[英]How to get second highest value in a pandas column for a certain ID?
[英]Pandas Rolling second Highest Value based on another column
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),我希望以滚动方式根据第二高的销售额计算价格,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。
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)
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