识别每列的序列

Question

我正在处理时间使用数据，并希望计算每个时间步长（每列）开始测量的持续时间和 select 每次测量的最长持续时间。测量编号从 1 到 27。长度用 1 (e.g increment is set to 1). 加权我不确定如果测量是零散的并且有多个持续时间，我不确定如何处理。

数据格式：

期望的输出（测量编号 1 的示例）：

Time     Measurement   Duration 
04:00    1             1
04:10    1             1
04:20    1             2
04:20    1             2
04:20    1             2

持续时间最长

Time  Measurement Duration 
04:20 1           2

示例数据：

df<-structure(list(id = c(1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 
13, 14), `04:00` = c(1, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 
11, 11, 11), `04:10` = c(1, 11, 11, 11, 11, 11, 11, 11, 11, 11, 
11, 11, 11, 11), `04:20` = c(1, 11, 1, 1, 11, 11, 11, 11, 11, 
1, 1, 11, 11, 11), `04:30` = c(1, 11, 1, 1, 3, 11, 11, 11, 11, 
1, 1, 13, 11, 11), `04:40` = c(1, 11, 1, 1, 3, 12, 11, 11, 4, 
1, 1, 13, 4, 11), `04:50` = c(4, 11, 11, 11, 3, 12, 11, 11, 4, 
11, 11, 13, 4, 11), `05:00` = c(4, 11, 11, 11, 3, 12, 11, 11, 
4, 13, 11, 13, 4, 11), `05:10` = c(4, 11, 11, 11, 3, 12, 11, 
11, 4, 13, 11, 13, 4, 11), `05:20` = c(4, 11, 11, 11, 11, 13, 
4, 11, 4, 13, 11, 13, 4, 11), `05:30` = c(4, 11, 11, 11, 11, 
13, 4, 13, 4, 13, 11, 1, 4, 13), `05:40` = c(4, 11, 3, 11, 11, 
13, 4, 13, 11, 13, 11, 1, 1, 13), `05:50` = c(11, 11, 3, 11, 
11, 13, 4, 13, 11, 13, 11, 1, 11, 13), `06:00` = c(11, 1, 3, 
11, 11, 13, 4, 13, 1, 11, 11, 11, 11, 13), `06:10` = c(11, 1, 
3, 11, 11, 13, 4, 13, 1, 11, 11, 11, 11, 13), `06:20` = c(11, 
1, 3, 11, 11, 11, 11, 13, 1, 11, 11, 11, 11, 13)), row.names = c(NA, 
-14L), spec = structure(list(cols = list(id = structure(list(), class = c("collector_double", 
"collector")), `04:00` = structure(list(), class = c("collector_double", 
"collector")), `04:10` = structure(list(), class = c("collector_double", 
"collector")), `04:20` = structure(list(), class = c("collector_double", 
"collector")), `04:30` = structure(list(), class = c("collector_double", 
"collector")), `04:40` = structure(list(), class = c("collector_double", 
"collector")), `04:50` = structure(list(), class = c("collector_double", 
"collector")), `05:00` = structure(list(), class = c("collector_double", 
"collector")), `05:10` = structure(list(), class = c("collector_double", 
"collector")), `05:20` = structure(list(), class = c("collector_double", 
"collector")), `05:30` = structure(list(), class = c("collector_double", 
"collector")), `05:40` = structure(list(), class = c("collector_double", 
"collector")), `05:50` = structure(list(), class = c("collector_double", 
"collector")), `06:00` = structure(list(), class = c("collector_double", 
"collector")), `06:10` = structure(list(), class = c("collector_double", 
"collector")), `06:20` = structure(list(), class = c("collector_double", 
"collector"))), default = structure(list(), class = c("collector_guess", 
"collector")), delim = ","), class = "col_spec"), class = c("spec_tbl_df", 
"tbl_df", "tbl", "data.frame"))

Answer 1

library(tidyverse)
library(lubridate)

df %>%
  pivot_longer(-id, names_to = "timepoint", values_to = "Measurement") %>%
  arrange(id, Measurement) %>%
  type_convert() %>%
  group_by(id) %>%
  # Duration to first time point for each id
  mutate(Duration = timepoint - min(timepoint)) %>%
  # get the longest duration
  filter(Duration == max(Duration))
#> 
#> ── Column specification ────────────────────────────────────────────────────────
#> cols(
#>   timepoint = col_time(format = "")
#> )
#> # A tibble: 14 × 4
#> # Groups:   id [14]
#>       id timepoint Measurement Duration 
#>    <dbl> <time>          <dbl> <drtn>   
#>  1     1 06:20              11 8400 secs
#>  2     2 06:20               1 8400 secs
#>  3     3 06:20               3 8400 secs
#>  4     4 06:20              11 8400 secs
#>  5     5 06:20              11 8400 secs
#>  6     6 06:20              11 8400 secs
#>  7     7 06:20              11 8400 secs
#>  8     8 06:20              13 8400 secs
#>  9     9 06:20               1 8400 secs
#> 10    10 06:20              11 8400 secs
#> 11    11 06:20              11 8400 secs
#> 12    12 06:20              11 8400 secs
#> 13    13 06:20              11 8400 secs
#> 14    14 06:20              13 8400 secs

^{由 reprex package (v2.0.0)}

于 2022-05-16 创建

Answer 2

这是一个主要使用 rle 的函数，它将为您提供特定测量所需的输出：

f <- function(n){
  l <- lapply(df[-1], \(x) with(rle(x), lengths[values == n]))
  enframe(l, name = "Time", value = "Duration") %>% 
    unnest("Duration") %>% 
    mutate(Measurement = n, .before = "Duration")
}

输出

> f(1)
# A tibble: 20 × 3
   Time  Measurement Duration
   <chr>       <dbl>    <int>
 1 04:00           1        1
 2 04:10           1        1
 3 04:20           1        1
 4 04:20           1        2
 5 04:20           1        2
 6 04:30           1        1
 7 04:30           1        2
 8 04:30           1        2
 9 04:40           1        1
10 04:40           1        2
11 04:40           1        2
12 05:30           1        1
13 05:40           1        2
14 05:50           1        1
15 06:00           1        1
16 06:00           1        1
17 06:10           1        1
18 06:10           1        1
19 06:20           1        1
20 06:20           1        1

用slice_max得到最大值：

f(1) %>% 
  slice_max(Duration, n = 1, with_ties = F)

# A tibble: 1 × 3
  Time  Measurement Duration
  <chr>       <dbl>    <int>
1 04:20           1        2

识别每列的序列

Identifying sequences per column

r

sequence

dataframe

rle