当 mutate 需要来自不同变量子集和循环的额外参数时——dplyr 和 tidyverse 应用于教育领域
When mutate needs extra arguments from different subset of variables and loops -- dplyr and tidyverse applied to educational field
假设我在三个时间点评估了一个测试分数。因此，每个参与者在 T1（基线）、T2（post-intervention）和 T3（3 个月 follow-up）都有一个单独的分数。下图显示了当前数据集。
如果第二次评估 (Post-intervention) 的结果比第一次评估（基线）少 0.5（或更多），我想添加一个新的二进制变量 (1/0)，依此类推.因此，第一个参与者 (id_1 == 1) 将收到此变量的 1（因为 1.5 - 0.8 = 0.7 而 0.7 > 0.5）。同一个参与者在 3 个月时 follow-up 将收到 0，因为 0.8 - 0.5 = 0.3 而这个结果不 > 0.5。
我的问题背后的基本原理建议我使用 mutate，但是一旦结果以两个变量为条件 "loops"，我就很难处理这个问题。
我想继续使用 tidyverse 环境，下面的代码使这个问题很容易重现。
谢谢
ds <-structure(list(id_1 = c(1, 1, 1, 2, 2, 2, 3, 3, 3, 4, 4, 4, 5,
5, 5, 6, 6, 6, 7, 7, 7, 8, 8, 8, 9, 9, 9, 10, 10, 10, 11, 11,
11, 12, 12, 12, 13, 13, 13, 14, 14, 14, 15, 15, 15, 16, 16, 16,
17, 17, 17, 18, 18, 18, 19, 19, 19, 20, 20, 20, 21, 21, 21, 22,
22, 22, 23, 23, 23, 24, 24, 24, 25, 25, 25, 26, 26, 26, 27, 27,
27, 28, 28, 28, 29, 29, 29, 30, 30, 30, 31, 31, 31, 32, 32, 32,
33, 33, 33, 34, 34, 34, 35, 35, 35, 36, 36, 36, 37, 37, 37, 38,
38, 38, 39, 39, 39, 40, 40, 40, 41, 41, 41, 42, 42, 42, 43, 43,
43, 44, 44, 44, 45, 45, 45, 46, 46, 46, 47, 47, 47, 48, 48, 48,
49, 49, 49, 50, 50, 50, 51, 51, 51, 52, 52, 52, 53, 53, 53, 54,
54, 54, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA,
NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA,
NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA,
NA, NA, NA, NA, NA, NA, NA, NA), time = structure(c(1L, 2L, 3L,
1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L,
2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L,
3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L,
1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L,
2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L,
3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L,
1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L,
2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L,
3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L,
1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, NA,
NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA,
NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA,
NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA,
NA, NA, NA, NA, NA), .Label = c("Baseline", "Post-intervention",
"3-month follow-up"), class = "factor"), acq_6 = c(1.5, 0.8,
0.5, 1, 0, NA, 0.5, 0.5, 0.3, 3.1, 1.5, NA, 3.1, 0.8, 1.2, 1,
0.3, 0, 3.8, 2.7, 0.3, 4, 2, NA, 0.5, 0.8, 1, 2.2, 2.1, NA, 1.5,
0.7, 0.7, 1.9, 0.5, 0.3, 3.7, 1.9, 2.5, 0.8, 1.3, 1, 2, 3.3,
3.3, 2.1, 1.6, 2.2, 2.2, 1.3, 2.3, 2, 0.5, 0.5, 1.7, 1.9, 1.5,
1.8, 1.4, NA, 2.9, 1.8, 0.3, 1.8, 1.3, 1.5, 1.3, 1, 0, 2.6, 0.7,
