Fix plural form for intword and improve performance

[dangillet]

Fixes #175

Changes proposed in this pull request:

• Fix the plural form produced by intword by correctly taking into account how rounding is done
• Improved code readability and performance
• Fix one small error in the French translation

[codecov]

Codecov Report

✅ All modified and coverable lines are covered by tests.
✅ Project coverage is 99.51%. Comparing base (50a413a) to head (00b0d3a).
⚠️ Report is 1 commits behind head on main.

Additional details and impacted files

@@           Coverage Diff           @@
##             main     #273   +/-   ##
=======================================
  Coverage   99.51%   99.51%           
=======================================
  Files          11       11           
  Lines         818      820    +2     
=======================================
+ Hits          814      816    +2     
  Misses          4        4           

Flag	Coverage Δ
macos-latest	97.80% <100.00%> (+<0.01%)	⬆️
ubuntu-latest	97.80% <100.00%> (+<0.01%)	⬆️
windows-latest	95.85% <100.00%> (+0.01%)	⬆️

Flags with carried forward coverage won't be shown. Click here to find out more.

☔ View full report in Codecov by Sentry.
📢 Have feedback on the report? Share it here.

🚀 New features to boost your workflow:

• ❄️ Test Analytics: Detect flaky tests, report on failures, and find test suite problems.

[hugovk]

Thanks!

• Improved code readability and performance

Yes, much nicer using bisect 👍

main:

❯ python -m timeit -n 10000 -s "from humanize import intword" 'for a in (1_000_000, 3_500_000, 1_000_000_000, 1_200_000_000, 1_000_000_000_000, 6_700_000_000_000, "1_000", "12_400", "12_490", "1_000_000", "-1_000_000", "1_200_000", "1_290_000", "999_999_999", "1_000_000_000", "-1_000_000_000", "2_000_000_000", "999_999_999_999", "1_000_000_000_000", "6_000_000_000_000", "-6_000_000_000_000", "999_999_999_999_999", "1_000_000_000_000_000", "1_300_000_000_000_000", "-1_300_000_000_000_000", "3_500_000_000_000_000_000_000", "8_100_000_000_000_000_000_000_000_000_000_000", "-8_100_000_000_000_000_000_000_000_000_000_000", "-8.1 quintilliards", 1_000_000_000_000_000_000_000_000_000_000_000_000, 1_100_000_000_000_000_000_000_000_000_000_000_000, 2_100_000_000_000_000_000_000_000_000_000_000_000): intword(a)'
Found existing alias for "python". You should use: "p"
10000 loops, best of 5: 42 usec per loop

PR@

❯ python -m timeit -n 10000 -s "from humanize import intword" 'for a in (1_000_000, 3_500_000, 1_000_000_000, 1_200_000_000, 1_000_000_000_000, 6_700_000_000_000, "1_000", "12_400", "12_490", "1_000_000", "-1_000_000", "1_200_000", "1_290_000", "999_999_999", "1_000_000_000", "-1_000_000_000", "2_000_000_000", "999_999_999_999", "1_000_000_000_000", "6_000_000_000_000", "-6_000_000_000_000", "999_999_999_999_999", "1_000_000_000_000_000", "1_300_000_000_000_000", "-1_300_000_000_000_000", "3_500_000_000_000_000_000_000", "8_100_000_000_000_000_000_000_000_000_000_000", "-8_100_000_000_000_000_000_000_000_000_000_000", "-8.1 quintilliards", 1_000_000_000_000_000_000_000_000_000_000_000_000, 1_100_000_000_000_000_000_000_000_000_000_000_000, 2_100_000_000_000_000_000_000_000_000_000_000_000): intword(a)'
Found existing alias for "python". You should use: "p"
10000 loops, best of 5: 35.3 usec per loop

---

This also 

[hugovk]

This isn't used:

Suggested change

power = powers[ordinal]

[dangillet]

There was a lot of refactoring. Thanks for noticing that this was still here. You're 100% right. It's not even needed anymore. :)

[hugovk]

Looks like this introduces a regression?

Before:

>>> intword(1234567, "%.0f")
1 million

PR:

>>> intword(1234567, "%.0f")
1.0 million

Try this:

Suggested change

	return f"{negative_prefix}{rounded_value} {unit}"
	return f"{negative_prefix}{format % rounded_value} {unit}"

And we could add some test cases like this to the main test_intword():

        (["1234567", "%.0f"], "1 million"),
        (["1234567", "%.1f"], "1.2 million"),
        (["1234567", "%.2f"], "1.23 million"),
        (["1234567", "%.3f"], "1.235 million"),
        (["999500", "%.0f"], "1 million"),
        (["999499", "%.0f"], "999 thousand"),

[dangillet]

Very good catch! I added what you suggested. Thank you!

[hugovk]

Thanks again!