python - 基于另一列的 Pandas 滚动第二个最高值

Question

对于以下示例数据:

data={'Person':['a','a','a','a','a','b','b','b','b','b','b'],
     'Sales':['50','60','90','30','33','100','600','80','90','400','550'],
     'Price':['10','12','8','10','12','10','13','16','14','12','10']}
data=pd.DataFrame(data)

对于每个人(组)我想要的价格基于 第二高销售额在滚动的基础上，但每个组的窗口会有所不同。结果应如下所示:

result={'Person':['a','a','a','a','a','b','b','b','b','b','b'],
     'Sales':['50','60','90','30','33','100','600','80','90','400','550'],
     'Price':['10','12','8','10','12','10','13','16','14','12','10'],
     'Second_Highest_Price':['','10','12','12','12','','10','10','10','12','10']}

我尝试使用 nlargest(2) 但不确定如何让它在滚动的基础上工作。

Answer 1

这不是最优雅的解决方案，但我会执行以下操作:
1- 加载数据集

import numpy as np
import pandas as pd

data={'Person':['a','a','a','a','a','b','b','b','b','b','b'],
     'Sales':['50','60','90','30','33','100','600','80','90','400','550'],
     'Price':['10','12','8','10','12','10','13','16','14','12','10']}

data=pd.DataFrame(data)

data['Sales'] = data['Sales'].astype(float)

2- 使用 Groupby 并一起扩展:

data['2nd_sales'] = data.groupby('Person')['Sales'].expanding(min_periods=2) \
                                  .apply(lambda x: x.nlargest(2).values[-1]).values

3- 计算 Second_Highest_Price :

data['Second_Highest_Price'] = np.where((data['Sales'].shift() == data['2nd_sales']), data['Price'].shift(),
                                (np.where((data['Sales'] == data['2nd_sales']), data['Price'], np.nan)))

data['Second_Highest_Price'] = data.groupby('Person')['Second_Highest_Price'].ffill()

输出:

data['Second_Highest_Price'].values

array([nan, '10', '12', '12', '12', nan, '10', '10', '10', '12', '10'],
      dtype=object)