Question

我有一个如下所示的数据框

Contract_ID    Place         Contract_Date      Price
1              Bangalore     2018-10-25         100
2              Bangalore     2018-08-25         200
3              Bangalore     2019-10-25         300
4              Bangalore     2019-11-25         200
5              Bangalore     2019-10-25         400
6              Chennai       2018-10-25         100
7              Chennai       2018-10-25         200
8              Chennai       2018-10-25         100
9              Chennai       2018-10-25         300
10             Chennai       2019-10-25         400
11             Chennai       2019-10-25         600

从上面我想使用熊猫生成下表。

预期输出：

Place       Year     Number_of_Contracts    Average_Price   
Bangalore   2018     2                      150
Bangalore   2019     3                      300
Chennai     2018     4                      175
Chennai     2019     2                      500

Answer 1

将GroupBy.agg与Series.dt.year创建的年份和元组一起用于新的列名称：

df['Contract_Date'] = pd.to_datetime(df['Contract_Date'])
df1 = (df.groupby(['Place', df['Contract_Date'].dt.year.rename('Year')])['Price']
         .agg([('Number_of_Contracts','size'),('Average_Price','mean')])
         .reset_index())
print (df1)
       Place  Year  Number_of_Contracts  Average_Price
0  Bangalore  2018                    2            150
1  Bangalore  2019                    3            300
2    Chennai  2018                    4            175
3    Chennai  2019                    2            500

解决方案named aggregation，但对于0.25+以上的熊猫是必需的：

df['Contract_Date'] = pd.to_datetime(df['Contract_Date'])
df1 = (df.groupby(['Place', df['Contract_Date'].dt.year.rename('Year')])
         .agg(Number_of_Contracts=('Contract_ID','size'),
              Average_Price=('Price','mean'))
         .reset_index())
print (df1)
       Place  Year  Number_of_Contracts  Average_Price
0  Bangalore  2018                    2            150
1  Bangalore  2019                    3            300
2    Chennai  2018                    4            175
3    Chennai  2019                    2            500

分组并计算平均值并计算熊猫的大小

1 个答案: