比较数据框中的行

Question

我的任务之一很麻烦。在我的第一种情况下，我需要比较数据框中的一些变量，然后如果它们相同，它将返回相同的标识符列值。

这是我的多个排序数据框看起来像

| no | age| gender | income_group | cars
| 1  | 15 |  male  |       0      | ford
| 2  | 15 |  male  |       0      | renault
| 3  | 15 |  female|       1      | bmw
| 4  | 16 |  female|       1      | bmw
| 5  | 16 |  female|       1      | mercedes
| 6  | 16 |  female|       1      | honda

我想要一些代码来比较此排序的数据帧上的每一行，如果某些行的[年龄，性别，收入组]相同，它将复制第一个[no]列值来替换其他

该代码将使我的数据框看起来像这样

| no | age| gender | income_group | cars
| 1  | 15 |  male  |       0      | ford
| 1  | 15 |  male  |       0      | renault
| 3  | 15 |  female|       1      | bmw
| 4  | 16 |  female|       1      | bmw
| 4  | 16 |  female|       1      | mercedes
| 4  | 16 |  female|       1      | honda

有没有可能在python中这样做呢？

编辑： 我的第二种情况变得更加复杂，因为我发现一些相同的[年龄，性别，收入组]变量但具有相同的[汽车]值，因此我希望在这种情况下将其视为不同的个体，而使用不同的[否]值

如果扩展数据框并获得列，则如下所示

| no | age| gender | income_group | cars
| 1  | 15 |  male  |       0      | ford
| 2  | 15 |  male  |       0      | renault
| 3  | 15 |  female|       1      | bmw
| 4  | 16 |  female|       1      | bmw
| 5  | 16 |  female|       1      | mercedes
| 6  | 16 |  female|       1      | honda

| 7  | 17 |  male  |       0      | bmw
| 8  | 17 |  male  |       0      | honda
| 9  | 17 |  male  |       0      | bmw
| 10 | 17 |  male  |       0      | honda
| 11 | 17 |  male  |       0      | renault

一个人不能拥有相同的汽车价值，该代码将使df：

| 7  | 17 |  male  |       0      | bmw
| 7  | 17 |  male  |       0      | honda
| 9  | 17 |  male  |       0      | bmw
| 9  | 17 |  male  |       0      | honda
| 9  | 17 |  male  |       0      | renault

白色jezrael解决方案：

df['a'] = df.duplicated(['age','gender','income_group', 'cars'], keep=False).cumsum()

df['no'] = df.groupby(['age','gender','income_group','a'], sort=False)['no'].transform('first')
df = df.drop('a', axis=1)

我得到：

no  age  gender  income_group      cars  a
 0   15    male             0      ford  0
 0   15    male             0   renault  0
 2   15  female             1       bmw  0
 3   16  female             1       bmw  0
 3   16  female             1  mercedes  0
 3   16  female             1     honda  0
 6   17    male             0       bmw  1
 7   17    male             0     honda  2
 8   17    male             0       bmw  3
 9   17    male             0     honda  4
 9   17    male             0   reanult  4

Answer 1

使用GroupBy.transform有GroupBy.first：

df['no'] = df.groupby(['age','gender','income_group'], sort=False)['no'].transform('first')
print (df)
   no  age  gender  income_group      cars
0   1   15    male             0      ford
1   1   15    male             0   renault
2   3   15  female             1       bmw
3   4   16  female             1       bmw
4   4   16  female             1  mercedes
5   4   16  female             1     honda

或者获取第一个值DataFrame.duplicated，然后向前填充缺少的值：

df['no'] = df.loc[(~df.duplicated(['age','gender','income_group'])), 'no']
df['no'] = df['no'].ffill().astype(int)
print (df)
   no  age  gender  income_group      cars
0   1   15    male             0      ford
1   1   15    male             0   renault
2   3   15  female             1       bmw
3   4   16  female             1       bmw
4   4   16  female             1  mercedes
5   4   16  female             1     honda

编辑：

df['a'] = df.duplicated(['age','gender','income_group', 'cars'])
mask = df.groupby(['age','gender','income_group'])['a'].transform('any')

df.loc[mask, 'no'] = df.groupby(df.loc[mask].groupby('cars').cumcount(ascending=False))['no'].transform('first')
df = df.drop('a', axis=1)              
print (df)
     no  age  gender  income_group      cars
0   1.0   15    male             0      ford
1   2.0   15    male             0   renault
2   3.0   15  female             1       bmw
3   4.0   16  female             1       bmw
4   5.0   16  female             1  mercedes
5   6.0   16  female             1     honda
6   7.0   17    male             0       bmw
7   7.0   17    male             0     honda
8   9.0   17    male             0       bmw
9   9.0   17    male             0     honda
10  9.0   17    male             0   reanult