Question

这似乎应该是一个常见的用例，但我没有找到任何好的指导。我有一个有效的解决方案，但我宁愿使用矢量化查找而不是使用Pandas apply()函数。

这是我正在做的一个例子：

import pandas as pd


example_dict = {
        "category1":{
                "field1": 0.0,
                "filed2": 5.0},
        "category2":{
                "field1": 5.0,
                "field2": 8.0}}

d = {"ids": range(10),
     "category": ["category1" if x % 2 == 0 else "category2" for x in range(10)]}

df = pd.DataFrame(d)
# The operation I am trying to vectorize
df['category_data'] = df.apply(lambda row: example_dict[row['category']], axis=1)

在最后一行，您可以看到我使用apply()函数执行字典查找的位置。我的直觉告诉我应该有一种方法来矢量化这个。我可能错了，但我也想知道。我经常遇到需要在字典中查找信息并将其添加为a DataFrame列的情况。

Answer 1

使用map

df['map']=df.category.map(example_dict)
df
Out[839]: 
    category  ids                   category_data  \
0  category1    0  {'field1': 0.0, 'filed2': 5.0}   
1  category2    1  {'field1': 5.0, 'field2': 8.0}   
2  category1    2  {'field1': 0.0, 'filed2': 5.0}   
3  category2    3  {'field1': 5.0, 'field2': 8.0}   
4  category1    4  {'field1': 0.0, 'filed2': 5.0}   
5  category2    5  {'field1': 5.0, 'field2': 8.0}   
6  category1    6  {'field1': 0.0, 'filed2': 5.0}   
7  category2    7  {'field1': 5.0, 'field2': 8.0}   
8  category1    8  {'field1': 0.0, 'filed2': 5.0}   
9  category2    9  {'field1': 5.0, 'field2': 8.0}   
                              map  
0  {'field1': 0.0, 'filed2': 5.0}  
1  {'field1': 5.0, 'field2': 8.0}  
2  {'field1': 0.0, 'filed2': 5.0}  
3  {'field1': 5.0, 'field2': 8.0}  
4  {'field1': 0.0, 'filed2': 5.0}  
5  {'field1': 5.0, 'field2': 8.0}  
6  {'field1': 0.0, 'filed2': 5.0}  
7  {'field1': 5.0, 'field2': 8.0}  
8  {'field1': 0.0, 'filed2': 5.0}  
9  {'field1': 5.0, 'field2': 8.0}

如果你需要他们进入不同的列

pd.DataFrame(df['map'].tolist())
Out[843]: 
   field1  field2  filed2
0     0.0     NaN     5.0
1     5.0     8.0     NaN
2     0.0     NaN     5.0
3     5.0     8.0     NaN
4     0.0     NaN     5.0
5     5.0     8.0     NaN
6     0.0     NaN     5.0
7     5.0     8.0     NaN
8     0.0     NaN     5.0
9     5.0     8.0     NaN

或

df['map'].apply(pd.Series)
Out[844]: 
   field1  field2  filed2
0     0.0     NaN     5.0
1     5.0     8.0     NaN
2     0.0     NaN     5.0
3     5.0     8.0     NaN
4     0.0     NaN     5.0
5     5.0     8.0     NaN
6     0.0     NaN     5.0
7     5.0     8.0     NaN
8     0.0     NaN     5.0
9     5.0     8.0     NaN

Answer 2

您可以从example_dict创建第二个DataFrame，然后merge创建两个Dataframe

d2 = pd.DataFrame(example_dict.keys(),columns=
             ['category']).assign(category_data=example_dict.values())

df.merge(d2,on='category',how='left')

    category  ids                     category_data
0  category1    0  {u'filed2': 5.0, u'field1': 0.0}
1  category2    1  {u'field2': 8.0, u'field1': 5.0}
2  category1    2  {u'filed2': 5.0, u'field1': 0.0}
3  category2    3  {u'field2': 8.0, u'field1': 5.0}
4  category1    4  {u'filed2': 5.0, u'field1': 0.0}
5  category2    5  {u'field2': 8.0, u'field1': 5.0}
6  category1    6  {u'filed2': 5.0, u'field1': 0.0}
7  category2    7  {u'field2': 8.0, u'field1': 5.0}
8  category1    8  {u'filed2': 5.0, u'field1': 0.0}
9  category2    9  {u'field2': 8.0, u'field1': 5.0}

将字典值分隔为列

d2 = pd.DataFrame(example_dict).T
df.merge(d2,how='left',left_on='category',right_index=True)

    category  ids  field1  field2  filed2
0  category1    0     0.0     NaN     5.0
1  category2    1     5.0     8.0     NaN
2  category1    2     0.0     NaN     5.0
3  category2    3     5.0     8.0     NaN
4  category1    4     0.0     NaN     5.0
5  category2    5     5.0     8.0     NaN
6  category1    6     0.0     NaN     5.0
7  category2    7     5.0     8.0     NaN
8  category1    8     0.0     NaN     5.0
9  category2    9     5.0     8.0     NaN

熊猫矢量化词典查找

2 个答案: