Question

我正在尝试合并到熊猫数据帧，尽管我想要的实际上可能不是合并。

我在两个框架中有两列相匹配，其中一列共享可用于连接的唯一值。另一列有一个空字段，一个填充了一个字段。

我想覆盖空字段，同时在唯一字段上进行匹配，但只保留被覆盖的列，我不希望第二个DataFrame中的其余列。

希望以下内容会进一步解释

>>> animals = [{"animal" : "dog", "name" : "freddy", "food" : ""},{"animal" : "cat", "name" : "dexter", "food" : ""},{"animal" : "dog", "name" : "lou lou", "food" : ""}]
>>> foods = [{"name" : "freddy", "food" : "dog mix", "brand" : "doggys dog"},{"name" : "dexter", "food" : "fussy cat mix", "brand" : "fish fishy"},{"name" : "lou lou", "food" : "bones", "brand" : "i was a cow"}]
>>> a_pd = pd.DataFrame(animals)
>>> a_pd
  animal food     name
0    dog        freddy
1    cat        dexter
2    dog       lou lou
>>> f_pd = pd.DataFrame(foods)
>>> f_pd
         brand           food     name
0   doggys dog        dog mix   freddy
1   fish fishy  fussy cat mix   dexter
2  i was a cow          bones  lou lou
>>>
>>>
>>> animal_data = a_pd.merge(f_pd, on='name', how='left')
>>> animal_data
  animal food_x     name        brand         food_y
0    dog          freddy   doggys dog        dog mix
1    cat          dexter   fish fishy  fussy cat mix
2    dog         lou lou  i was a cow          bones
>>>

我应该只吃食物，而不想要这个品牌（还要注意，这是示例数据，实时数据有很多列

所需结果

>>> animal_data
  animal        name            food
0    dog      freddy         dog mix
1    cat      dexter   fussy cat mix
2    dog     lou lou           bones

Answer 1

使用：

animal_data = a_pd.merge(f_pd, on='name', how='left', suffixes=('_x','')).drop('food_x', axis=1)

输出：

  animal     name        brand           food
0    dog   freddy   doggys dog        dog mix
1    cat   dexter   fish fishy  fussy cat mix
2    dog  lou lou  i was a cow          bones

或

a_pd[['animal','name']].merge(f_pd, how='left')

输出：

  animal     name        brand           food
0    dog   freddy   doggys dog        dog mix
1    cat   dexter   fish fishy  fussy cat mix
2    dog  lou lou  i was a cow          bones

Answer 2

您可以使用update

a_pd.set_index('name',inplace=True)
a_pd.update(f_pd.set_index('name'))
a_pd
Out[68]: 
        animal           food
name                         
freddy     dog        dog mix
dexter     cat  fussy cat mix
lou lou    dog          bones
a_pd.reset_index()
Out[69]: 
      name animal           food
0   freddy    dog        dog mix
1   dexter    cat  fussy cat mix
2  lou lou    dog          bones

或者我们使用map

a_pd.food=a_pd.name.map(f_pd.set_index('name').food)
a_pd
Out[74]: 
  animal           food     name
0    dog        dog mix   freddy
1    cat  fussy cat mix   dexter
2    dog          bones  lou lou

Answer 3

我会尝试drop或只是选择要保留的列：

animal_data.drop(['food_x', 'brand'], axis=1, inplace=True)

或

animal_data = animal_data[['animal', 'name', 'food']]

Answer 4

最好合并数据框的视图，这些视图不包含合并数据框中不需要的列。例如：

point

这可能会更快，并且如果处理非常大的数据帧，可能会节省一些内存，因为在合并中仅结转了相关列。

大熊猫-合并两个数据帧覆盖并指定要保留的列

4 个答案: