本文介绍了在 pandas 中按组分配模式(处理NaN的组模式)的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
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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