如何使用dplyr在3天时间间隔内找到中位数？

Question

我的数据是长格式的id，day和记录的测量。我想要一个新的变量，它在每3天间隔内生成中值(不是滚动，而是第1-3天、4-6天、7-9天等等)。

到目前为止，我使用dplyr作为总体中位数，但不确定如何每隔3天用id编码一次：

test%>%group_by(id)%>%mutate(m=median(o2))

以下是一些数据：

structure(list(id = c("1A", "1A", "1A", "1A", "1A", "1A", "1A", 
"1A", "1A", "1A"), day = 1:10, o2 = c(40L, 70L, 100L, 100L, 30L, 
35L, 30L, 30L, 40L, 40L)), row.names = c(NA, -10L), class = c("tbl_df", 
"tbl", "data.frame"), spec = structure(list(cols = list(id = structure(list(), class = c("collector_character", 
"collector")), day = structure(list(), class = c("collector_integer", 
"collector")), o2 = structure(list(), class = c("collector_integer", 
"collector"))), default = structure(list(), class = c("collector_guess", 
"collector"))), class = "col_spec"))

Answer 1

组按id和3天间隔，然后计算中位数.

library(dplyr)

test %>%
  group_by(id, interval_id = (day-1) %/% 3) %>%
  mutate(m = median(o2))

#  id      day    o2 interval_id     m
#  <chr> <int> <int>       <dbl> <int>
# 1A        1    40           0    70
# 1A        2    70           0    70
# 1A        3   100           0    70
# 1A        4   100           1    35
# 1A        5    30           1    35
# 1A        6    35           1    35
# 1A        7    30           2    30
# 1A        8    30           2    30
# 1A        9    40           2    30
# 1A       10    40           3    40

Answer 2

我们可以使用gl创建为期3天的组，并计算每个组的median。

library(dplyr)

test %>%
  group_by(id) %>%
  mutate(group = gl(n()/3, 3), 
         group = cumsum(group != lag(group, default = first(group)))) %>%
  group_by(id, group) %>%
  summarise(med = median(o2))



#  id    group   med
#  <chr> <int> <int>
#1 1A        0    70
#2 1A        1    35
#3 1A        2    30
#4 1A        3    40

Answer 3

由于这是对data.table::rleid的一个很好的使用，下面是data.table的答案，

library(data.table)

setDT(dd)[, grp := gl(.N, 3, length = .N), by = id][, .(med = median(o2)), .(id, rleid(grp))]

#   id rleid med
#1: 1A     1  70
#2: 1A     2  35
#3: 1A     3  30
#4: 1A     4  40