如何根据列中的特定值序列在 df 中标记组答案

【问题标题】：How to label groups in df based on specific sequence of values in a column如何根据列中的特定值序列在 df 中标记组
【发布时间】：2020-06-23 14:48:07
【问题描述】：

我有一个数据框，其 id 和 value 列如下所示，但想根据 value 列中的值按 id 组确定状态列。

  
x <- data.frame(id = c(rep(1,10), rep(2,10), rep(3,10)),
                serial = rep(1:10,3),
                value = c(rep(1,4), rep(0,3), rep(1,3),
                          rep(1,4), rep(0,1), rep(-1,2), rep(1,3),
                          rep(c(1,0),5)),
                status = c(rep("Fluctuating", 10),
                           rep("Fluctuating", 10),
                           rep("Not fluctuating", 10)))

   id serial value          status
1   1      1     1     Fluctuating
2   1      2     1     Fluctuating
3   1      3     1     Fluctuating
4   1      4     1     Fluctuating
5   1      5     0     Fluctuating
6   1      6     0     Fluctuating
7   1      7     0     Fluctuating
8   1      8     1     Fluctuating
9   1      9     1     Fluctuating
10  1     10     1     Fluctuating
11  2      1     1     Fluctuating
12  2      2     1     Fluctuating
13  2      3     1     Fluctuating
14  2      4     1     Fluctuating
15  2      5     0     Fluctuating
16  2      6    -1     Fluctuating
17  2      7    -1     Fluctuating
18  2      8     1     Fluctuating
19  2      9     1     Fluctuating
20  2     10     1     Fluctuating
21  3      1     1 Not fluctuating
22  3      2     0 Not fluctuating
23  3      3     1 Not fluctuating
24  3      4     0 Not fluctuating
25  3      5     1 Not fluctuating
26  3      6     0 Not fluctuating
27  3      7     1 Not fluctuating
28  3      8     0 Not fluctuating
29  3      9     1 Not fluctuating
30  3     10     0 Not fluctuating

这里，如果三个或更多的 1 后面跟着 3 个或更多（0 或 -1），再跟着 3 个或更多的 1，则认为一个组在波动。如果三个或更多交替的 0s-1s-0s、-1s-0s-1s 等也被认为是波动的。

想知道分配状态列的最佳方法是什么，最好使用dplyr？

谢谢！

【问题讨论】：

Fluctuating 序列是否应该始终以 1 开头，或者它也可以以 0 开头，只要 1 和 0 至少交替 3 乘 3？
只要他们交替！

标签： r conditional-statements dplyr rle

【解决方案1】：

使用rle 函数和dplyr 库

x %>% 
  mutate(value_new = ifelse(value == -1, 0, value)) %>% 
  group_by(id) %>% 
  mutate(status = ifelse(all(rle(value_new)$lengths >= 3), "Fluctuating", "Not fluctuating")) %>% 
  select(-value_new)

输出

# A tibble: 30 x 4
# Groups:   id [3]
      id serial value status         
   <dbl>  <int> <dbl> <chr>          
 1     1      1     1 Fluctuating    
 2     1      2     1 Fluctuating    
 3     1      3     1 Fluctuating    
 4     1      4     1 Fluctuating    
 5     1      5     0 Fluctuating    
 6     1      6     0 Fluctuating    
 7     1      7     0 Fluctuating    
 8     1      8     1 Fluctuating    
 9     1      9     1 Fluctuating    
10     1     10     1 Fluctuating    
11     2      1     1 Fluctuating    
12     2      2     1 Fluctuating    
13     2      3     1 Fluctuating    
14     2      4     1 Fluctuating    
15     2      5     0 Fluctuating    
16     2      6    -1 Fluctuating    
17     2      7    -1 Fluctuating    
18     2      8     1 Fluctuating    
19     2      9     1 Fluctuating    
20     2     10     1 Fluctuating    
21     3      1     1 Not fluctuating
22     3      2     0 Not fluctuating
23     3      3     1 Not fluctuating
24     3      4     0 Not fluctuating
25     3      5     1 Not fluctuating
26     3      6     0 Not fluctuating
27     3      7     1 Not fluctuating
28     3      8     0 Not fluctuating
29     3      9     1 Not fluctuating
30     3     10     0 Not fluctuating

【讨论】：

这将触发"Fluctuating"，即使它的序列包含 0s/1s/0s 而不是 1s/0s/1s。
你说得对，我认为 1 和 0 的顺序无关紧要，只要它们在 3 上至少相差 3 即可
“三个或更多的 1 后面跟着 3 个或更多（0 或 -1），然后又是 3 个或更多的 1。” 我看到了你假设中的逻辑，但如果是这种情况，那么 OP 应该更新他们的问题。
是的，我明白这一点。我推断出一些可能是错误的。我确实要求澄清。
为不全面的例子道歉，澄清问题。

【解决方案2】：

library(dplyr)
# library(zoo) # rollapply
threes <- function(z, minlen = 3L, ptn = c(TRUE, FALSE, TRUE)) {
  r <- rle(z > 0)
  starts <- zoo::rollapply(r$lengths >= minlen, minlen, all, fill = FALSE, align = "left")
  for (st in which(starts)) {
    if (all(r$values[st + seq_len(minlen) - 1L] == ptn)) return(TRUE)
  }
  return(FALSE)
}

x %>%
  group_by(id) %>%
  mutate(status2 = paste0(if (threes(value)) "" else "Not ", "Fluctuating")) %>%
  ungroup() %>%
  print(n = 99)
# # A tibble: 30 x 5
#       id serial value status          status2        
#    <dbl>  <int> <dbl> <chr>           <chr>          
#  1     1      1     1 Fluctuating     Fluctuating    
#  2     1      2     1 Fluctuating     Fluctuating    
#  3     1      3     1 Fluctuating     Fluctuating    
#  4     1      4     1 Fluctuating     Fluctuating    
#  5     1      5     0 Fluctuating     Fluctuating    
#  6     1      6     0 Fluctuating     Fluctuating    
#  7     1      7     0 Fluctuating     Fluctuating    
#  8     1      8     1 Fluctuating     Fluctuating    
#  9     1      9     1 Fluctuating     Fluctuating    
# 10     1     10     1 Fluctuating     Fluctuating    
# 11     2      1     1 Fluctuating     Fluctuating    
# 12     2      2     1 Fluctuating     Fluctuating    
# 13     2      3     1 Fluctuating     Fluctuating    
# 14     2      4     1 Fluctuating     Fluctuating    
# 15     2      5     0 Fluctuating     Fluctuating    
# 16     2      6    -1 Fluctuating     Fluctuating    
# 17     2      7    -1 Fluctuating     Fluctuating    
# 18     2      8     1 Fluctuating     Fluctuating    
# 19     2      9     1 Fluctuating     Fluctuating    
# 20     2     10     1 Fluctuating     Fluctuating    
# 21     3      1     1 Not fluctuating Not Fluctuating
# 22     3      2     0 Not fluctuating Not Fluctuating
# 23     3      3     1 Not fluctuating Not Fluctuating
# 24     3      4     0 Not fluctuating Not Fluctuating
# 25     3      5     1 Not fluctuating Not Fluctuating
# 26     3      6     0 Not fluctuating Not Fluctuating
# 27     3      7     1 Not fluctuating Not Fluctuating
# 28     3      8     0 Not fluctuating Not Fluctuating
# 29     3      9     1 Not fluctuating Not Fluctuating
# 30     3     10     0 Not fluctuating Not Fluctuating

【讨论】：