string – 由R中的组连接列
发布时间:2020-12-12 13:10:14 所属栏目:百科 来源:网络整理
导读:假设我有这个员工名单: Dept Date Name ----- --------- --------------- 30 07-DEC-02 Raphaely 30 18-MAY-03 Khoo 40 07-JUN-02 Mavris 50 01-MAY-03 Kaufling 50 14-JUL-03 Ladwig 70 07-JUN-02 Baer 90 13-JAN-01 De Haan 90 17-JUN-03 King 100 16-AUG
假设我有这个员工名单:
Dept Date Name ----- --------- --------------- 30 07-DEC-02 Raphaely 30 18-MAY-03 Khoo 40 07-JUN-02 Mavris 50 01-MAY-03 Kaufling 50 14-JUL-03 Ladwig 70 07-JUN-02 Baer 90 13-JAN-01 De Haan 90 17-JUN-03 King 100 16-AUG-02 Faviet 100 17-AUG-02 Greenberg 110 07-JUN-02 Gietz 110 07-JUN-02 Higgins 我希望按照R部门(类似于Oracle PL/SQL’s Dept Date Name Emp_list ----- --------- --------------- --------------------------------------------- 30 07-DEC-02 Raphaely Raphaely; Khoo 30 18-MAY-03 Khoo Raphaely; Khoo 40 07-JUN-02 Mavris Mavris 50 01-MAY-03 Kaufling Kaufling; Ladwig 50 14-JUL-03 Ladwig Kaufling; Ladwig 70 07-JUN-02 Baer Baer 90 13-JAN-01 De Haan De Haan; King 90 17-JUN-03 King De Haan; King 100 16-AUG-02 Faviet Faviet; Greenberg 100 17-AUG-02 Greenberg Faviet; Greenberg 110 07-JUN-02 Gietz Gietz; Higgins 110 07-JUN-02 Higgins Gietz; Higgins 任何建议? 你可以使用ave和paste:within(mydf,{ Emp_list <- ave(Name,Dept,FUN = function(x) paste(x,collapse = "; ")) }) # Dept Date Name Emp_list # 1 30 07-DEC-02 Raphaely Raphaely; Khoo # 2 30 18-MAY-03 Khoo Raphaely; Khoo # 3 40 07-JUN-02 Mavris Mavris # 4 50 01-MAY-03 Kaufling Kaufling; Ladwig # 5 50 14-JUL-03 Ladwig Kaufling; Ladwig # 6 70 07-JUN-02 Baer Baer # 7 90 13-JAN-01 De Haan De Haan; King # 8 90 17-JUN-03 King De Haan; King # 9 100 16-AUG-02 Faviet Faviet; Greenberg # 10 100 17-AUG-02 Greenberg Faviet; Greenberg # 11 110 07-JUN-02 Gietz Gietz; Higgins # 12 110 07-JUN-02 Higgins Gietz; Higgins (编辑:李大同) 【声明】本站内容均来自网络,其相关言论仅代表作者个人观点,不代表本站立场。若无意侵犯到您的权利,请及时与联系站长删除相关内容! |