Missing option to display years and months when years > 1

[adonig]

What did you do?

In [18]: humanize.naturaldelta(timedelta(days=4.6*365), months=True)
Out[18]: '4 years'

In [19]: humanize.naturaldelta(timedelta(days=4.6*365), months=False)
Out[19]: '4 years'

What did you expect to happen?

In [18]: humanize.naturaldelta(timedelta(days=4.6*365), months=True)
Out[18]: '4 years, 7 months'

In [19]: humanize.naturaldelta(timedelta(days=4.6*365), months=False)
Out[19]: '4 years'

What actually happened?

see above

What versions are you using?

• OS: x86_64 GNU/Linux
• Python: 3.11.3
• Humanize: 4.6.0

How to fix?

 # Excerpt from humanize/time.py 

 # ... more code here
 elif years == 1:
 if not num_months and not days:
 return _("a year")

 if not num_months:
 return _ngettext("1 year, %d day", "1 year, %d days", days) % days

 if use_months:
 if num_months == 1:
 return _("1 year, 1 month")

 return (
 _ngettext("1 year, %d month", "1 year, %d months", num_months)
 % num_months
 )

 return _ngettext("1 year, %d day", "1 year, %d days", days) % days

 # TODO: Here we should check whether use_months is true and format accordingly.
 return _ngettext("%d year", "%d years", years).replace("%d", "%s") % intcomma(years)

[hugovk]

What it's doing is rounding to years, when years >= 2:

>>> import humanize
>>> from datetime import timedelta
>>> for year in range(5):
... humanize.naturaldelta(timedelta(days=(year+0.6) * 365), months=True)
...
'7 months'
'1 year, 7 months'
'2 years'
'3 years'
'4 years'

And similarly with months=True, except the granularity is in days not months:

>>> for year in range(5):
... humanize.naturaldelta(timedelta(days=(year+0.6) * 365), months=False)
...
'219 days'
'1 year, 219 days'
'2 years'
'3 years'
'4 years'

And likewise for naturaltime:

>>> for year in range(5):
... humanize.naturaltime(timedelta(days=(year+0.6) * 365), months=True)
...
'7 months ago'
'1 year, 7 months ago'
'2 years ago'
'3 years ago'
'4 years ago'
>>> for year in range(5):
... humanize.naturaltime(timedelta(days=(year+0.6) * 365), months=False)
...
'219 days ago'
'1 year, 219 days ago'
'2 years ago'
'3 years ago'
'4 years ago'

So months=True doesn't control whether to show a more granular time, but what units to use.

[adonig]

Hi @hugovk! Thank you for the explanation. My use case is that we need to display the amortization of an investment. Currently it says "4.6 years" and I thought I could use humanize to turn it into something like "4 years, 7 months" because that's more human friendly. There's a way to do this using a hack:

In [12]: def naturaldelta(delta):
 ...: if delta < timedelta(days=2*365):
 ...: return humanize.naturaldelta(delta)
 ...: return f"{humanize.naturaldelta(delta)},{humanize.naturaldelta(timedelta(days=delta.days%365))}"
 ...: 

In [13]: naturaldelta(timedelta(days=4.6*365))
Out[13]: '4 years,7 months'

In [14]: naturaldelta(timedelta(days=1.6*365))
Out[14]: '1 year, 7 months'

In [15]: naturaldelta(timedelta(days=0.6*365))
Out[15]: '7 months'

The problem with this is that it might break translations.

[hugovk]

I guess one idea is to add a new option to make the cutoff configurable, and (I expect) use it to replace the 1 in elif years == 1.

That would also require a lot more handling in that block to deal with singular and plural years, especially for the translations.

[dharapvj]

Would like to chime in here. I would like to see naturaldelta behavior to be..

1. allow 1 or 2 units in output.. i.e.

>>> delta = dt.timedelta(seconds=36, minutes=3) 
>>> humanize.naturaldelta(delta)

should allow for either 3 minutes OR 3m36s

2. May be >= 10m, we can drop the seconds part without option.
3. Similarly for hours, output should be allowed to get configured between 3 hours vs 3h43m and then again, after may be >=10h, we drop the minutes part altogether.

This is a default output format in kubernetes client to show the age. It always shows 2 units (unless we are in seconds)

❯ kubectl get deployments -A
NAMESPACE NAME READY UP-TO-DATE AVAILABLE AGE
kube-system coredns 2/2 2 2 6d14h
local-path-storage local-path-provisioner 1/1 1 1 6d14h

❯ kubectl get pod
NAME READY STATUS RESTARTS AGE
temp 1/1 Running 0 18s
❯ kubectl get pod
NAME READY STATUS RESTARTS AGE
temp 1/1 Running 0 105s
❯ kubectl get pod
NAME READY STATUS RESTARTS AGE
temp 1/1 Running 0 2m4s

Logic for above calculation is here.

[dangillet]

My use case is that we need to display the amortization of an investment.

You already know that you would like to have the precision down to the months. Maybe you could use precisedelta for your use case?

>>> humanize.precisedelta(timedelta(days=4.6*365), minimum_unit="months")
'4 years and 7.18 months'

>>> humanize.precisedelta(timedelta(days=4.6*365), minimum_unit="months", format="%d")
'4 years and 7 months'

>>> humanize.precisedelta(timedelta(days=4*365), minimum_unit="months", format="%d")
'4 years'

Using minimum_units="months" and format="%d" will truncate any decimals for the months.

ℹ️ Using format="%.0f" would instead round to the nearest month.

>>> humanize.precisedelta(timedelta(days=4.65*365), minimum_unit="months")
'4 years and 7.77 months'

>>> humanize.precisedelta(timedelta(days=4.65*365), minimum_unit="months", format="%d")
'4 years and 7 months'

>>> humanize.precisedelta(timedelta(days=4.65*365), minimum_unit="months", format="%.0f")
'4 years and 8 months'