我正在尝试使用 tidyverse 中的传播函数来处理以下数据的不同方法，但没有成功。目的是为变量中的值的每个 id 1 和 0 提供一个新列:health、ci_high、ci_low。
id unemployment health ci_high ci_low
1 5 100 110 90
1 10 80 90 70
1 15 70 80 60
0 5 90 100 80
0 10 50 60 40
0 15 40 50 30
structure(list(id = structure(c(1, 1, 1, 0, 0, 0), format.stata = "%9.0g"),
unemployment = structure(c(5, 10, 15, 5, 10, 15), format.stata = "%9.0g"),
health = structure(c(100, 80, 70, 90, 50, 40), format.stata = "%9.0g"),
ci_high = structure(c(110, 90, 80, 100, 60, 50), format.stata = "%9.0g"),
ci_low = structure(c(90, 70, 60, 80, 40, 30), format.stata = "%9.0g")), row.names = c(NA,
-6L), class = c("tbl_df", "tbl", "data.frame"))
我想得到这样的输出:
unemployment health_id1 health_id0 ci_high_id1 ci_high_id0 ci_low_id1 ci_low_id0
5 100 90 110 100 90 80
10 80 50 90 60 70 40
15 70 40 80 50 60 30
有人可以指导我吗？
最佳答案
使用pivot_wider
pivot_wider(df, unemployment, names_from = id, values_from = c("health", "ci_high", "ci_low"), names_prefix = "id")
# A tibble: 3 x 7
unemployment health_id1 health_id0 ci_high_id1 ci_high_id0 ci_low_id1 ci_low_id0
<dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
1 5 100 90 110 100 90 80
2 10 80 50 90 60 70 40
3 15 70 40 80 50 60 30
使用data.table
dt <- as.data.table(df)
dcast(dt, unemployment ~ id, value.var = c("health", "ci_high", "ci_low"))
unemployment health_0 health_1 ci_high_0 ci_high_1 ci_low_0 ci_low_1
1: 5 90 100 100 110 80 90
2: 10 50 80 60 90 40 70
3: 15 40 70 50 80 30 60
关于r - 如何使用 tidyverse 中使用多个变量的传播函数？，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/65006939/