ENH: Add `area_limit` to `fillna

[joshdunnlime]

Feature Type

• [X] Adding new functionality to pandas

• [X] Changing existing functionality in pandas

• [ ] Removing existing functionality in pandas

Problem Description

The pandas methods interpolate, ffill and bfill all have the area_limit options, however, fillna does not. It would be nice to add this.

Feature Description

DataFrame.fillna(value=None, *, method=None, axis=None, inplace=False, limit=None, area_limit=None, downcast=<no_default>)

Alternative Solutions

Interpolate with method='constant'. The somewhat obvious downsides to this are that constant isn't included in the scipy interpolation API.

Additional Context

See https://github.com/pandas-dev/pandas/issues/56492 for this functionality added to ffill and bfill. It would be nice to have better API consistency between thee methods and also interpolate.

[rhshadrach]

interpolate, ffill, and bfill all fill values using values near the given location. With the method argument being deprecated, fillna does not. It doesn't seem appropriate to have limit_area because fillna does not work with nearby values.

[joshdunnlime]

I get that it doesn't fill with nearby values but I don't see why it couldn't fill missing values with some knowledge of the rows around it? After all it has the limit kwarg.

What about my other suggestion of having a constant term on interpolate? (I still think it makes more sense to have limit_area on fillna as its a very valid use case as shown by ffill etc)

[rhshadrach]

I get that it doesn't fill with nearby values but I don't see why it couldn't fill missing values with some knowledge of the rows around it?

I'm not saying it couldn't. But this increases the scope of the function which I think is undesirable.

What about my other suggestion of having a constant term on interpolate?

I don't think that fits the definition of "interpolate", so I find this undesirable from an API design perspective.

One can implement this behavior as follows:

df = pd.DataFrame({"a": [np.nan, 1, np.nan, 2], "b": [1, np.nan, 2, np.nan]})
isna = df.isna()

# inside
mask = isna & (~isna).cummax() & (~isna).loc[::-1].cummax()
print(df.mask(mask, 5.0))
#      a    b
# 0  NaN  1.0
# 1  1.0  5.0
# 2  5.0  2.0
# 3  2.0  NaN

# outside
mask = isna & (isna.cummin() | isna.loc[::-1].cummin())
print(df.mask(mask, 5.0))
#      a    b
# 0  5.0  1.0
# 1  1.0  NaN
# 2  NaN  2.0
# 3  2.0  5.0

Perhaps this is too technical to be expected from users though. cc @pandas-dev/pandas-core for any thoughts.