Question

我有以下数据示例

df<-data.frame(ID=c(1,1,1,1,1,1,1,1,1,1,2,2,2,2,2,2,2,2,2,2,3,3,3,3,3,3,3,3,3,3),           CODE=c("A","A","A","A","A","A","B","B","B","B","B","B","B","B","B","A","A","A","A","A","A","A","A","A","B","B","B","B","B","B"),
              DAT_NUM=c(20180101,20180101,20180105,20180107,20180107,20180108,20180203,20180203,20180201,20180205,
                        20180501,20180501,20180505,20180507,20180425,20180408,20180403,20180403,20180401,20180405,
                        20180105,20180105,20180105,20180107,20180107,20180110,20180206,20180203,20180201,20180205))

我需要一个新列（测试），该列基于DAT_NUM（例如1-6）分配一个连续的日期值，但是必须根据ID和CODE列的唯一组合来重置序列。测试列中的第1天指的是按顺序排列的1A，1B，2B，2A等的第一天，但对DAT_NUM中潜在天数的长度没有限制。

所需的输出：

df1<-data.frame(ID=c(1,1,1,1,1,1,1,1,1,1,2,2,2,2,2,2,2,2,2,2,3,3,3,3,3,3,3,3,3,3), 
              CODE=c("A","A","A","A","A","A","B","B","B","B","B","B","B","B","B","A","A","A","A","A","A","A","A","A","B","B","B","B","B","B"),
              DAT_NUM=c(20180101,20180101,20180105,20180107,20180107,20180108,20180203,20180203,20180201,20180205,
                        20180501,20180501,20180505,20180507,20180425,20180408,20180403,20180403,20180401,20180405,
                        20180105,20180105,20180105,20180107,20180107,20180110,20180206,20180203,20180201,20180205),
              test=c(1,1,2,3,3,4,2,2,1,3,
                     2,2,3,4,1,4,2,2,1,3,
                     1,1,1,2,1,2,6,4,3,5))

Answer 1

按“ ID”和“ CODE”分组后，我们可以使用match

library(dplyr)
df %>% 
   group_by(ID, CODE) %>% 
   mutate(test = match(DAT_NUM, sort(unique(DAT_NUM))))
# A tibble: 30 x 4
# Groups:   ID, CODE [6]
#      ID CODE   DAT_NUM  test
#   <dbl> <fct>    <dbl> <int>
# 1     1 A     20180101     1
# 2     1 A     20180101     1
# 3     1 A     20180105     2
# 4     1 A     20180107     3
# 5     1 A     20180107     3
# 6     1 A     20180108     4
# 7     1 B     20180203     2
# 8     1 B     20180203     2
# 9     1 B     20180201     1
#10     1 B     20180205     3
# … with 20 more rows

r-根据三列的条件填充新列

1 个答案: