Question

样本DF：

ID   Name        Price     Condition   Fit_Test
1    Apple         10      Good        Super_Fit
2    Apple         10      OK          Super_Fit
3    Apple         10      Bad         Super_Fit
4    Orange        12      Good        Not_Fit
5    Orange        12      OK          Not_Fit
6    Banana        15      OK          Medium_Fit
7    Banana        15      Bad         Medium_Fit
8    Pineapple     25      OK          Medium_Fit
9    Pineapple     25      OK          Medium_Fit
10   Cherry        30      Bad         Medium_Fit

预期DF：

ID   Name        Price     Condition   Fit_Test
1    Apple         10      Good        Super_Fit
2    Apple         10      OK          Super_Fit
3    Apple         10      Bad         Super_Fit

4    Orange        12      Good        Not_Fit

6    Banana        15      OK          Medium_Fit

8    Pineapple     25      OK          Medium_Fit
9    Pineapple     25      OK          Medium_Fit
10   Cherry        30      Bad         Medium_Fit

问题陈述：

我想用group-by和Name Price，然后根据Condition进行过滤。

如果在Name和Price中都存在良好，不良和确定这3个条件，则仅保留好一个条件，而Fit_Test不是Super_Fit
如果在“名称”和“价格”中存在“良好”和“确定”的条件，则仅保留“良好”一项（ID 4,5仅是预期的ID 4），而Fit_Test不是Super_Fit
如果在Name和Price的情况下存在“正常”和“不良”的条件，则仅保留“确定”一个（ID 6,7仅是预期的ID 6）并且Fit_Test不是Super_Fit
如果在Name和Price内存在OK和OK的条件，则存在Good和Good存在或仅存在Bad，则不执行任何操作，则仅保留OK （ ID 8,9,10是预期的ID 8,9,10），而Fit_Test不是Super_Fit

更新答案

您没有条件df的所有Fit_Test的第一答案和测试编辑。在此答案中，预期DF 将不包含第2行和第3行，如答案中所示
更新更新答案在您需要添加另一列Fit_Test时有效，并且仅在值不为Super_Fit时有效。

在两种解决方案中，基于Condition列和2列分组的行过滤是相同的。

我在数字列上找到了带有filter + group by的东西，但在String列上找不到了。

Answer 1

创建的想法set用于比较：

a = df.join(df.groupby(['Price','Name'])['Condition'].apply(set).rename('m'),
             on=['Price','Name'])['m']
print (a)
0    {Bad, Good, OK}
1    {Bad, Good, OK}
2    {Bad, Good, OK}
3         {Good, OK}
4         {Good, OK}
5          {Bad, OK}
6          {Bad, OK}
7               {OK}
8               {OK}
9              {Bad}
Name: m, dtype: object

m1 = (a == set({'Bad', 'Good', 'OK'})) | (a == set({'Good', 'OK'}))
m2 = a == set({'Bad', 'OK'})
#check if unique value - length of set is 1
m3 = a.str.len() == 1
m4 = df['Condition'] == 'Good'
m5 = df['Condition'] == 'OK'

df = df[(m1 & m4) | (m2 & m5) | m3]
print (df)
   ID       Name  Price Condition
0   1      Apple     10      Good
3   4     Orange     12      Good
5   6     Banana     15        OK
7   8  Pineapple     25        OK
8   9  Pineapple     25        OK
9  10     Cherry     30       Bad

编辑测试：

要进行测试，请使用assign：

print (df.assign(sets=a, m1 = m1, m2=m2, m3=m3, m4=m4, m5=m5, m=m))
   ID       Name  Price Condition             sets     m1     m2     m3  \
0   1      Apple     10      Good  {Bad, Good, OK}   True  False  False   
1   2      Apple     10        OK  {Bad, Good, OK}   True  False  False   
2   3      Apple     10       Bad  {Bad, Good, OK}   True  False  False   
3   4     Orange     12      Good       {Good, OK}   True  False  False   
4   5     Orange     12        OK       {Good, OK}   True  False  False   
5   6     Banana     15        OK        {Bad, OK}  False   True  False   
6   7     Banana     15       Bad        {Bad, OK}  False   True  False   
7   8  Pineapple     25        OK             {OK}  False  False   True   
8   9  Pineapple     25        OK             {OK}  False  False   True   
9  10     Cherry     30       Bad            {Bad}  False  False   True   

      m4     m5      m  
0   True  False   True  
1  False   True  False  
2  False  False  False  
3   True  False   True  
4  False   True  False  
5  False   True   True  
6  False  False  False  
7  False   True   True  
8  False   True   True  
9  False  False   True

编辑更新：

对于新条件，请使用：

    m6 = df['Fit_Test'] == 'Super_Fit'
    df = df[((m1 & m4) | (m2 & m5) | m3) | m6]
    print (df)
       ID       Name  Price Condition    Fit_Test
    0   1      Apple     10      Good   Super_Fit
    1   2      Apple     10        OK   Super_Fit
    2   3      Apple     10       Bad   Super_Fit
    3   4     Orange     12      Good     Not_Fit
    5   6     Banana     15        OK  Medium_Fit
    7   8  Pineapple     25        OK  Medium_Fit
    8   9  Pineapple     25        OK  Medium_Fit
    9  10     Cherry     30       Bad  Medium_Fit

Answer 2

如果您没有太多条件（例如这里只有3个条件），则以下是一种简单的解决方法：

df.loc[df["Condition"] == 'Good',"Condition"] = 3
df.loc[df["Condition"] == 'OK',"Condition"] = 2
df.loc[df["Condition"] == 'Bad',"Condition"] = 1

df = df.groupby(['Name','Price']).max()

df.loc[df["Condition"] == 3] = "Good"
df.loc[df["Condition"] == 2] = "OK"
df.loc[df["Condition"] == 1] = "Bad"

2列上的Groupby以及String列上的过滤器

2 个答案: