跨列交叉表

时间:2018-11-05 03:49:56

标签: python pandas

我有一个带有名称,日期和位置的数据框。对于每个名称-日期-位置三元组,我想知道具有该名称-日期的行中有多少比例具有该位置。

在代码中,我从df开始并寻找expected

import pandas as pd

df = pd.DataFrame(
    [
        {"name": "Alice", "day": "friday", "location": "left"},
        {"name": "Alice", "day": "friday", "location": "right"},
        {"name": "Bob", "day": "monday", "location": "left"},
    ]
)

print(df)



expected = pd.DataFrame(
    [
        {"name": "Alice", "day": "friday", "location": "left", "row_percent": 50.0},
        {"name": "Alice", "day": "friday", "location": "right", "row_percent": 50.0},
        {"name": "Bob", "day": "monday", "location": "left", "row_percent": 100.0},
    ]
).set_index(['name', 'day', ])
print(expected)

打印:

In [13]: df                                                                                                                                                                                  
Out[13]: 
      day location   name
0  friday     left  Alice
1  friday    right  Alice
2  monday     left    Bob




In [12]: expected                                                                                                                                                                            
Out[12]: 
             location  row_percent
name  day                         
Alice friday     left         50.0
      friday    right         50.0
Bob   monday     left        100.0

1 个答案:

答案 0 :(得分:3)

使用groupbyvalue_counts

df.groupby(['name', 'day']).location.value_counts(normalize=True).mul(100)

name   day     location
Alice  friday  left         50.0
               right        50.0
Bob    monday  left        100.0
Name: location, dtype: float64

为您所需的输出进行更多清洁:

out = (df.groupby(['name', 'day']).location.value_counts(normalize=True).mul(100)
          .rename('row_percent').reset_index(2))

             location  row_percent
name  day
Alice friday     left         50.0
      friday    right         50.0
Bob   monday     left        100.0

out == expected

              location  row_percent
name  day
Alice friday      True         True
      friday      True         True
Bob   monday      True         True
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