Question

鉴于以下DataFrame，

df = pd.DataFrame({'Student':['Siri','Alexa'], 'Class':['6', '7'], 'Section':['A','B'], 'Teacher':['Ravi','Mark'], 'School':['C','D']})

我想返回一个包含所有列值的所有可能组合的表。我使用下面的代码实现了这个目标

df['key'] = 1

df1 = pd.merge(df.loc[:, ['key','Student']], df.loc[:,['key','Class']], how='outer')

df2 = df1.merge(df.loc[:,['key','Section']], how='outer')

df3 = df2.merge(df.loc[:,['key','Teacher']], how='outer')

df4 = df3.merge(df.loc[:,['key','School']], how='outer')

df4.drop(columns='key', inplace=True)

完成此操作的最简单方法是什么，因为我有15列，通过上面的方法，会导致14次合并和效率低下的代码？

Answer 1

您正在寻找笛卡尔积。这可以通过itertools.product：

实现

from itertools import product

prod = product(df['Class'].unique(), df['Section'].unique())
student_cols = [x for x in df.columns if x not in ('Class', 'Section')]
students = df[student_cols].drop_duplicates().values.tolist()

res = pd.DataFrame([s + list(p) for p in prod for s in students],
                   columns=list(student_cols+['Class', 'Section']))\
        .sort_values(list(student_cols+['Class', 'Section']))

print(res)

#   School Student Teacher Class Section
# 0      C    Siri    Ravi     6       A
# 2      C    Siri    Ravi     6       B
# 4      C    Siri    Ravi     7       A
# 6      C    Siri    Ravi     7       B
# 1      D   Alexa    Mark     6       A
# 3      D   Alexa    Mark     6       B
# 5      D   Alexa    Mark     7       A
# 7      D   Alexa    Mark     7       B

Answer 2

您的意思是：

from itertools import product

df = pd.DataFrame({'Student':['Siri','Alexa'], 'Class':['6', '7'], 'Section':['A','B'], 'Teacher':['Ravi','Mark'], 'School':['C','D']})

uniques = [df[i].unique().tolist() for i in df.columns ]
pd.DataFrame(product(*uniques), columns = df.columns)

生成每列中所有唯一项的笛卡尔乘积。

   Student Class Section Teacher School
0     Siri     6       A    Ravi      C
1     Siri     6       A    Ravi      D
2     Siri     6       A    Mark      C
3     Siri     6       A    Mark      D
4     Siri     6       B    Ravi      C
5     Siri     6       B    Ravi      D
6     Siri     6       B    Mark      C
7     Siri     6       B    Mark      D
8     Siri     7       A    Ravi      C
9     Siri     7       A    Ravi      D
10    Siri     7       A    Mark      C
11    Siri     7       A    Mark      D
12    Siri     7       B    Ravi      C
13    Siri     7       B    Ravi      D
14    Siri     7       B    Mark      C
15    Siri     7       B    Mark      D
16   Alexa     6       A    Ravi      C
17   Alexa     6       A    Ravi      D
18   Alexa     6       A    Mark      C
19   Alexa     6       A    Mark      D
20   Alexa     6       B    Ravi      C
21   Alexa     6       B    Ravi      D
22   Alexa     6       B    Mark      C
23   Alexa     6       B    Mark      D
24   Alexa     7       A    Ravi      C
25   Alexa     7       A    Ravi      D
26   Alexa     7       A    Mark      C
27   Alexa     7       A    Mark      D
28   Alexa     7       B    Ravi      C
29   Alexa     7       B    Ravi      D
30   Alexa     7       B    Mark      C
31   Alexa     7       B    Mark      D

在Pandas DataFrame中创建多个列的所有可能组合

2 个答案: