本文介绍了在 R 中按组连接列的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！
问题描述
假设我有这个员工列表:
Suppose I've got this employee list:
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
我想要在 R 中按部门聚合列表(类似于 Oracle PL/SQL 的 LISTAGG 函数) 将生成最后一列:
I want a list aggregation by department in R (similar to Oracle PL/SQL's LISTAGG function) that would product this last column:
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
有什么建议吗?
推荐答案
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
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