如何正确构造具有单个出口的递归函数？

Question

我编写了一个递归函数来增强 pandas.DataFrame.describe。它将峰度和偏斜添加为行。它还创建了第二个描述表来转置第一个描述表，以便您获得汇总统计数据的摘要。它工作得很好，只是我不喜欢编写具有多个出口的函数。我尝试用一个 return 语句编写它（请参阅注释掉的部分），但它会在一个表中创建两个转置表。正确，但太多了。def get_better_desc(df, recursions: int = 1):    '''Adds kurtosis and skew to pandas.DataFrame.describe output. And, creates    second transposed version of this table called on itself for summary stats    of summary stats.    Parameters:        df: pandas.DataFrame, or Series but its super_desc isn't so meaningful.        recursions: integer number of times to apply recursively to create            super_desc. Default value of 1 is all that is necessary.    Returns:        better_desc: pandas.Dataframe (or Series) with kurtosis and skew added.        super_desc: pandas.DataFrame (or Series) of better_desc transposed and            made into a better_desc itself.'''        kurt = df.kurtosis()    kurt.name = 'kurt'    skew = df.skew()    skew.name = 'skew'    better_desc = df.describe().append([kurt, skew])        if recursions > 0:        super_desc = get_better_desc(better_desc.transpose(),                                     recursions=(recursions - 1))        return better_desc, super_desc    else:        return better_desc#     if recursions > 0:#         super_desc = get_better_desc(better_desc.transpose(),#                                      recursions=(recursions - 1))#     else:#         super_desc = better_desc    #     return better_desc, super_desc

Answer 1

该函数返回两个数据帧，如果您只将其分配给一个变量，而不是忽略第二个返回值，它会将两者分配给该变量。在递归调用中，您只需指定第一个表，即 returns 列表中的第一个元素[0]。

...

if recursions > 0:

        super_desc = get_better_desc(better_desc.transpose(),

                                     recursions=(recursions - 1))[0] #<=

    else:

        super_desc = better_desc

    

    return better_desc, super_desc