[patch] add precision argument to floatformat from default filters

[real.human@…]

updated floatformat and related docs to allow for a precision argument (which defaults to 1).

[Chris Beaven]

The following line is incorrect:

return re.sub(r'\.?0+$', '', ('%%.%sf' % precision) % f)

With both precision and f set to 0, you'll return '' instead of '0'

[Chris Beaven]

Updated patch, with tests

[Chris Beaven]

Ok, my patch fixes the above issue, but I noticed when running against existing tests that this patch changes the existing behaviour.

floatformat(0.007) used to return 0.0, now returns 0. I'm not sure if the original test is correct - if it is, this patch will need to be modified.

[real.human@…]

whoops, i just updated my original patch, too (floatformat.2.diff). i also noticed a rounding error in the old version, so now i'm using str() and round() and then stripping trailing periods and zeroes.

[Chris Beaven]

Yep, that's even a better solution :)

Put my tests in your diff too. My comments still stand about this changing existing behaviour.

I also noticed rounding errors using string %s but they still came up for me using str(round( (eg: str(round(1.015, 2)))

[real.human@…]

floatformat.3.diff includes your regression test changes.

i also noticed that in the old behaviour, floatformat(0.0) would return '0', but floatformat(0.007) would return '0.0'. as the behaviour so far has been to strip trailing zeroes, i think we should also to strip trailing periods to remove this inconsistency. if anyone wants to keep trailing periods and zeroes, they can use the stringformat filter.

about the rounding error, the old behaviour which uses python's string formatting rounds up for .6 but down for .5, while round() goes up for .5 and down for .4, which i believe is correct. if that's true, is this a bug in python's string formatting?

>>> '%.1f' % 0.15
'0.1'
>>> '%.1f' % 0.16
'0.2'
>>> round(0.14, 1)
0.10000000000000001
>>> round(0.15, 1)
0.20000000000000001
>>> str(round(0.14, 1))
'0.1'
>>> str(round(0.15, 1))
'0.2'

[Chris Beaven]

Yes, it's a rounding error in Python's string formatting. Like I said in my example, str(round()) doesn't fix everything:

>>> str(round(1.015, 2))
'1.01'

[Chris Beaven]

And you still haven't included the new tests in your diff!

[mrmachine <real.human@…>]

6th time's a charm. tests should be in the diff now, and i've used the decimal module to round the input accurately.

[Chris Beaven]

Hate to tell you this, but decimal is new in 2.4. Django needs Python 2.3 compatibility.

[mrmachine <real dot human at mrmachine dot net>]

bummer. well, we can just use floatformat.4.diff and live with the rounding errors (unless there is another alternative).

[Eric Floehr <eric@…>]

Created ticket #3176 as an alternative implementation to this.

[Chris Beaven]

Implemented in [4274]

[Mark Riedesel <mriedesel@…>]

I see this bug was fixed almost a year ago, but it seems like either it's broken again or I'm using it improperly. The value I'm passing to the filter is 18.125, if I use floatformat:2 the result is 18.12 instead of the expected 18.13. Tested with SVN revision 6483 (Sept 13th, 2007)

[Chris Beaven]

This ticket was specifically to introduce a precision argument and that hasn't broken.

The bug you're seeing is a Python rounding problem with float, but we can get around it by using decimal. I opened a new ticket (#5748) now a backwards compatible decimal has been added to Django.