如何合并两个 Pandas DataFrame 并聚合一个特定列？

Question

我有两个数据框：

         city  count    school
0    New York      1  school_3
1  Washington      1  School_4
2  Washington      1  School_5
3          LA      1  School_1
4          LA      1  School_4

         city  count    school
0    New York      1  School_3
1  Washington      1  School_1
2          LA      1  School_3
3          LA      2  School_4

我希望获得这个结果：

         city  count    school
0    New York      2  school_3
1  Washington      1  School_1
2  Washington      1  School_4
3  Washington      1  School_5
4          LA      1  School_1
5          LA      1  School_3
6          LA      3  School_4

以下是代码。

d1 = [{'city':'New York', 'school':'school_3', 'count':1},
      {'city':'Washington', 'school':'School_4', 'count':1},
      {'city':'Washington', 'school':'School_5', 'count':1},
      {'city':'LA', 'school':'School_1', 'count':1},
      {'city':'LA', 'school':'School_4', 'count':1}]


d2 = [{'city':'New York', 'school':'School_3', 'count':1},
      {'city':'Washington', 'school':'School_1', 'count':1},
      {'city':'LA', 'school':'School_3', 'count':1},
      {'city':'LA', 'school':'School_4', 'count':2}]

x1 = pd.DataFrame(d1)
x2 = pd.DataFrame(d2)
#just get empty DataFrame
print pd.merge(x1, x2)

如何获取聚合结果？

- wangke1020

Answer 1

您可以做以下事情：

>>> pd.concat([x1, x2]).groupby(["city", "school"], as_index=False)["count"].sum()
       city    school        count
0          LA  School_1      1
1          LA  School_3      1
2          LA  School_4      3
3    New York  School_3      1
4    New York  school_3      1
5  Washington  School_1      1
6  Washington  School_4      1
7  Washington  School_5      1

请注意，由于数据中出现了拼写错误（school_3与School_3），纽约出现了两次。

- elyase

Answer 2

这是与 @elyase 的解决方案略有不同的实现，使用了 pandas.DataFrame.merge(...)。

x1.merge(x2,on=['city', 'school', 'count'], how='outer').groupby(['city', 'school'], as_index=False)['count'].sum()

在ipython笔记本%timeit中计时，这种方法比@elyase的方法稍微快一点（<1ms）

100 loops, best of 3: 6.25 ms per loop  #using concat(...) with @elyase's solution
100 loops, best of 3: 5.49 ms per loop #using merge(...) in this solution

另外，如果您想使用 pandas aggregate 功能，您也可以这样做：

x1.merge(x2,on=['city', 'school', 'count'], how='outer').groupby(['city', 'school'], as_index=False).agg(numpy.sum)

唯一需要说明的是，使用agg(...)是三种方案中最慢的。

显然，这3种方法都提供了正确的结果：

         city    school  count
0          LA  School_1      1
1          LA  School_3      1
2          LA  School_4      3
3    New York  School_3      1
4    New York  school_3      1
5  Washington  School_1      1
6  Washington  School_4      1
7  Washington  School_5      1

- alacy