Python：将元组作为函数中的参数传递

debugcn 发表于 Dev

ilovewt

我正在为Pandas DataFrameusing创建新列np.select(condition, choices)。我想将我的代码模块化为一个函数，这样做的麻烦方式如下：

def selection(
    df: pd.DataFrame,
    conditions: Optional[List] = None,
    choices: Optional[List] = None,
    column_names: Optional[List] = None,
):
    if conditions is not None:  # if its none, then don't run this, implies choices and column names are none too
        for condition, choice, col_name in zip(conditions, choices, column_names):
            df[col_name] = np.select(condition, choice, default=" ")
    return df

要运行此功能，我只需要这样做：

conditions = [...]
choices = [...]
column_names = [...]
my_tuple = (conditions, choices, column_names)
df = selection(df, *my_tuple)

我想提高自己的编码技能，并且我觉得这种方式不是最优的，特别是，我觉得涉及到的论点conditions, choices, column_names可以是要作为论点传递的元组。我欢迎任何有关改进此代码的建议。

广晃

如果您真的想将它们作为元组传递，可以，您可以：

def selection(
    df: pd.DataFrame,
    cond_choice_col: Optional[Tuple] = None
):
    if cond_choice_col is not None:  # if its none, then don't run this, implies choices and column names are none too

        # unpack the tuple here
        for condition, choice, col_name in zip(*cond_choice_col):
            df[col_name] = np.select(condition, choice, default=" ")
    return df

conditions = [...]
choices = [...]
column_names = [...]
my_tuple = (conditions, choices, column_names)

df = selection(df, cond_choice_col=my_tuple)

但老实说，我确实认为您的功能比上述功能更加用户友好。

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