ENH: Make ExcelFormatter.header_style a class attribute instead of a property

[terazus]

Feature Type

• Adding new functionality to pandas

• Changing existing functionality in pandas

• Removing existing functionality in pandas

Problem Description

When I'm trying to remove the default ExcelFormatter header style, my IDE complains that the property cannot be set.
Pycharm suggest adding a setter but I don't think it's the correct behaviour.

This is very minor improvement as code runs perfectly well.

Feature Description

Change the ExcelFormatter header_style to be a class attribute instead of a property:

class ExcelFormatter:
    """
    Class for formatting a DataFrame to a list of ExcelCells,

    Parameters
    ----------
    ...
    """

    max_rows = 2 ** 20
    max_cols = 2 ** 14
    header_style: dict[str, dict[str, str | bool]] = {
        "font": {"bold": True},
        "borders": {
            "top": "thin",
            "right": "thin",
            "bottom": "thin",
            "left": "thin",
        },
        "alignment": {"horizontal": "center", "vertical": "top"},
    }

    def __init__(
            self,
            df,
            na_rep: str = "",
            float_format: str | None = None,
            cols: Sequence[Hashable] | None = None,
            header: Sequence[Hashable] | bool = True,
            index: bool = True,
            index_label: IndexLabel | None = None,
            merge_cells: bool = False,
            inf_rep: str = "inf",
            style_converter: Callable | None = None,
    ) -> None:...

Alternative Solutions

Pycharm setter suggestion (which i find less relevant):

class ExcelFormatter:
    """
    Class for formatting a DataFrame to a list of ExcelCells,

    Parameters
    ----------
    ...
    """

    max_rows = 2 ** 20
    max_cols = 2 ** 14

    def __init__(
            self,
            df,
            na_rep: str = "",
            float_format: str | None = None,
            cols: Sequence[Hashable] | None = None,
            header: Sequence[Hashable] | bool = True,
            index: bool = True,
            index_label: IndexLabel | None = None,
            merge_cells: bool = False,
            inf_rep: str = "inf",
            style_converter: Callable | None = None,
    ) -> None:...

    @property
    def header_style(self) -> dict[str, dict[str, str | bool]]:
        return {
            "font": {"bold": True},
            "borders": {
                "top": "thin",
                "right": "thin",
                "bottom": "thin",
                "left": "thin",
            },
            "alignment": {"horizontal": "center", "vertical": "top"},
        }

    @header_style.setter
    def header_style(self, value):
        self._header_style = value

Additional Context

Just thanks for your hard work. 🐼

[rhshadrach]

Thanks for the request; can you shorten the code posted? I'm guessing you do not need the entire code of the class.

[terazus]

Thanks for the request; can you shorten the code posted? I'm guessing you do not need the entire code of the class.

Hello. Thx for the quick reply. It's done. Let me know if I shorthened too much.

[rhshadrach]

Thanks - some relevant issues: #22758, #22773

cc @WillAyd @attack68

[WillAyd]

Would it suffice to make a setter for the property instead? The advantage of a property is we could theoretically control the assignment and raise if for example particular expectations are not met as to what the header_style needs to provide

[terazus]

Hello,
I'm personally fine with both. It just feels weird to call a property setter over an object that hasn't been instantianted yet. My suggestion came from the logic that we are modifying the class before we instantiate the object so it would made more sense to use a class attribute than an object property.
I do agree that using a setter would provide extra validation the attribute doesn't.
Thankx :)

[rhshadrach]

In either case, is there a risk of errors due to the changing of this property while an ExcelWriter is open? I think the answer is no, but we should be certain before entertaining the idea.

[attack68]

df = pd.DataFrame([[1,2],[3,4]], index=["one", "two"], columns=["A", "B"])
df.to_excel("file.xlsx")

The above is the default. Using Styler with no styles applied also yields the same, but when adding styles you can control this.

df.style \
  .applymap_index(lambda v: "color:red; text-align:center;", axis=0) \
  .applymap_index(lambda v: "color:blue;", axis=1) \
  .to_excel("df_styler.xlsx")

There is a bug that was fixed for v1.5.0 that corrected adding borders as well.

Personally, I would prefer df.to_excel to produce a completely un-styled excel sheet, and let Styler be used to add and define all the custom styles, without setting properties directly on the ExcelFormatter, which are not generally consistent with the way the rest of the API works and is documented for users.

[rmhowe425]

take