计数相等时的熊猫条件选择

Question

以下数据集表示购买行为：

user_id, product_code, bought_date, time_spent, store_id, product_type, refurbished, unqiue_visit_id

001, e.12, 20120102, 104, 101, computer, yes, 1010

002, e.24, 20120201, 100, 101, infant-dress, no, 2001

003, s.32, 20130302, 230, 101, shoes, no, 2121

004, y.23, 20130404, 212, 103, computer, yes, 2422

005, s.43, 20130803, 104, 101, laptop, yes, 2342

001, a.12, 20120102, 104, 101, computer, yes, 1011

002, b.24, 20120201, 100, 101, infant-dress, no, 2001

003, c.32, 20130302, 230, 101, shoes, no, 2122

004, e.23, 20130404, 212, 103, computer, yes, 2424

005, f.43, 20130803, 104, 101, laptop, yes, 2340

001, g.12, 20120102, 104, 101, computer, yes, 1013

002, h.24, 20120201, 100, 101, infant-dress, no, 2031

003, l.32, 20130302, 230, 101, shoes, no, 2000

004, m.23, 20130404, 212, 103, computer, yes, 1422

005, d.43, 20130803, 104, 101, laptop, yes, 1142

001, d.12, 20120102, 104, 101, desk, yes, 1110

002, f.24, 20120201, 100, 101, glass, no, 1111

003, n.32, 20130302, 230, 101, liquid, no, 2021

004, t.23, 20130404, 212, 103, liquid, yes, 22

005, u.43, 20130803, 104, 101, dress, yes, 2942

001, d.12, 20120102, 104, 101, desk, yes, 1910

002, f.24, 20120201, 100, 101, glass, no, 2901

003, n.32, 20130302, 230, 101, liquid, no, 2921

004, t.23, 20130404, 212, 103, liquid, yes, 2922

005, u.43, 20130803, 104, 101, dress, yes, 2942

001, kk.12, 20120103, 105, 101, desk, yes, 410

003, n.32, 20130303, 230, 101, liquid, no, 2621

最终目标是使用以下步骤为用户分配产品类型。

首先，我对进行分组user_id，product_type并获得用户访问过的访问次数（次数）product_type。

如果组（）中的计数相等user_id，则product_id选择用户最近访问的产品类型，并将其分配给用户。如果访问日期相等，那么我们通过查看refurbished值来打破平局(yes > no)。

visit_counts = merged_visits_df.groupby(['user_id','product_type'],                                            as_index=False).agg({'unique_visits_id': 'nunique'})

上面给出了访问次数，试图找出其余的过程。