1, 2.1, 1.8, 1.8, 3.3, 2.6, 3, 1.2, NA, NA, 0.7, NA, NA, 1, NA,
NA, 0.7, 0.9, 0.5, 1, 0.6, NA, 3.3, 0.2, NA, 1, 0.7, 1.3, 1.7,
2.3, NA, 1.7, 1.6, 1.6, 2.5, 1.2, 2, 2.5, 2.8, 3.8, 0.7, 0, 0.2,
1.2, 2.2, NA, 1.2, 1.8, 2, 2.5, 1.8, 2.2, 2, 1.7, NA, 1.7, 2.5,
2, 1.8, 2.5, 1.8, 1.8, 3.2, 3.3, 0.6, 0.5, 1, 2.9, 1.8, 2.6,
1, 0.5, 1.2, 0.8, 0.7, 1, 1.8, 1.5, 1.8, 1.7, 0.7, 1, 1.2, 1,
0.9, 1.8, NA, NA, 2.6, 2.3, NA, NA, NA, NA, NA, NA, NA, NA, NA,
NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA,
NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA,
NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA)), class = c("tbl_df",
"tbl", "data.frame"), row.names = c(NA, -216L))
当我的ypu正确时，可以这样实现。按 id 分组，添加滞后值，检查差异是否大于 .5:
library(dplyr)
ds <- structure(list(id_1 = c(
1, 1, 1, 2, 2, 2, 3, 3, 3, 4, 4, 4, 5,
5, 5, 6, 6, 6, 7, 7, 7, 8, 8, 8, 9, 9, 9, 10, 10, 10, 11, 11,
11, 12, 12, 12, 13, 13, 13, 14, 14, 14, 15, 15, 15, 16, 16, 16,
17, 17, 17, 18, 18, 18, 19, 19, 19, 20, 20, 20, 21, 21, 21, 22,
22, 22, 23, 23, 23, 24, 24, 24, 25, 25, 25, 26, 26, 26, 27, 27,
27, 28, 28, 28, 29, 29, 29, 30, 30, 30, 31, 31, 31, 32, 32, 32,
33, 33, 33, 34, 34, 34, 35, 35, 35, 36, 36, 36, 37, 37, 37, 38,
38, 38, 39, 39, 39, 40, 40, 40, 41, 41, 41, 42, 42, 42, 43, 43,
43, 44, 44, 44, 45, 45, 45, 46, 46, 46, 47, 47, 47, 48, 48, 48,
49, 49, 49, 50, 50, 50, 51, 51, 51, 52, 52, 52, 53, 53, 53, 54,
54, 54, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA,
NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA,
NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA,
NA, NA, NA, NA, NA, NA, NA, NA
), time = structure(c(
1L, 2L, 3L,
1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L,
2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L,
3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L,
1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L,
2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L,
3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L,
1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L,
2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L,
3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L,
1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, NA,
NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA,
NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA,
NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA,
NA, NA, NA, NA, NA
), .Label = c(
"Baseline", "Post-intervention",
"3-month follow-up"
), class = "factor"), acq_6 = c(
1.5, 0.8,
0.5, 1, 0, NA, 0.5, 0.5, 0.3, 3.1, 1.5, NA, 3.1, 0.8, 1.2, 1,
0.3, 0, 3.8, 2.7, 0.3, 4, 2, NA, 0.5, 0.8, 1, 2.2, 2.1, NA, 1.5,
0.7, 0.7, 1.9, 0.5, 0.3, 3.7, 1.9, 2.5, 0.8, 1.3, 1, 2, 3.3,
