在 pandas 中按组分配模式(处理NaN的组模式)-Python问题

mode imputation by groups in pandas (handling group modes that are NaN)(在 pandas 中按组分配模式(处理NaN的组模式))

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问题描述

我有一个分类列&WALLSMATERIAL_MODE&QOOT；其中包含NAN，我希望通过以下组[‘NAME_RECOVICATION_TYPE’，‘AGE_GROUP’]将其归因于NAN：

    NAME_EDUCATION_TYPE             AGE_GROUP   WALLSMATERIAL_MODE
20  Secondary / secondary special   45-60       Stone, brick
21  Secondary / secondary special   21-45       NaN
22  Secondary / secondary special   21-45       Panel
23  Secondary / secondary special   60-70       Mixed
24  Secondary / secondary special   21-45       Panel
25  Secondary / secondary special   45-60       Stone, brick
26  Secondary / secondary special   45-60       Wooden
27  Secondary / secondary special   21-45       NaN
28  Higher education                21-45       NaN
29  Higher education                21-45       Panel

可再生性代码

df = pd.DataFrame({'NAME_EDUCATION_TYPE': {20: 'Secondary / secondary special',
  21: 'Secondary / secondary special',
  22: 'Secondary / secondary special',
  23: 'Secondary / secondary special',
  24: 'Secondary / secondary special',
  25: 'Secondary / secondary special',
  26: 'Secondary / secondary special',
  27: 'Secondary / secondary special',
  28: 'Higher education',
  29: 'Higher education'},
 'AGE_GROUP': {20: '45-60',
  21: '21-45',
  22: '21-45',
  23: '60-70',
  24: '21-45',
  25: '45-60',
  26: '45-60',
  27: '21-45',
  28: '21-45',
  29: '21-45'},
 'WALLSMATERIAL_MODE': {20: 'Stone, brick',
  21: np.nan,
  22: 'Panel',
  23: 'Mixed',
  24: 'Panel',
  25: 'Stone, brick',
  26: 'Wooden',
  27: np.nan,
  28: np.nan,
  29: 'Panel'}})

我尝试从这个post改编以下函数，该函数适用于中位数推算并处理非中位数的组中值

输入：

def mode(s):
    if pd.isnull(s.mode()):
        return df['WALLSMATERIAL_MODE'].mode()
    return s.mode()
        
df['WALLSMATERIAL_MODE'] = df['WALLSMATERIAL_MODE'].groupby([df['NAME_EDUCATION_TYPE'], df['AGE_GROUP']], dropna=False).apply(lambda x: x.fillna(mode(x)))

out：调用pd.isull时引发以下错误

The truth value of a Series is ambiguous. Use a.empty, a.bool(), a.item(), a.any() or a.all()

我不明白，我已尝试在所有组模式上应用pd.isull，但没有引发此错误。请参阅下面的群组模式

输入：

df['WALLSMATERIAL_MODE'].groupby([df['NAME_EDUCATION_TYPE'], df['AGE_GROUP']]).agg(pd.Series.mode).to_dict()

输出：

{('Higher education', '60-70'): nan,
 ('Higher education', '45-60'): nan,
 ('Higher education', '21-45'): 'Panel',
 ('Higher education', '0-21'): nan,
 ('Secondary / secondary special', '60-70'): 'Mixed',
 ('Secondary / secondary special', '45-60'): 'Stone, brick',
 ('Secondary / secondary special', '21-45'): 'Panel',
 ('Secondary / secondary special', '0-21'): nan}

如果有人能指出错误在哪里，或者是否有有效的方法对本专栏进行分组归罪，我将不胜感激！

推荐答案

下面的代码似乎使用了Try Except来完成此操作。我宁愿避免使用Try，除非我想不出一种更干净的方法。

def mode_cats(s):
        try:
            if pd.isnull(s.mode().any()): # check if the mode of the subgroup is NaN or contains NaN 
                                          # (mode() may indeed return a list of several modes)
                m = app_train_dash['WALLSMATERIAL_MODE'].mode().iloc[0] # returns the mode of the column
            else:
                m = s.mode().iloc[0]  # returns the mode of the subgroup
            return m
        except IndexError: # mode returns an empty series if the subgroup consists of a single NaN value
                           # this causes s.mode().iloc[0] to raise an index error
            return app_train_dash['WALLSMATERIAL_MODE'].mode().iloc[0]

正如@Ben.T指出的那样，我必须使用.iloc[0]和.mode() 但是当.mode().iloc[0]有一个空数组作为输入时，我得到IndexError: single positional indexer is out-of-bounds。错误回溯：

模式()在一行的子组上被调用，值=NaN。.mode()返回单个NaN的这个子组的空数组
对传递的空数组调用pd.isull并返回空数组
对空数组调用.iloc[0]会引发索引错误

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