R dplyr 定期总结

Question

这可能是一个

R

dplyr

summarise

的问题。 我有一个

data.frame

，其中以 5 分钟的时间间隔记录受试者的值，它具有以下三列：

id

：受试者 ID、

value

：该时间点的记录值，以及

cum_time

：累积时间每个

id

的值：

library(dplyr)
set.seed(1)
df <- data.frame(id = c(rep("id1", 100), rep("id2", 100), rep("id3", 100)),
                 value = runif(300, 10, 20)) %>%
  dplyr::group_by(id) %>%
  dplyr::mutate(cum_time = 5 * (dplyr::row_number()-1))

我想在 60 分钟的时间间隔内用

data.frame

的

median

来计算

value

，以给出结果

data.frame

:

rbind(data.frame(id = "id1", median_value = c(median(dplyr::filter(df, id == "id1" & cum_time >= 0 & cum_time <= 60)$value),
                                              median(dplyr::filter(df, id == "id1" & cum_time >= 65 & cum_time <= 120)$value),
                                              median(dplyr::filter(df, id == "id1" & cum_time >= 125 & cum_time <= 180)$value),
                                              median(dplyr::filter(df, id == "id1" & cum_time >= 185 & cum_time <= 240)$value),
                                              median(dplyr::filter(df, id == "id1" & cum_time >= 245 & cum_time <= 300)$value)),
                 cum_time = c(60, 120, 180, 240, 300)),
      data.frame(id = "id2", median_value = c(median(dplyr::filter(df, id == "id2" & cum_time >= 0 & cum_time <= 60)$value),
                                              median(dplyr::filter(df, id == "id2" & cum_time >= 65 & cum_time <= 120)$value),
                                              median(dplyr::filter(df, id == "id2" & cum_time >= 125 & cum_time <= 180)$value),
                                              median(dplyr::filter(df, id == "id2" & cum_time >= 185 & cum_time <= 240)$value),
                                              median(dplyr::filter(df, id == "id2" & cum_time >= 245 & cum_time <= 300)$value)),
                 cum_time = c(60, 120, 180, 240, 300)),
      data.frame(id = "id3", median_value = c(median(dplyr::filter(df, id == "id3" & cum_time >= 0 & cum_time <= 60)$value),
                                              median(dplyr::filter(df, id == "id3" & cum_time >= 65 & cum_time <= 120)$value),
                                              median(dplyr::filter(df, id == "id3" & cum_time >= 125 & cum_time <= 180)$value),
                                              median(dplyr::filter(df, id == "id3" & cum_time >= 185 & cum_time <= 240)$value),
                                              median(dplyr::filter(df, id == "id3" & cum_time >= 245 & cum_time <= 300)$value)),
                 cum_time = c(60, 120, 180, 240, 300)))

    id median_value cum_time
1  id1     15.72853       60
2  id1     15.74687      120
3  id1     14.87811      180
4  id1     16.00048      240
5  id1     14.57858      300
6  id2     15.98761       60
7  id2     14.65317      120
8  id2     15.36035      180
9  id2     15.16835      240
10 id2     13.90954      300
11 id3     12.68951       60
12 id3     15.79852      120
13 id3     14.03968      180
14 id3     14.29187      240
15 id3     15.11250      300

Answer 1

也许这种方法适合你？

df %>% 
  filter(cum_time<=300) %>% 
  group_by(id, grp=cut(cum_time, seq(0, max(cum_time),60), include.lowest = T)) %>% 
  summarize(median_value = median(value), .groups = "drop")

输出：

    id       grp median_value
1  id1    [0,60]     15.72853
2  id1  (60,120]     15.74687
3  id1 (120,180]     14.87811
4  id1 (180,240]     16.00048
5  id1 (240,300]     14.57858
6  id2    [0,60]     15.98761
7  id2  (60,120]     14.65317
8  id2 (120,180]     15.36035
9  id2 (180,240]     15.16835
10 id2 (240,300]     13.90954
11 id3    [0,60]     12.68951
12 id3  (60,120]     15.79852
13 id3 (120,180]     14.03968
14 id3 (180,240]     14.29187
15 id3 (240,300]     15.11250