3.3, 2.1, 1.6, 2.2, 2.2, 1.3, 2.3, 2, 0.5, 0.5, 1.7, 1.9, 1.5,
1.8, 1.4, NA, 2.9, 1.8, 0.3, 1.8, 1.3, 1.5, 1.3, 1, 0, 2.6, 0.7,
1, 2.1, 1.8, 1.8, 3.3, 2.6, 3, 1.2, NA, NA, 0.7, NA, NA, 1, NA,
NA, 0.7, 0.9, 0.5, 1, 0.6, NA, 3.3, 0.2, NA, 1, 0.7, 1.3, 1.7,
2.3, NA, 1.7, 1.6, 1.6, 2.5, 1.2, 2, 2.5, 2.8, 3.8, 0.7, 0, 0.2,
1.2, 2.2, NA, 1.2, 1.8, 2, 2.5, 1.8, 2.2, 2, 1.7, NA, 1.7, 2.5,
2, 1.8, 2.5, 1.8, 1.8, 3.2, 3.3, 0.6, 0.5, 1, 2.9, 1.8, 2.6,
1, 0.5, 1.2, 0.8, 0.7, 1, 1.8, 1.5, 1.8, 1.7, 0.7, 1, 1.2, 1,
0.9, 1.8, NA, NA, 2.6, 2.3, NA, NA, NA, NA, NA, NA, NA, NA, NA,
NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA,
NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA,
NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA
)), class = c(
"tbl_df",
"tbl", "data.frame"
), row.names = c(NA, -216L))
ds %>%
group_by(id_1) %>%
mutate(acq_6_lag = lag(acq_6),
get_better = ifelse(acq_6_lag - acq_6 > .5, 1, 0))
#> # A tibble: 216 x 5
#> # Groups: id_1 [55]
#> id_1 time acq_6 acq_6_lag get_better
#> <dbl> <fct> <dbl> <dbl> <dbl>
#> 1 1 Baseline 1.5 NA NA
#> 2 1 Post-intervention 0.8 1.5 1
#> 3 1 3-month follow-up 0.5 0.8 0
#> 4 2 Baseline 1 NA NA
#> 5 2 Post-intervention 0 1 1
#> 6 2 3-month follow-up NA 0 NA
#> 7 3 Baseline 0.5 NA NA
#> 8 3 Post-intervention 0.5 0.5 0
#> 9 3 3-month follow-up 0.3 0.5 0
#> 10 4 Baseline 3.1 NA NA
#> # ... with 206 more rows
由 reprex package (v0.3.0)
于 2020-03-16 创建
假设我在三个时间点评估了一个测试分数。因此，每个参与者在 T1（基线）、T2（post-intervention）和 T3（3 个月 follow-up）都有一个单独的分数。下图显示了当前数据集。
如果第二次评估 (Post-intervention) 的结果比第一次评估（基线）少 0.5（或更多），我想添加一个新的二进制变量 (1/0)，依此类推.因此，第一个参与者 (id_1 == 1) 将收到此变量的 1（因为 1.5 - 0.8 = 0.7 而 0.7 > 0.5）。同一个参与者在 3 个月时 follow-up 将收到 0，因为 0.8 - 0.5 = 0.3 而这个结果不 > 0.5。
我的问题背后的基本原理建议我使用 mutate，但是一旦结果以两个变量为条件 "loops"，我就很难处理这个问题。
我想继续使用 tidyverse 环境，下面的代码使这个问题很容易重现。
谢谢
ds <-structure(list(id_1 = c(1, 1, 1, 2, 2, 2, 3, 3, 3, 4, 4, 4, 5,
5, 5, 6, 6, 6, 7, 7, 7, 8, 8, 8, 9, 9, 9, 10, 10, 10, 11, 11,
11, 12, 12, 12, 13, 13, 13, 14, 14, 14, 15, 15, 15, 16, 16, 16,
17, 17, 17, 18, 18, 18, 19, 19, 19, 20, 20, 20, 21, 21, 21, 22,
22, 22, 23, 23, 23, 24, 24, 24, 25, 25, 25, 26, 26, 26, 27, 27,
27, 28, 28, 28, 29, 29, 29, 30, 30, 30, 31, 31, 31, 32, 32, 32,
33, 33, 33, 34, 34, 34, 35, 35, 35, 36, 36, 36, 37, 37, 37, 38,
38, 38, 39, 39, 39, 40, 40, 40, 41, 41, 41, 42, 42, 42, 43, 43,
43, 44, 44, 44, 45, 45, 45, 46, 46, 46, 47, 47, 47, 48, 48, 48,
49, 49, 49, 50, 50, 50, 51, 51, 51, 52, 52, 52, 53, 53, 53, 54,
54, 54, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA,
NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA,
NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA,
NA, NA, NA, NA, NA, NA, NA, NA), time = structure(c(1L, 2L, 3L,
1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L,
2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L,
3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L,
1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L,
2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L,
3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L,
1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L,
2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L,
3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L,
1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, NA,
NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA,
NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA,
NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA,
NA, NA, NA, NA, NA), .Label = c("Baseline", "Post-intervention",
"3-month follow-up"), class = "factor"), acq_6 = c(1.5, 0.8,
0.5, 1, 0, NA, 0.5, 0.5, 0.3, 3.1, 1.5, NA, 3.1, 0.8, 1.2, 1,
0.3, 0, 3.8, 2.7, 0.3, 4, 2, NA, 0.5, 0.8, 1, 2.2, 2.1, NA, 1.5,
0.7, 0.7, 1.9, 0.5, 0.3, 3.7, 1.9, 2.5, 0.8, 1.3, 1, 2, 3.3,
3.3, 2.1, 1.6, 2.2, 2.2, 1.3, 2.3, 2, 0.5, 0.5, 1.7, 1.9, 1.5,
1.8, 1.4, NA, 2.9, 1.8, 0.3, 1.8, 1.3, 1.5, 1.3, 1, 0, 2.6, 0.7,
1, 2.1, 1.8, 1.8, 3.3, 2.6, 3, 1.2, NA, NA, 0.7, NA, NA, 1, NA,
NA, 0.7, 0.9, 0.5, 1, 0.6, NA, 3.3, 0.2, NA, 1, 0.7, 1.3, 1.7,
2.3, NA, 1.7, 1.6, 1.6, 2.5, 1.2, 2, 2.5, 2.8, 3.8, 0.7, 0, 0.2,
1.2, 2.2, NA, 1.2, 1.8, 2, 2.5, 1.8, 2.2, 2, 1.7, NA, 1.7, 2.5,
2, 1.8, 2.5, 1.8, 1.8, 3.2, 3.3, 0.6, 0.5, 1, 2.9, 1.8, 2.6,
1, 0.5, 1.2, 0.8, 0.7, 1, 1.8, 1.5, 1.8, 1.7, 0.7, 1, 1.2, 1,
0.9, 1.8, NA, NA, 2.6, 2.3, NA, NA, NA, NA, NA, NA, NA, NA, NA,
NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA,
NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA,
NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA)), class = c("tbl_df",
"tbl", "data.frame"), row.names = c(NA, -216L))
当我的ypu正确时，可以这样实现。按 id 分组，添加滞后值，检查差异是否大于 .5:
library(dplyr)
ds <- structure(list(id_1 = c(
1, 1, 1, 2, 2, 2, 3, 3, 3, 4, 4, 4, 5,
5, 5, 6, 6, 6, 7, 7, 7, 8, 8, 8, 9, 9, 9, 10, 10, 10, 11, 11,
11, 12, 12, 12, 13, 13, 13, 14, 14, 14, 15, 15, 15, 16, 16, 16,
17, 17, 17, 18, 18, 18, 19, 19, 19, 20, 20, 20, 21, 21, 21, 22,
22, 22, 23, 23, 23, 24, 24, 24, 25, 25, 25, 26, 26, 26, 27, 27,
27, 28, 28, 28, 29, 29, 29, 30, 30, 30, 31, 31, 31, 32, 32, 32,
33, 33, 33, 34, 34, 34, 35, 35, 35, 36, 36, 36, 37, 37, 37, 38,
38, 38, 39, 39, 39, 40, 40, 40, 41, 41, 41, 42, 42, 42, 43, 43,
43, 44, 44, 44, 45, 45, 45, 46, 46, 46, 47, 47, 47, 48, 48, 48,
49, 49, 49, 50, 50, 50, 51, 51, 51, 52, 52, 52, 53, 53, 53, 54,
54, 54, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA,
NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA,
NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA,
NA, NA, NA, NA, NA, NA, NA, NA
), time = structure(c(
1L, 2L, 3L,
1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L,
2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L,
3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L,
1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L,
2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L,
3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L,
1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L,
2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L,
3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L,
1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, NA,
NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA,
NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA,
NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA,
NA, NA, NA, NA, NA
), .Label = c(
"Baseline", "Post-intervention",
"3-month follow-up"
), class = "factor"), acq_6 = c(
1.5, 0.8,
0.5, 1, 0, NA, 0.5, 0.5, 0.3, 3.1, 1.5, NA, 3.1, 0.8, 1.2, 1,
0.3, 0, 3.8, 2.7, 0.3, 4, 2, NA, 0.5, 0.8, 1, 2.2, 2.1, NA, 1.5,
0.7, 0.7, 1.9, 0.5, 0.3, 3.7, 1.9, 2.5, 0.8, 1.3, 1, 2, 3.3,
3.3, 2.1, 1.6, 2.2, 2.2, 1.3, 2.3, 2, 0.5, 0.5, 1.7, 1.9, 1.5,
1.8, 1.4, NA, 2.9, 1.8, 0.3, 1.8, 1.3, 1.5, 1.3, 1, 0, 2.6, 0.7,
1, 2.1, 1.8, 1.8, 3.3, 2.6, 3, 1.2, NA, NA, 0.7, NA, NA, 1, NA,
NA, 0.7, 0.9, 0.5, 1, 0.6, NA, 3.3, 0.2, NA, 1, 0.7, 1.3, 1.7,
2.3, NA, 1.7, 1.6, 1.6, 2.5, 1.2, 2, 2.5, 2.8, 3.8, 0.7, 0, 0.2,
1.2, 2.2, NA, 1.2, 1.8, 2, 2.5, 1.8, 2.2, 2, 1.7, NA, 1.7, 2.5,
2, 1.8, 2.5, 1.8, 1.8, 3.2, 3.3, 0.6, 0.5, 1, 2.9, 1.8, 2.6,
1, 0.5, 1.2, 0.8, 0.7, 1, 1.8, 1.5, 1.8, 1.7, 0.7, 1, 1.2, 1,
0.9, 1.8, NA, NA, 2.6, 2.3, NA, NA, NA, NA, NA, NA, NA, NA, NA,
NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA,
NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA,
NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA
)), class = c(
"tbl_df",
"tbl", "data.frame"
), row.names = c(NA, -216L))
ds %>%
group_by(id_1) %>%
mutate(acq_6_lag = lag(acq_6),
get_better = ifelse(acq_6_lag - acq_6 > .5, 1, 0))
#> # A tibble: 216 x 5
#> # Groups: id_1 [55]
#> id_1 time acq_6 acq_6_lag get_better
#> <dbl> <fct> <dbl> <dbl> <dbl>
#> 1 1 Baseline 1.5 NA NA
#> 2 1 Post-intervention 0.8 1.5 1
#> 3 1 3-month follow-up 0.5 0.8 0
#> 4 2 Baseline 1 NA NA
#> 5 2 Post-intervention 0 1 1
#> 6 2 3-month follow-up NA 0 NA
#> 7 3 Baseline 0.5 NA NA
#> 8 3 Post-intervention 0.5 0.5 0
#> 9 3 3-month follow-up 0.3 0.5 0
#> 10 4 Baseline 3.1 NA NA
#> # ... with 206 more rows
由 reprex package (v0.3.0)
于 2020-03-16 创建