Pandas 当列值匹配时，Dataframe 从行中替换 Nan

Question

我有数据框，即

Input Dataframe

      class  section  sub  marks  school  city
0     I      A        Eng  80     jghss   salem
1     I      A        Mat  90     jghss   salem 
2     I      A        Eng  50     Nan     salem 
3     III    A        Eng  80     gphss   Nan
4     III    A        Mat  45     Nan     salem
5     III    A        Eng  40     gphss   Nan
6     III    A        Eng  20     gphss   salem
7     III    A        Mat  55     gphss   Nan

当 "class" 和 "section" 列中的值匹配时，我需要替换 "school" 和 "city" 中的 "Nan"。结果应该是，输入数据帧

      class  section  sub  marks  school  city
0     I      A        Eng  80     jghss   salem
1     I      A        Mat  90     jghss   salem 
2     I      A        Eng  50     jghss   salem 
3     III    A        Eng  80     gphss   salem
4     III    A        Mat  45     gphss   salem
5     III    A        Eng  40     gphss   salem
6     III    A        Eng  20     gphss   salem
7     III    A        Mat  55     gphss   salem

谁能帮我解决这个问题？

Answer 1

假设每对class和section对应一对唯一的school和city，我们可以使用groupby:

# create a dictionary of class and section with school and city
# here we assume that for each pair and class there's a row with both school and city
# if that's not the case, we can separate the two series 
school_city_dict = df[['class', 'section','school','city']].dropna().\
                     groupby(['class', 'section'])[['school','city']].\
                     max().to_dict()
# school_city_dict = {'school': {('I', 'A'): 'jghss', ('III', 'A'): 'gphss'},
#                     'city': {('I', 'A'): 'salem', ('III', 'A'): 'salem'}}

# set index, prepare for map function
df.set_index(['class','section'], inplace=True)

df.loc[:,'school'] = df.index.map(school_city_dict['school'])
df.loc[:,'city'] = df.index.map(school_city_dict['city'])

# reset index to the original
df.reset_index()

Answer 2

在 DataFrame.groupby 列表中指定的列中使用 lambda function 每组正向和反向填充缺失值 - 对于每组相同值的每个组合是必要的：

cols = ['school','city']
df[cols] = df.groupby(['class','section'])[cols].apply(lambda x: x.ffill().bfill())
print (df)
  class section  sub  marks school   city
0     I       A  Eng     80  jghss  salem
1     I       A  Mat     90  jghss  salem
2     I       A  Eng     50  jghss  salem
3   III       A  Eng     80  gphss  salem
4   III       A  Mat     45  gphss  salem
5   III       A  Eng     40  gphss  salem
6   III       A  Eng     20  gphss  salem
7   III       A  Mat     55  gphss  salem

Pandas 当列值匹配时，Dataframe 从行中替换 Nan

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