我感兴趣的是去除具有时间固定和时间变量值的敏感数据集。我想(a)按社会安全号码对所有案件进行分组,(b)为这些案件分配一个唯一的ID,然后(c)删除社会安全号码。
以下是一个示例数据集:
personal_id gender temperature
111-11-1111 M 99.6
999-999-999 F 98.2
111-11-1111 M 97.8
999-999-999 F 98.3
888-88-8888 F 99.0
111-11-1111 M 98.9
非常感谢任何解决方案。
答案 0 :(得分:27)
dplyr
具有group_indices
功能,可用于创建唯一的群组ID
library(dplyr)
data <- data.frame(personal_id = c("111-111-111", "999-999-999", "222-222-222", "111-111-111"),
gender = c("M", "F", "M", "M"),
temperature = c(99.6, 98.2, 97.8, 95.5))
data$group_id <- data %>% group_indices(personal_id)
data <- data %>% select(-personal_id)
data
gender temperature group_id
1 M 99.6 1
2 F 98.2 3
3 M 97.8 2
4 M 95.5 1
或在同一管道(https://github.com/tidyverse/dplyr/issues/2160)内:
data %>%
mutate(group_id = group_indices(., personal_id))
答案 1 :(得分:9)
dplyr::group_indices()
起,不推荐使用 dplyr 1.0.0
。应改为使用dplyr::cur_group_id()
:
df %>%
group_by(personal_id) %>%
mutate(group_id = cur_group_id())
personal_id gender temperature group_id
<chr> <chr> <dbl> <int>
1 111-11-1111 M 99.6 1
2 999-999-999 F 98.2 3
3 111-11-1111 M 97.8 1
4 999-999-999 F 98.3 3
5 888-88-8888 F 99 2
6 111-11-1111 M 98.9 1
答案 2 :(得分:0)
使用dplyr包:
library(dplyr)
data <- data.frame(personal_id = c("111-111-111", "999-999-999", "222-222-222", "111-111-111"),
gender = c("M", "F", "M", "M"),
temperature = c(99.6, 98.2, 97.8, 95.5))
首先提取personal_id以创建唯一ID:
cases <- data.frame(levels = levels(data$personal_id))
使用rownames,您将获得唯一标识符:
cases <- cases %>%
mutate(id = rownames(cases))
结果:
levels id
1 111-111-111 1
2 222-222-222 2
3 999-999-999 3
然后您将案例数据框与您的数据一起加入:
data <- left_join(data, cases, by = c("personal_id" = "levels"))
通过粘贴使用性别生成的ID来创建更加唯一的ID:
mutate(UID = paste(id, gender, sep=""))
最后删除了personal_id和简单的id:
select(-personal_id, -id)
然后你去:):
data <- left_join(data, cases, by = c("personal_id" = "levels")) %>%
mutate(UID = paste(id, gender, sep="")) %>%
select(-personal_id, -id)
结果:
gender temperature UID
1 M 99.6 1M
2 F 98.2 3F
3 M 97.8 2M
4 M 95.5 1M