当列值匹配时，Pandas Dataframe 从一行中替换 Nan

Posted 2023-03-11

【中文标题】当列值匹配时，Pandas Dataframe 从一行中替换 Nan

【英文标题】：Pandas Dataframe replace Nan from a row when a column value matches

【发布时间】：2019-08-18 19:02:08

【问题描述】：

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”。结果假设是， 输入数据框

      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？

【参考方案1】：

使用lambda function 在列表中指定的列中使用DataFrame.groupby 正向和反向填充每个组的缺失值 - 对于每个组合，每个组都需要相同的值：

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

【讨论】：

我已经尝试过你的建议，但我无法得到结果

@MahamuthaM - 不确定是否理解，这是创建DataFrame 的解决方案？还有什么问题？

@MahamuthaM - 你能解释更多吗？不换？在我的解决方案之前尝试使用df = df.replace(['Nan', 'NaN'], np.nan)。

【参考方案2】：

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()

【讨论】：

AttributeError: 'list' 对象没有属性 'dropna'