Question

我的数据集如下：

ID   VISIT_ID  DATE  DV
1001 112233    12-23 3
1001 112233    12-23 4
1001 112244    12-23 5
1001 112244    12-23 6
1001 112244    12-23 7
1001 112244    12-23 8
1002 112254    12-23 3
1002 112254    12-23 4
1002 112254    12-23 5
1002 112264    12-23 6
1002 112264    12-23 7
1002 112264    12-23 8

我想要的结果如下;它为每个唯一的VISIT_ID分配增量遭遇值。对于每个ID，序列将从1重新开始。将非常感谢帮助。

ID   VISIT_ID  DATE  DV  ENCOUNTER
1001 112233    12-23 3   1
1001 112233    12-23 4   1
1001 112244    12-23 5   2
1001 112244    12-23 6   2
1001 112244    12-23 7   2
1001 112244    12-23 8   2
1002 112254    12-23 3   1
1002 112254    12-23 4   1
1002 112254    12-23 5   1
1002 112264    12-23 6   2
1002 112264    12-23 7   2
1002 112264    12-23 8   2

Answer 1

我们可以使用match在按“ID”分组后找到唯一'VISIT_ID'的索引

library(dplyr)
df1 %>% 
   group_by(ID) %>% 
   mutate(ENCOUNTER = match(VISIT_ID, unique(VISIT_ID))) 
#      ID VISIT_ID  DATE    DV ENCOUNTER
#    <int>    <int> <chr> <int>     <int>
#1   1001   112233 12-23     3         1
#2   1001   112233 12-23     4         1
#3   1001   112244 12-23     5         2
#4   1001   112244 12-23     6         2
#5   1001   112244 12-23     7         2
#6   1001   112244 12-23     8         2
#7   1002   112254 12-23     3         1
#8   1002   112254 12-23     4         1
#9   1002   112254 12-23     5         1
#10  1002   112264 12-23     6         2
#11  1002   112264 12-23     7         2
#12  1002   112264 12-23     8         2

或另一个选项是duplicated

df1 %>%
   group_by(ID) %>% 
   mutate(ENCOUNTER = cumsum(!duplicated(VISIT_ID)))

或使用data.table

library(data.table)
setDT(df1)[, ENCOUNTER := match(VISIT_ID, unique(VISIT_ID), by = ID]

或base R

with(df1, ave(VISIT_ID, ID, FUN = function(x) cumsum(!duplicated(x))))

Answer 2

使用base R ave，我们可以将VISIT_ID转换为factor，然后将numeric转换为VISIT_ID ID的唯一编号1}}

df$ENCOUNTER <- ave(df$VISIT_ID, df$ID,FUN = function(x) as.numeric(as.factor(x)))
df

#     ID VISIT_ID  DATE DV ENCOUNTER
#1  1001   112233 12-23  3         1
#2  1001   112233 12-23  4         1
#3  1001   112244 12-23  5         2
#4  1001   112244 12-23  6         2
#5  1001   112244 12-23  7         2
#6  1001   112244 12-23  8         2
#7  1002   112254 12-23  3         1
#8  1002   112254 12-23  4         1
#9  1002   112254 12-23  5         1
#10 1002   112264 12-23  6         2
#11 1002   112264 12-23  7         2
#12 1002   112264 12-23  8         2

如何根据R中的两列分配增量值？

2 个答案: