enter image description here df_6m_sum = df_6m.pivot_table(index='ACC_NBR', columns='class', values='TRANS_CHARGE', aggfunc=np.sum)
df_6m_sum.head(10)
class bus enter busi campus online offline drink buy change finance
ACC_NBR
1
我正试图根据画家的身份证号码,在扣除画廊佣金后,将画家的总收入显示出来。到目前为止,这就是我所拥有的:
SELECT painter.ptr_num, painter.ptr_firstname, (SUM() -
(gallery.gal_rate*painting.ptng_price)) AS Earnings
FROM painting,painter,gallery
GROUP BY painter.ptr_num
正如您可能已经注意到的,和是空的,这是因为我能想到的实现上述目标的唯一方法是有一个子查询。
下面是必须包含在和中的绘画表:
绘画
Ptng_Price Ptr_Num