Pandas groupby，在另一列中标识出最大值的元素 [英] Pandas groupby with identification of an element with max value in another column
问题描述
我有一个数据框，其中包含具有不同定价规则的商品的销售结果:
I have a dataframe with sales results of items with different pricing rules:
import pandas as pd
from datetime import timedelta
df_1 = pd.DataFrame()
df_2 = pd.DataFrame()
df_3 = pd.DataFrame()
# Create datetimes and data
df_1['item'] = [1, 1, 2, 2, 2]
df_1['date'] = pd.date_range('1/1/2018', periods=5, freq='D')
df_1['price_rule'] = ['a', 'b', 'a', 'b', 'b']
df_1['sales']= [2, 4, 1, 5, 7]
df_1['clicks']= [7, 8, 9, 10, 11]
df_2['item'] = [1, 1, 2, 2, 2]
df_2['date'] = pd.date_range('1/1/2018', periods=5, freq='D')
df_2['price_rule'] = ['b', 'b', 'a', 'a', 'a']
df_2['sales']= [2, 3, 4, 5, 6]
df_2['clicks']= [7, 8, 9, 10, 11]
df_3['item'] = [1, 1, 2, 2, 2]
df_3['date'] = pd.date_range('1/1/2018', periods=5, freq='D')
df_3['price_rule'] = ['b', 'a', 'b', 'a', 'b']
df_3['sales']= [6, 5, 4, 5, 6]
df_3['clicks']= [7, 8, 9, 10, 11]
df = pd.concat([df_1, df_2, df_3])
df = df.sort_values(['item', 'date'])
df.reset_index(drop=True)
df
结果为:
item date price_rule sales clicks
0 1 2018-01-01 a 2 7
0 1 2018-01-01 b 2 7
0 1 2018-01-01 b 6 7
1 1 2018-01-02 b 4 8
1 1 2018-01-02 b 3 8
1 1 2018-01-02 a 5 8
2 2 2018-01-03 a 1 9
2 2 2018-01-03 a 4 9
2 2 2018-01-03 b 4 9
3 2 2018-01-04 b 5 10
3 2 2018-01-04 a 5 10
3 2 2018-01-04 a 5 10
4 2 2018-01-05 b 7 11
4 2 2018-01-05 a 6 11
4 2 2018-01-05 b 6 11
我的目标是:
1.按天对所有项目进行分组(以获取给定日期的每个项目的一行)
2.用"sum"汇总点击次数"
3. 生成"winning_pricing_rule"列，如下所示:
-对于给定的项目和给定的日期，采用具有最高销售"价值的定价规则
-如果是绘制"(例如，参见上面示例中2018年1月3日的项目2):仅选择其中之一(这在我的数据集中很少见，因此可以是随机的...)
My goal is to:
1. group all items by day (to get a single row for each item and given day)
2. aggregate 'clicks' with "sum"
3. generate a "winning_pricing_rule" columns as following:
- for a given item and given date, take a pricing rule with the highest 'sales' value
- in case of 'draw' (see eg: item 2 on 2018-01-03 in a sample above): choose just one of them (that's rare in my dataset, so it can be random...)
我想象结果看起来像这样:
I imagine the result to look like this:
item date winning_price_rule clicks
0 1 2018-01-01 b 21
1 1 2018-01-02 a 24
2 2 2018-01-03 b 27 <<remark: could also be a (due to draw)
3 2 2018-01-04 a 30 <<remark: could also be b (due to draw)
4 2 2018-01-05 b 33
我尝试过:
a.groupby(['item', 'date'], as_index = False).agg({'sales':'sum','revenue':'max'})
但未能确定获胜的定价规则.
but failed to identify a winning pricing rule.
有什么想法吗?非常感谢您的帮助:)
Any ideas? Many Thanks for help :)
安迪
推荐答案
First convert column price_rule to index by DataFrame.set_index, so for winning_price_rule is possible use DataFrameGroupBy.idxmax - get index value by maximum sales in GroupBy.agg, because also is necessary aggregate sum:
df1 = (df.set_index('price_rule')
.groupby(['item', 'date'])
.agg({'sales':'idxmax', 'clicks':'sum'})
.reset_index())
对于熊猫0.25.+可以使用:
For pandas 0.25.+ is possible use:
df1 = (df.set_index('price_rule')
.groupby(['item', 'date'])
.agg(winning_pricing_rule=pd.NamedAgg(column='sales', aggfunc='idxmax'),clicks=pd.NamedAgg(column='clicks', aggfunc="sum'))
.reset_index())
这篇关于Pandas groupby，在另一列中标识出最大值的元素的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！
