根据给定变量识别连续序列

Question

我确实坚持这一点。 df1 有以下变量：

1. serial = 一群人

2. id1 = 群组中的人（例如，12 (serial) 1 (id1) =group 12 person 1; 12 2 = group 12 person 2, etc。）

3. 'Day'第一次（或开始）录制时。

这些天由相同数量的观察组成（例如 95 个）

        day1 (Monday)  =  day11-day196 
        day2 (Tuesday) = day21-day296     
        day3 (Wednesday) =  day31-day396   
        day4 (Thursday) =  day41-day496   
        day5 (Friday) = day51-day596      
        day6 (Saturday) = day61-day696   
        day7 (Sunday) =  day71-day796  

df1 示例

serial id1  Day     day1 day2 day3 day4 day5 day6 day7
12      1   Monday    2    1    2    1    1    3    1
123     1   Tuesday   0    3    0    3    3    0    3
10      1   Wednesday 0    3    3    3    3    3    3

我想确定连续记录（每日记录之间没有间隙）和记录总数。

连续录制的开始日期是“Day”变量。例如，连续记录将是序列号 12。记录从星期一开始，并且在一周内有记录（至少有一个来自 95 变量）。一周内（7 x 95 变量）有 11 条记录

不连续的记录将是 id 123，因为在第 3 天和第 6 天有一个间隔日。记录从周二开始，周三和周六有间隔。

最后我想记录下连续录音的时长。

示例输出：

 serial  id1   Duration Occurance        Days
12       1      11        7        day1 day2 day3 day4 day5 day6 day7
123      1      12        0        0
10       1      18        5        day3 day4 day5 day6 day7

样本数据

structure(list(serial = c(12, 123, 10), id1 = c(1, 1, 1), Day = structure(1:3, .Label = c("Monday",
"Tuesday", "Wednesday"), class = "factor"), day1 = c(2, 0, 0),
day2 = c(1, 3, 3), day3 = c(2, 0, 3), day4 = c(1, 3, 3),
day5 = c(1, 3, 3), day6 = c(3, 0, 3), day7 = c(1, 3, 3)), row.names = c(NA,
3L), class = "data.frame")

类似帖子R - identify consecutive sequences

Answer 1

我们可以使用data.table 中的rleid 来获得正确的“发生”

library(data.table)
wkdays <- c("Monday", "Tuesday", "Wednesday", "Thursday", 
"Friday", "Saturday", "Sunday")

out1 <-  do.call(rbind, Map(function(x, y) {
              i1 <- match(y, wkdays): length(x)
              i2 <- x[i1] != 0
              i3 <- all(i2)
              grp1 <- rleid(i2)
              Days <- if(i3) tapply(names(x)[i1][i2], grp1[i2], FUN = paste, collapse= ' ') else ''
             Occurance <- if(i3) length(grp1[i2]) else 0
             data.frame(Occurance, Days)
            }, asplit(df[-(1:3)], 1), df$Day))

 out1$Duration <- rowSums(df1[startsWith(names(df1), 'day')])
 out1
 # Occurance                               Days Duration
 #1         7 day1 day2 day3 day4 day5 day6 day7       11
 #2         0                                          12
 #3         5           day3 day4 day5 day6 day7       18

Answer 2

你可以使用lead和lag的dplyr，

我试过了，结果如下：

library(dplyr)

df %>% 
    select(serial, contains("day", ignore.case = FALSE)) %>% 
    group_by(serial) %>% 
    tidyr::gather(day, val, -serial) %>% 
    # convert to binary 
    mutate(occur = ifelse(val > 0, 1, 0)) %>% 
    # if detect a seq, add cumulative, else 0
    mutate(cums = ifelse(lead(occur) > 0 & lag(occur) > 0 & occur > 0, 
                         occur + cumsum(occur), 0)) %>% 
    summarise(occurance = max(cums, na.rm = T), duration = sum(val))

# A tibble: 3 x 3
  serial occurance duration
   <dbl>     <dbl>    <dbl>
1     10         6       18
2     12         7       11
3    123         0       12