Missing option to display years and months when years > 1 #108

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1. hugovk commented on Apr 13, 2023

   

   hugovk

   on Apr 13, 2023

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   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.
2. adonig commented on Apr 13, 2023

   

   adonig

   on Apr 13, 2023

   Author

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   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.
3. hugovk commented on Apr 13, 2023

   

   hugovk

   on Apr 13, 2023

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   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.
4. dharapvj commented on Mar 5, 2024

   

   dharapvj

   on Mar 5, 2024

   Last edited by hugovk

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   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.
5. dangillet commented on Nov 2, 2025

   

   dangillet

   on Nov 2, 2025

   Contributor

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   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'
6. hugovk commented on Sep 16, 2026

   

   hugovk

   on Sep 16, 2026

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   Let's close, use precisedelta(minimum_unit="months") instead.
7. hugovk

   closed this as not plannedon Sep 16, 2026