR5iM nn[UR5n[ X5nA[ X5nUR n URn URn [[5uppnnnn[U SUXU5unn [U SUXU5unn U S-S-U -n [U SUXU5up[U SUXU5unn [U S XXs5unn U S -U -n [U S XXs5up[U S XXs5unn US :aX;a US -nU S - n U S :aX;a U S -n US - nUS :aUU;a US -nUS - nUS:aUU;a US-nU S - n U S :aUU;a U S -n US - nUS:aUU;a US-nUS - nSSU4SSU4SSU 4SSU4SSU4SSU 4SSU4SSU 4/n/n[[[5U5HunnUunnnUS:d U(dUU:XaS Us=:aS :aO OS O [U5n [UUU 5n!SSKn"UU:Xa+U"R!U5SS:aU!R#S!U5n!OZUU:XaTU"R!U5SS:Xa [U5nU!R#S!S"5n!UR%U!['U5-5 MUR%U!U-5 UU:XdM O [)U5S :XaUS$S#R+USS$5n#US$n$[-S%5U#U$4-$s snf)&aReturn a precise representation of a timedelta or number of seconds. ```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' ``` T)r<NrHrIrVrSiQrPg.ArJrr[r^r_r\r]rYrZrWrXrTrUrQrRrMrNrKrLrrr`raz, rz %s and %s)rFrerrbrrr9r?rglistrzipreversedrcrmathmodfriappendr lenjoinrh)%rrkrrrEr:s suppress_setrnr9secsusecsr*r+r,r-r.r/r0r1rprjrtrsmsecsfmtstextsrfmt singular_txt plural_txt fmt_value _fmt_valuefmt_txtrheadtails% r rrsP"%6KD |5z-56XDOXL6L&&()H%h=H)@L ::D ==D   EOS PLL%vu * c5(&KE4+ dFHFLFD "9t d "D( ixvJD* dE86KE4, b'\MGT 3J E) sG|KD+ t\\LE5 ~'5    rzg1  1 "}l22    {t</     rzfL0  !  |\1"    J& [&) 9d# J& lG, lD) ,e4 ,e4 DE$. c.1+ j) q=48+;)/a/S^J j*EG xDIIi$8$;a$?!//$799Y'*a/ #II!//$5 Wx '::; LL9, - 8  '/* 5zQQx 99U3BZ D 9D [>T4L ((y7s!L*c^X-n[U5nU$![a [U5nU$f=f)aTRound a number according to the string format provided. The string format is the old printf-style string formatting. If we are using a format which truncates the value, such as "%d" or "%i", the returned value will be of type `int`. If we are using a format which rounds the value, such as "%.2f" or even "%.0f", we will return a float. )rcrCrd)rrresults r rrs?^FF  L f  Ls ,,)r%z dt.datetime)r: dt.timedeltar%r)rr r5dt.datetime | Noner<boolr%ztuple[Any, Any])Tr?)rzdt.timedelta | floatrjrrkrer%re)FTr?N) rz"dt.datetime | dt.timedelta | floatrwrrjrrkrerxrr%re)rz)dt.datetime | dt.timedelta | float | Noner%r )z%b %d)rdt.date | dt.datetimerrer%re)rrr%re)rrdrrdrrrkrrIterable[Unit]rrer%ztuple[float, float])rnrrrr%r)rnrrrr%z set[Unit])r?r$z%0.2f) rzdt.timedelta | float | Nonerkrerz Iterable[str]rrer%re)rrerrdr%z float | int)#__doc__ __future__renumr functoolsri18nrrhrnumberr TYPE_CHECKINGr4r6collections.abcr typingr __all__rr7r;rFrrrvr r rrrrrr$r#r rs #$ (  4     .25' '*'<@'':!WY WY WYWY WYx!# 0 -0 0 0 0  0  0f 4"@$* * * * *  *  **Z: $"  H) &H)H)H)  H)  H)Vr#