python - Pandas Groupby : Aggregations on the same column but totals based on two different critera/dataframes

Question

我的数据框:

display_name    security_type1  currency_str     state
         A            GOVT           USD         Done
         B            CORP           NZD         Passed
         B            CORP           USD         Done
         C            CORP           EUR         Done
         C            CORP           EUR         Traded Away
         C            CORP           GBP         Done
         C            CORP           GBP         Done
         C            CORP           USD         Done

我想要的结果是:

a.按 display_name、security_type1 和 currency_str 进行分组

b.然后计算 column state 包含 Done 的行数并更新列 Done_RFQ

c.显示每个 display_name、security_type1 和 currency_str 组合的总行数并更新列 Total_RFQ

d.最后显示完成占总计数的百分比，即 Done_Pct = Done_RFQ/Total_RFQ

display_name    security_type1  currency_str   Done_RFQ Total_RFQ Done_Pct
A               GOVT             USD           1           1      100%
B               CORP             USD           1           2      50%
C               CORP             EUR           1           5      20%
C               CORP             GBP           2           5      40%
C               CORP             USD           1           5      20%

我的代码适用于除 Total_RFQ 之外的情况，因此 Done_Pct 也适用

d = [('Done_RFQ', 'size')]
df_Done_Client = df[
                    df['state'].str.contains('Done')
                ][['display_name','security_type1','currency_str','state']].copy()

df_Done_Client =    
    df_Done_Client.groupby(['display_name','security_type1','currency_str'])['state'].agg(d).reset_index()
    # Sum of all Done RFQ's per display_name
    Sum_of_Done_For_Month = df_Done_Client.groupby('display_name')['Done_RFQ'].transform('sum')
    df_Done_Client['Total_Done_RFQ'] = Sum_of_Done_For_Month
    df_Done_Client['Done_Pct'] = df_Done_Client['Done_RFQ_For_Month'].div(Sum_of_Done_For_Month).round(5)
    display(df_Done_Client)

我不清楚如何计算这个总数，因为它需要来自另一个数据帧，即相同的字段，但没有“完成”标准。

df_All_Client = df[['display_name','security_type1','currency_str','state']].copy()

Answer 1

我认为需要 Total_RFQ 列，其中 size - 总计数和 Done_RFQ 按 bool 掩码计数 - 与 Done 进行比较code> 和 True 的 sum:

d = [('Total_RFQ', 'size'), ('Done_RFQ', lambda x: x.eq('Done').sum())]
df=df.groupby(['display_name','security_type1','currency_str'])['state'].agg(d).reset_index()
df['Done_Pct'] = df['Done_RFQ'] / df['Total_RFQ'] * 100
print (df)
  display_name security_type1 currency_str  Total_RFQ  Done_RFQ  Done_Pct
0            A           GOVT          USD          1         1     100.0
1            B           CORP          NZD          1         0       0.0
2            B           CORP          USD          1         1     100.0
3            C           CORP          EUR          2         1      50.0
4            C           CORP          GBP          2         2     100.0
5            C           CORP          USD          1         1     100.0

如果需要检查子字符串:

d = [('Total_RFQ', 'size'), ('Done_RFQ', lambda x: x.str.contains('Done').sum())]
df=df.groupby(['display_name','security_type1','currency_str'])['state'].agg(d).reset_index()
df['Done_Pct'] = df['Done_RFQ'] / df['Total_RFQ'] * 100
print (df)
  display_name security_type1 currency_str  Total_RFQ  Done_RFQ  Done_Pct
0            A           GOVT          USD          1         1     100.0
1            B           CORP          NZD          1         0       0.0
2            B           CORP          USD          1         1     100.0
3            C           CORP          EUR          2         1      50.0
4            C           CORP          GBP          2         2     100.0
5            C           CORP          USD          1         1     100.0